ELECTRO-MAGNETIC INDUCTION AND ALTERNATING CURRENT

1 MARKS QUESTIONS:

1. In an a.c. circuit, the instantaneous voltage and current are V = 200 sin 300 t volt and I = 8 cos 300 t ampere respectively. Is the nature of the circuit capacitive or inductive? Give reason.

Ans: Yes, it is capacitive as V is the function of sine and I is the function of cosine with the same angle (current leads).

2. Define the term ‘self-inductance’ of a coil. Write its S.I. unit. Ans: It is the ratio of magnetic flux through a coil to that of electric current. SI unit is Henery.

3. Draw a graph to show variation of capacitive-reactance with frequency in an a.c. circuit.

4. Define the term ‘quality factor’ of resonance in series LCR circuit. What is its S.I. unit?

Ans: It is a measure of the sharpness of resonance. Alternatively, Q = 1/WOCR /w0L/R It has no unit. 5. Define the term ‘wattless current’. Ans: It is the component of electric current perpendicular to the voltage and due to which power loss does not take place.

6. Draw a graph to show variation of capacitive-reactance with frequency in an a.c. circuit. 7. What is the average value of a.c. over a cycle and why? Ans: Zero because a.c. is positive during the half cycle and negative during next half. 8. When a.c. passes through a galvanometer, it shows no deflection. Why? Ans: A moving coil galvanometer measure average value of current, which is zero at every cycle. 9. Which value of current do you read with a.c. ammeter? Ans: rms value of current. 10. When are the voltage and current in LCR series circuit in phase? Ans: When XL = XC

11. What is the minimum value of power factor? When does it occur? Ans: Zero, it occurs when Ф = + π/2

12. Why is the core of transformer laminated? Ans: Minimum energy losses due to eddy circuit.

13. Is there any device by which d.c. can be controlled without any loss of energy? Ans: No 14. What is the basic cause of induced emf? Ans: Change in magnetic flux. 15. Induced emf is also called back emf. Why? Ans: it always opposes any change in emf. 16. Why is spark produced in the switch of a fan, when it is switched off? Ans: Large amount of emf is induced in gap of switch. 17. The inductance coils are made of copper. Why? Ans: Copper has small resistance. So induced current set up is large. 18. A solenoid with an iron core and a bulb are connected to a d.c. source. How does the brightness of the bulb change when the iron core is removed from the solenoid? Ans: No change in brightness because reactance of inductor is zero. 19. Why is the induce emf in a coil zero when its plane is normal to the magnetic field even though maximum magnetic flux is liked with the coil in this position? Ans: Rate of change of magnetic flux is zero. 20. A wire kept along east-west direction is allowed to fall freely. Will an emf be induced across the ends of the wire? Ans: Yes 21. Will an induced emf develop in a conductor, when moved in a direction parallel to the magnetic field? Ans: No

22. How would you detect the presence of magnetic field on an unknown planet? Ans: By deflection in galvanometer. 23. Define one Henry? Ans: 1 Henry is a self inductance of that coil in which 1 V emf is produced, when the rate of change of current in that coil is 1 A/s.

24. If the number of turns in solenoid is doubled, keeping other factors constant, how does the self inductance of the coil change? Ans: The self inductance becomes 4 times. (L α N2). 25. What is phase difference between the voltage across the inductance and a capacitor in an a.c. circuit? Ans: 180o.

26. The coil in certain galvanometers have fixed core made of a non-magnetic material. Why does the oscillating coil come to rest so quickly in such a core? Ans: Due to large restoring couple developed in the coil. 27. The power factor of an a.c. circuit is 0.5. What will be the phase difference between voltage and current in this circuit? Ans: Power factor cos Ф = 0.5 and phase difference between voltage and current Ф = 600

28. An electrical element X when connected to an alternating voltage source, has a current through it leading the voltage by π/2 rad. Identify X and write an expression for its reactance. Ans: Pure capacitor, XC = 1/ὼC. 29. What is the phase difference between voltage and current in a LCR series circuit at resonance? Ans: Zero. 30. Mention two characteristic properties of the material suitable for making core of a transformer. Ans: (i) Low hysteresis loss (ii) low coercivity. 2 MARKS QUESTIONS: Q1. Define the term ‘power loss’ in a conductor of resistance R carrying a current I. In what form does this power loss appear? Show that to minimise the power loss in the transmission cables connecting the power stations to homes, it is necessary to have the connecting wires carrying current at enormous high values of voltage. Q2. Why does a metallic piece become very hot when it is surrounded by a coil carrying high frequency a.c.? Ans: Heavy eddy current produces in it.

Q3. A coil is removed from a magnetic field (i) rapidly (ii) slowly. In which case more work will be done? Ans: In 1st case

Q4. Why is a transformer not used to step up d.c. voltage? Ans: In d.c. there is no change in magnetic flux liked with secondary coil. Q5. What is the frequency of domestic a.c. supply? How many times does it become zero in one second? Ans: 50 Hz, 100 times. Q6. Distinguish between the terms effective value and peak value of ac. Ans: ac change in magnitude and direction both. The maximum value of ac is called peak value. The RMS value of current is called effective value. Q7. Show that lenz’s law is accordance with the law of conservation of energy. Ans. The direction of induced current in a closed circuit is always such to oppose the cause that produces it. When the north pole of a coil is brought near a closed coil the direction of current induced in the coil is such as to oppose the approach of North Pole and vice-versa. Q8. State the principle of working of a transformer. Can a transformer be used to step up or step down a dc voltage? Ans. The working of transformer is based on the principle of mutual inductance. No, a transformer cannot be used to step up or step down dc. Q9. Both ac and dc are measured in amperes but how is the ampere defined for an ac? Ans. An ac change direction with the source frequency and the attractive force would average to zero. Thus, the ac ampere must be defined in some property i.e. independent of the direction of current. Joule’s heating effect is such property and hence it is used to define rms value of ac. Q10. An air coil solenoid is connected to an ac source and a bulb. If an iron core is inserted in a solenoid, how does the brightness of the bulb change? Give reason. Ans. Self inductance of the coil increases, so reactance of solenoid increases and the brightness of the bulb will decrease. Q11. In India domestic power supply is at 220V, 50Hz while in USA it is 110V and 50Hz. Give one advantage and one disadvantage of 220V supply over 110V supply. Ans. Advantage- line loss is low Disadvantage- high voltage is dangerous. Q12. A coil of inductance L, a capacitor of capacitance C and a register of resistance R are all put in series with an ac source of emf E =Eo sin ωt. Write an expression for (1) Total impedance of the circuit (2) Frequency of the source emf for which current will show resonance. Ans. (i) Z = √R2 + (1/ωc - ωL)2

(ii) fr= 1/2π√LC

Q13. Explain why the reactants provided by a capacitor to an ac decreases with increasing frequency? Ans. A capacitor does not allow flow of dc through it as the resistance across the gap is infinite. When an alternating voltage applied across the capacitor, the plates are alternately charged and discharged. This current through the capacitor is a result of changing voltage. Thus a capacitor will pass more current through it this implies that the reactants offered by the capacitor will become less with increase in frequency. Q14. A bar magnet falls through a closed ring of wire. State with reason, will the acceleration of

magnet will be equal to the acceleration due to gravity?

Ans. No, it will be less than the gravity because induced current opposes the motion of magnet. Q15. A bar magnet falls through a open ring of wire. State with reason, will the acceleration of magnet will be equal to the acceleration due to gravity? Ans. Yes

Q16. A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply (a) what is the rms value of current in the circuit? (b) What is the net power consumed over a full cycle? Ans. (a) Irms = Vrms/R = 220/100 = 2.2 A

(b) P = I2rms R = 2.2 X 2.2 X100 = 484 W

Q17. A pair of adjacent coils has a mutual inductance of 1.5H. If the current in one coil changes from 0 to 20 ampere in 0.5 s, what is the change in magnetic flux linkage with the other coil? Ans. Ф2 = MI1

ΔФ2 = MΔI1 =1.5 X 20 = 30 wb

Q18. As soon the current is switched on in a high voltage wire, the bird sitting on it flies away. Why? Ans. Because the induced current flows in the bird’s body. Its wings due to opposite current experiences mutual repulsion hence birds fly away. Q19. Two coils of wire A and B are placed mutually perpendicular. When current is changed in any one coil will the current induced in other coil? Ans. No, because the magnetic field will be parallel to the plain to of other coil. Q20. The instantaneous voltage from an ac source is given by E = 300 sin 314 t; what is the rms voltage of the source. Ans: Erms = Eo/√2 = 300/√2 = 150√2 = 150 X 1.414 = 212 V.

Q21. An ac from a source is represented by, I = 10 sin 314 t. Write the corresponding values of (i) its effective value (ii) frequency of the source.

Ans: (i) Ieff = Io/√2 = 5√2 units (ii) f = ω/2π = 314/2 X 3.14 = 50 Hz

Q22. A wire in the form of a tightly wound solenoid is connected to a dc source, and carries a current. If the coil is stretched so that there are gaps between successive elements of the spiral coil, will the current increase or decrease? Explain. Ans: Increase by Lenz’s law Q23. What will be the effect on inductive reactance XL and capacitive reactance XC, if frequency of the source increased? Ans: XL = ωL = 2πfL (increase)

Xc = 1/ωC = 1/2πfC (decrease) Q24. A rectangular coil of N-turn and area of cross-sections A is held in a time varying magnetic field. Deduce an expression for the emf induced in the coil. Ans: Ф = NBA cos Ө

e = -dФ/dt = d(NBA COS Ө)/dt = NA cos Ө (dB/dt)

Q25. What is meant by impedance? Ans: The effective resistance offered by a circuit. 03 MARKS QUESTIONS: Q1. An inductor L of inductance XL is connected in series with a bulb B and an ac source. How would brightness of the bulb change when (i) number of turn in the inductor is reduced, (ii) an iron rod is inserted in the inductor and (iii) a capacitor of reactance XC = XL is inserted in series in the circuit. Justify your answer in each case. Q2. (a) Show that the average power consumed in an inductor L connected to an a.c. source is zero.

(b) In a series LR circuit, XL = R and the power factor of the circuit is P1. When a capacitor with capacitance C such that XC = XL is put in series, the power factor becomes P2. Find out P1 / P2. ANS: For an ideal inductor connected to ac source, V = VO Sin ωt and I = Io Sin ( ωt – π/2)

Pav = 1/ T ʃoT VoIo Sin ωt Cos ωt dt =0

(Also accept any other correct method) Power factor cos∅ = R/Z For LR circuit, at XL = R

Z = √ R2 + R2, Z = R √2

P1 = cos Ф = R / R √2 = 1/√2 For LCR circuit, Z = √ R2 = R Power factor P2 = R/R = 1 P1/P2 = 1/√2

Q3. You are given three circuit elements X, Y and Z. When the element X is connected across an a.c. source of a given voltage, the current and the voltage are in the same phase. When the element Y is connected in series with X across the source, voltage is ahead of the current in phase by λ/4. But the current is ahead of the voltage in phase by λ/4 when Z is connected in series with X across the source.

(i) Identify the circuit elements X, Y and Z. (ii) When all the three elements are connected in series across the same source, determine

the impedance of the circuit. (iii) Draw a plot of the current versus the frequency of applied source and mention the

significance of this plot.

Q4. State Lenz’s law. Illustrate, by giving an example, how this law helps in predicting the direction of the current in a loop in the presence of a changing magnetic flux. In a given coil of self-inductance of 5 mH, current changes from 4 A to 1 A in 30 ms. Calculate the emf induced in the coil. Q5. A series LCR circuit is connected across an a.c. source of variable angular frequency ‘ω’. Plot a graph showing variation of current I as a function of ‘ω’ for two resistances R1 and R2 (R1 > R2). Answer the following questions using this graph:

(a) In which case is the resonance sharper and why? (b) In which case is the power dissipation more and why?

Q6. State the condition for resonance to occur in series LCR ac circuit and derive an expression for resonance frequency. Ans: Z = R

ωL = 1/ωC ω =1/√LC f = ω/2π = 1/2π √LC

Q7. What are eddy currents? How are they produced? Give two applications of eddy currents. Ans: When a metallic plate is placed in a time varying magnetic field, the magnetic flux liked with the plate changes, the induced current are set up in plate, these current are called eddy currents. Applications:

(i) Induction furnace (ii) Induction motor.

Q8. Mention two reasons for energy losses in an actual transformer. Ans: (i) Joule’s heating, (ii) Flux leakage

ELECTRO-MAGNETIC WAVES

1 MARK QUESTIONS:

Q1. Which part of EM spectrum has the largest penetrating power? Ans: ϒ- rays. Q2. Which part of EM spectrum is used in operating a RADAR? Ans: Microwaves. Q3. How are X-rays produced? Ans: X-rays are produced when high energetic electron beam is made incident on a metallic target of high melting point and high atomic weight. Q4. What is the origin of the displacement current? Ans: Time variation of electric flux. Q5. Which of the following has the shortest wave length: microwaves, UV rays or X- rays? Ans: X-rays 2 MARKS QUESTIONS: Q1. Experimental observations have shown that X- rays

(i) travels in vacuum with a speed of 3 X 108 m/s. (ii) exhibit the phenomenon of diffraction and can be polarized. Ans: (i) X-rays are electromagnetic waves.

(iii) X-rays are transverse in nature.

Q2. The small ozone layer on the top of the stratosphere is crucial for human survival, why? Ans: The ozone layer absorbs UV and other low wavelength radiation which are harmful to living cells of human bodies and plants; hence ozone layer is crucial for human survival. Q3. Explain briefly how EM waves are produced by an oscillating charge. How frequency of the EM waves is produced related to that of the oscillating charge? Ans: An oscillation or accelerated is supposed to be source of an EM wave. An oscillating charge produces an oscillating electric field in space which further produces an oscillating magnetic field which in turn is a source of electric field. These oscillating electric and magnetic field, hence, keep on regenerating each other and an EM wave is produced. The frequency of EM wave = Frequency of oscillating charge. 3 MARKS QUESTIONS: Q1. Name the parts of the electromagnetic spectrum which is

(a) suitable for radar systems used in aircraft navigation. (b) used to treat muscular strain. (c) used as a diagnostic tool in medicine. Write in brief, how these waves can be produced.

Ans: (a) Microwave, Production : Klystron/magnetron/Gunn diode (any one) (b) Infrared Radiation, Production : Hot bodies / vibrations of atoms and molecules (any one) (c) X-Rays, Production: Bombarding high energy electrons on metal target/ x-ray tube/inner shell electrons(any one).

Q2. How are electromagnetic waves produced? What is the source of the energy carried by a propagating electromagnetic wave? Identify the electromagnetic radiations used

(i) in remote switches of household electronic devices; and (ii) as diagnostic tool in medicine.

Ans: Electromagnetic waves are produced by accelerated / oscillating charges which produces oscillating electric field and magnetic field (which regenerate each other). Source of the Energy: Energy of the accelerated charge. (or the source that accelerates the charges) Identification: (1) Infra red radiation (2) X - rays Q3. (a) Which one of the following electromagnetic radiations has least frequency: UV radiations, X-rays, Microwaves

(b) How do you show that electromagnetic waves carry energy and momentum?

(c) Write the expression for the energy density of an electromagnetic wave propagating in free space. Ans: (a) Microwaves (b) Electric charges can acquire energy and momentum from EM waves. (c) U = UE + UB = ½ εo E2 + B2/2µo. Q4. Arrange the following electromagnetic waves in the order of their increasing wavelength:

(a) ϒ-rays (b) Microwaves (c) X-rays (d) Radio waves How are infra-red waves produced? What role does infra-red radiation play in (i) maintaining the Earth’s warmth and (ii) physical therapy?

Ans: Gamma(훾) rays, X-rays, Microwaves, Radio waves Infrared rays are produced by hot bodies / vibration of atoms and molecules Infrared rays: (i) Maintain Earth’s warmth through green house effect (ii) Produce heat Q5. How are the following types of electromagnetic waves produced:

(i) Microwaves (ii) Infra-red waves

Write two important uses of each of the above electromagnetic radiations. Ans: (i) Microwaves are produced by special vacuum tubes called Klystrons / Magnetrons / Gun diodes / Point contact diodes. (any one) Uses: Radar system, Ovens, Communication (any two) (ii) Infrared waves are produced by vibration of atoms and hot bodies. Uses: Physical therapy, remote switches in household electronic systems, detectors in earth satellites

Q6. How are electromagnetic waves produced ? What is the source of energy of these waves? Draw a schematic sketch of the electromagnetic waves propagating along the + x-axis. Indicate the directions of the electric and magnetic fields. Write the relation between the velocity of propagation and the magnitudes of electric and magnetic fields. Ans: Production: Electromagnetic waves are produced by “accelerated Charges” The battery/ Electric field that accelerates the charge carriers is the source of energy of EM waves. Directions of 퐸 Along 푦 axis/ Along 푧 axis Directions of 퐵 Along 푧 axis/ Along 푦 axis Relation: 푐=퐸퐵

OPTICS

1 MARK QUESTIONS:

Q1. When light travels from an optically denser medium to a rarer medium, why does the critical angle of incidence depend on the colour of light?

Ans: Critical angle depends upon the refractive index (n) of the medium and refractive index is different for different colours of light.

Q2. A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens?

Q3. The radii of curvature of both the surfaces of a lens are equal. If one of the surfaces is made plane by grinding, how will the focal length and power of the lens change?

Q4. For the same angle of incidence, the angle of refraction in two media A and B are 25o and 35o respectively. In which medium is the speed of light less?

Q5. The focal length of an equiconvex lens is equal to the radius of curvature of either face. What is the refractive index of the material of the lens?

Q6. Why does bluish colour predominate in a clear sky?

Q7. For the same angle of incidence, the angle of refraction in two media A and B are 25° and 35° respectively. In which medium is the speed of light less?

Q8. The radii of curvature of both the surfaces of a lens are equal. If one of the surfaces is made plane by grinding, how will the focal length and power of the lens change?

Q9. To which wavelength of light our eye most sensitive? In which region does this wavelength lie? Ans: Our eye is most sensitive to wavelength λ = 5500 A0. This wavelength lies in the yellow-green region. Q10. A ray of light falls normally on a mirror. What are the values of angle of incidence and angle of refraction? Ans: I and r = 0o Q11. A person moves with velocity v towards a plane mirror. With what velocity does his image moves toward him? Ans: 2 v Q12. A mirror turns about 10o. By what angle will the reflected ray turn? Ans: 20o Q13. An object is held between two plane parallel mirrors inclined at 45o to each other. How many images do you expect to see? Ans: n = (360/45) – 1 = 7 images Q14. One wants to see an enlarged image of an object in a mirror. Which type of mirror one should use? Ans: Concave mirror Q15. Define refractive index. Ans: The refractive index of a medium for a light of a given wavelength is the ratio of the speed of light in vacuum to its speed in that medium.

Q16. State the factors on which the refractive index of a medium depends. Ans: (i) Nature of medium (ii) wavelength of light (iii) temperature (iv) Nature of surrounding medium. Q17. For which material the value of refractive index is (i) minimum (ii) maximum? Ans: (i) Vacuum (ii) Diamond. Q18. What is lateral shift in refraction? Ans: The side wise shift in the path of the light on emerging from a refracting medium with parallel faces is called lateral shift. Q19. Define critical angle for total internal reflection. Ans: The angle of incidence in the denser medium for which the angle of refraction in the rarer medium is 90o is called critical angle of the denser medium. Q20. State the conditions under which total internal reflection occurs. Ans: (i) Light must travel from denser to rarer medium.

(ii) The angle of incidence in the denser medium must be greater than the critical angle between two media. Q21. Name the physical principle on which the working of optical fibres is based. Ans: Total internal reflection. Q22. State the factors on which dispersive power of a prism depends. Ans: (i) Nature of the prism material

(ii) Choice of extreme colours for which dispersive power is to be measured. Q23. How does the angle of minimum deviation of a glass prism vary, if the incident violet light is replaced with the red light? Ans: As δ = (µ - 1) A and µR < µV, so the angle of minimum deviation decrease which incident violet light is replaced with red light. Q24. What is monochromatic light? Give one example of the source of monochromatic light. Ans: A light of single wavelength is called monochromatic light. The commonly used source is sodium lamp. Q25. What can we say about the length of a compound microscope if the final image is formed at infinity? Ans: The length of compound microscope is greater than fo + fe. Q26. Can two Wave fronts cross one another? Give reason.

Ans: No, if they intersect, then there will be two rays or two direction of propagation of energy at the point of intersection which is not possible. Q27. Define the term coherence for light wave. Ans: Two light waves interfering with each other are said to show coherence if the initial phase difference between them remains constant with time. Q28. What happen to the interference pattern if the phase difference between two sources varies continuously? Ans: The position of bright and dark fringes will change rapidly. A uniform illumination is seen on the screen. Q29. If the separation between two slits is decreased in Young’s double slit experiment keeping the screen position fixed, what will happen to the fringes width? Ans: β = Dλ/d, β increases. Q30. State the essential condition for diffraction of light to occur. Ans: It occurs when size of the obstacle or the aperture is comparable to the wavelength of the light. Q31. Which type of wave show the property of polarization? Ans: Transverse. Q32. Name three properties, which are mutually perpendicular to each other in a plane, polarized light wave. Ans: electric field vector, magnetic field vector and direction of propagation of the light. Q33. What is the geometrical shape of the wave front of light diverging from a point source. Ans: Spherical Q34. What are secondary wavelets? Ans: Every point on a wave front becomes a source of fresh disturbance which is called a secondary wavelet. Q35. What is interference of light? Ans: When two wave of same frequency and having zero or constant phase difference traveling in same direction superpose on each other, intensity in the region of super position gets redistributed, becoming maximum at some points minimum at other . This phenomenon is called interference of waves.

Q36. How can the fringe width be increased in the Young’s double slits experiments? Ans: β α λ. Q37. Is the central fringe bright or dark in the Young’s double slits experiments? Ans: yes Q38. What is the ratio of the fringe width for bright and dark fringes in the Young’s double slits experiments? Ans: 1 : 1 Q39. What is the ratio of the slit width when amplitudes of light waves from them have a ratio √2 : 1? Ans: w1/w2 = a1

2/a22 = 2:1

Q40. What is diffraction of light? Ans: The phenomenon of bending of light around the corners of small obstacles or aperture and its spreading into the regions of geometrical shadow is called diffraction of light. Q41. State Brewster law for polarization of light. Ans: it states that the tangent of the polarizing angle of incidence for a transparent refracting medium is equal to the refractive index of the medium. tan ip = µ Q42. Give two examples of commonly used devices which make use of Polaroid. Ans: Sun glasses and LCDs Q43. What percentage of the incident light is transmitted if the angle between polarizer and analyser is 30o? Ans: 75%. Q44. Why do we see two images of an object when looked through a calcite crystal? Ans: due to double refraction of light. Q45. State one feature by which the phenomenon of interference can be distinguished from that of the diffraction? Ans: In interference all bright fringes are of same intensity. But in diffraction it is not. Q46. Which phenomenon leads us to conclude that light has transverse wave nature? Ans: Polarization of light. Q47. Is light from the sodium lamp plane polarized? Ans: No

Q48. A polarizer and analyser are also obtained that no light is transmitted. What is the angle between the axes of polarizer and analyser? Ans 90o Q49. A small piece of stone is dropped into a pond of still water. What is the shape of wave front? Ans: Circular. Q50. What is the shape of wave front emitted by a light source in the form of a narrow slit? Ans: Cylindrical shape. Q51. Write a relationship between angle of incidence I, angle of prism A and angle of minimum deviation for a triangular prism. Ans: 2 I = A + δm Q52. If the polarizing angle for air glass interface is 56.3o, what is the angle of refraction in glass? Ans: rp = 90o – ip = 900 – 56.3o = 33.7o

Q53. When monochromatic light travels from one medium to another, its wavelength changes but frequency remains the same. Explain. Ans. Frequency is a fundamental property and does not depend upon the medium. Q54. Does the apparent depth of a tank of water change if viewed obliquely? Ans. Yes Q55. For the same angle of incidence the angle of refraction in three different media A, B and C are 150, 25o and 35o respectively. In which medium the velocity of light is minimum? Ans. A Q56. A glass lens of refractive index 1.5 is placed in a trough of liquid. What must be the refractive index of the liquid in order to make the lens disappear? Ans. 1.5 Q57. Why does the bluish colour predominate in the clear sky? Ans. I α 1/λ4 Q58. What type of wavefront will emerge from a (i) point source and (ii) distant light source? Ans. (i) Spherical wavefront (ii) plane wavefront Q59. Can two identical and independent sodium lamps act as coherent sources? Ans. No

Q60. In what way the diffraction from each slit related to interference pattern in double slit experiment? Ans. The intensity of interference fringes in a double slit arrangement is modulated by the diffraction pattern of each slit. Q61. Does the polarizing angle of any transparent medium depend on the wavelength of the light. Ans. Yes Q62. A person moves with velocity v towards a plane mirror. With what velocity does his image moves toward him? Ans: 2 v Q63. What is shape of the wavefront on the earth for sunlight? Ans. Plane wavefront. Q64. For which colour the magnification power of simple microscope is highest? For which colour is it lowest? Ans. Highest- violet and lowest- red Q65. Among the following waves which can be polarized? Sound wave, radio waves, X-rays, cathode rays. Ans. Radio waves and X- rays Q66. Can two Wave fronts cross one another? Give reason. Ans: No, if they intersect, then there will be two rays or two direction of propagation of energy at the point of intersection which is not possible. 2 MARKS QUESTIONS: Q1. Use the mirror equation to show that an object placed between f and 2f of a concave mirror produces a real image beyond 2f. Q2. Find an expression for intensity of transmitted light when a polaroid sheet is rotated between two crossed polaroids. In which position of the polaroid sheet will the transmitted intensity be maximum ? Q3. Define a wavefront. Using Huygens’ Wave Principle, draw the shape of a refracted wavefront, when a plane wave is incident on a convex lens. Ans:

Wave front: It is the locus of points which oscillate in phase. Or It is a surface of constant phase.

Q4. (a) When a wave is propagating from a rarer to a denser medium, which characteristic of the wave does not change and why?

(b) What is the ratio of the velocity of the wave in the two media of refractive indices µ1 and µ2? Ans: (a) Frequency does not change, as frequency is a characteristic of the source of waves. (Alternatively: 풗ퟏ/흀ퟏ= 풗ퟐ/흀ퟐ=푛) (b) The ratio of velocities of wave in two media of refractive indices μ1 and μ2 is μ2/μ1. (Alternatively: 풗ퟏ/풗ퟐ= μퟏ/μퟐ) Q5. A ray of light incident on an equilateral glass prism propagates parallel to the base line of the prism inside it. Find the angle of incidence of this ray. Given refractive index of material of glass prism is 1.5. Q6. A biconvex lens of glass of refractive index 1·5 having focal length 20 cm is placed in a medium of refractive index 1·65. Find its focal length. What should be the value of the refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass? Q7. Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern. Ans:

Interference Diffraction 1 All the bright bands are of same intensity. Intensity of bright bands goes on

decreasing with increasing order.

2 All the bright bands are of same width. Not of same width. 3 Dark bands may be completely dark. Not completely dark.

4 Number of fringes is more. Less in number.

Q8. Explain the basic differences between the construction and working of a telescope and a microscope.

Microscope Telescope

Construction Objective is of very short focal length and short aperture and eye piece of short focal length and large aperture. 푓푒 > 푓표

Objective is of large focal length and large aperture but eye piece of short focal length and short aperture.

Working It will form magnified image of a small nearby object. (Object is placed close to focus of objective which forms real and magnified image.)

It will form magnified image of distant object. (Objective will form the image of distant object at its focus and image is diminished.)

Q9. A screen is placed 90 cm away from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens.

Q10. A ray of light incident on one of the faces of a glass prism of angle ‘A’ has angle of incidence 2A. The refracted ray in the prism strikes the opposite face which is silvered, the reflected ray from it retracing its path. Trace the ray diagram and find the relation between the refractive index of the material of the prism and the angle of the prism.

Q11. An object is placed 40 cm from a convex lens of focal length 30 cm. If a concave lens of focal length 50 cm is introduced between the convex lens and the image formed such that it is 20 cm from the convex lens, find the change in the position of the image.

Q12. A ray of light passes through an equilateral glass prism such that the angle of incidence is equal to the angle of emergence and each of these angles is equal to 3/4 of angle of prism. Find the angle of deviation.

Q13. Calculate the speed of light in a medium whose critical angle is 45o. Does critical angle for a given pair of media depend on the wavelength of incident light? Give reason. Q14. You are given two converging lenses of focal lengths 1·25 cm and 5 cm to design a compound microscope. If it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece. Q15. A small telescope has an objective lens of focal length 150 cm and eyepiece of focal length 5 cm. What is the magnifying power of the telescope for viewing distant objects in normal adjustment? If this telescope is used to view a 100 m tall tower 3 km away, what is the height of the image of the tower formed by the objective lens?

Q16. Define a wavefront. Using Huygens’ wave principle, draw the shape of a refracted wavefront, when a plane wave is incident on a convex lens. Q17. A biconvex lens of glass of refractive index 1·5 having focal length 20 cm is placed in a medium of refractive index 1·65. Find its focal length. What should be the value of the refractive index of the medium in which the lens should be placed so that it acts as a plane sheet of glass? Q18. Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern. Q19. Write the important characteristic features by which the interference can be distinguished from the observed diffraction pattern. Q20. Explain the basic differences between the construction and working of a telescope and a microscope. Q21. A screen is placed 90 cm away from an object. The image of the object on the screen is formed by a convex lens at two different locations separated by 20 cm. Determine the focal length of the lens. Q22. State the essential conditions for two light waves to be coherent. Ans: (i) The two wave must be continuous.

(ii) The two wave must be of same frequency or wavelength. (iii) They should have a constant or zero phase difference.

(iv) Preferably, they should have equal amplitudes. Q23. Write two uses of polaroids. Ans: (i) Polaroids are used in sunglass and camera filters to reduce glare of light.

(ii) In window panes of aero planes to control the amount of light coming in. Q24. Why longitudinal wave cannot be polarized? Ans: In polarization, vibrations perpendicular to the direction of propagation are restricted to just one direction. This is possible in transverse wave which has such vibrations. Q25. In the wave picture of light, intensity of light is determined by the square of the amplitude of the wave. What determines the intensity of light in the photon picture of light? Ans: Number of photons incidence per unit area. Q26. Astronomers prefer to use telescopes with large objective diameter to observe astronomical objects. Why? Ans: It increases the resolving power. Because it as large gathering capacity. Q27. What is the effect on the interference pattern observed in young’s double slit experiment in the following cases:

(i) Screen is moved away from the plane of the slits. (ii) Separation between the slits is increased. (iii) Widths of the slits are doubled.

Ans: (i) As β α D (ii) β α 1/d (iii)When widths of slits are doubled, contrast between maxima and minima decrease due to the overlapping of interference patterns formed by various narrow pairs of the two slits.

Q28. What change in the interference pattern in Young’s double slit experiment will be observed when:

(i) light of smaller frequency is used? (ii) the apparatus is immersed in water?

Ans: (i) When light of smaller frequency is used, Fringe width increases. (ii) Wavelength of light in water decreases, since β α λ.

Q29. What change will occur in diffraction pattern if

(i) light of smaller wavelength is used (ii) slit is made narrower, and (iii) another slit is placed near and parallel to the first slit?

Ans: (i) When light of smaller wavelength λ is used, the diffraction pattern become narrower. (ii) When d decreases, Өn increases and the diffraction pattern spreads out. (iii) Interference pattern replaces the diffraction pattern.

Q30. Define the term magnification power and resolving power of a telescope. Ans: The magnification power of a telescope in normal adjustment is the ratio of the angle subtended by the image at the eye as seen through the telescope to the angle subtended by the object directly, when both the object and image lie at infinity. m = fo/fe When the final image is formed at the least distance of distinct vision, m = fo/fe(1 + fe/D) Q31. A person standing before a concave mirror cannot see his inverted image unless he stands beyond the centre of curvature. Why? Ans: A concave mirror forms real inverted image when the object is placed beyond F. When the person stands between F and C, the image is formed beyond C i.e. behind the man and the man is not able to see his image. When he stands beyond C, real and inverted image is formed between F and C i.e. in front of him and so he can see his image.

Q32 A right angled prism made from a material of refracted index µ is kept in air. A ray PQ is incident normally on the side AB of the prism as shown. Find the maximum value of Q upto which this incident ray necessarily undergoes total internal reflection at the face AC of the prism. Ans: A P Ө Q B C Angle of incidence on face AC = i = A Clearly, Ө = π/2 – A =π/2 – i The minimum value of i so that Total Internal Reflection at face AC, is equal to critical angle iC = Sin-1 (1/µ) Hence, the maximum value of Ө corresponding to the minimum value of i is Өmax = π/2 – ic = π/2 – Sin-1 (1/µ) Q33. If the wavelength of incident light on a (i) concave mirror, and (ii) convex lens is increased, how will the focal length of each of these change? Ans: (i) Focal length of concave mirror remains unchanged. (ii) AS f = 1/ (µ-1) For λ2 > λ1, µ2 < µ1 f2 > f1, i.e. when wavelength of incident light is increased, focal length of the convex lens increases.

Q34. Give reason for the following observation on the surface of the moon: (i) Sunrise and sunset are abrupt. (ii) Sky appears dark. Ans. (i) Moon has no atmosphere. There is no scattering of light. (ii) There is no atmosphere refraction. Q35. How will the magnifying power of a refracting type astronomical telescope be affected on increasing for its eyepiece (i) the focal length (ii) the aperture? Justify your answer. Ans. (i) As m = fo/fe so m decrease on increasing fe. (ii) m is not affected. Q36. Why should we have a narrow source to produce good interference fringes?

Ans. It is because a broad source is equivalent to the large number of narrow sources lying close to each other. Different pair of narrow sources will produce their own interference pattern which will overlap each other. So the fringe system is lost. Q37. Why are diffraction effects more prominent trough a slit formed by two blades than through a slit formed by two fringes? Ans. Diffraction is prominent when we use a narrow slit having parallel edges. Such a slit can be obtained by using two blades and not by using two fringes. Q38. Light wave can be polarized while sound wave cannot. Why? Ans. Light wave is transverse and sound wave is longitudinal in nature. Q39. Why does a soap bubble show beautiful colours when illuminated by a white light? Ans. Light waves reflected from upper and lower surfaces of a thin film interfere. Since the condition of bright and dark fringes are satisfied at the different wavelength. Q40. Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? Ans . When µL = µG Q41. In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back. Ans. Yes Q42. A virtual image, we always say, cannot be caught on the screen. Yet when we see a virtual image we are obviously bringing it on to the screen of our eye. Is there a contradiction? Ans. When the refracted or reflected rays are divergent, image formed are virtual which can to taken on the screen. 3 MARKS QUESTIONS: Q1. (i) A giant refracting telescope has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope?

`(ii) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens ? The diameter of the moon is 3.48 × 106 m and the radius of lunar orbit is 3.8 × 108 m. Q2. Answer the following questions:

(a) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1 . Find the spacing between the two slits.

(b) Light of wavelength 5000 Å propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected? Q3. In Young’s double slit experiment, the two slits are separated by a distance of 1·5 mm and the screen is placed 1 m away from the plane of the slits. A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes. Find (a) the distance of the third bright fringe = 520 nm on the screen from the central maximum. (b) the least distance from the central maximum where the bright fringes due to both the wavelengths coincide. Q4. Answer the following: (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment? (b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain, why. (c) How does the resolving power of a microscope depend on (i) the wavelength of the light used and (ii) the medium used between the object and the objective lens? Ans: (a) The intensity of interference fringes in double slit arrangement is modulated by the diffraction pattern of each slit. Alternatively, In double slit experiment the interference pattern on the screen is actually superposition of single slit diffraction for each slit. (b) Waves diffracted from the edges of the circular obstacle interfere constructively at the centre of the shadow producing a bright spot. (c) Resolving power = 2µ sin Ө/1.22λ, Resolving power is inversely proportional to wavelength and directly proportional to the refractive index. Q5. (a) The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. Describe, with the help of a suitable diagram, the basic phenomenon/process which occurs to explain this observation. (b) Show how light reflected from a transparent medium gets polarised. Hence deduce Brewster’s law. Q6. What does a polaroid consist of? Show, using a simple polaroid, that light waves are transverse in nature. Intensity of light coming out of a polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why. Q7. You are given three circuit elements X, Y and Z. When the element X is connected across an a.c. source of a given voltage, the current and the voltage are in the same phase. When the element Y is connected in series with X across the source, voltage is ahead of the current in phase by λ/4. But the current is ahead of the voltage in phase by λ/4 when Z is connected in series with X across the source. Identify the circuit elements X, Y and Z.

Q8. An object is placed 15 cm in front of a convex lens of focal length 10 cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image is formed at the position of the object itself?

Q9. Compare and explain three distinguishing features observed in Young’s double slit interference pattern with those seen for a coherently illuminated single slit producing diffraction pattern.

Q10. (a) Assume that the light of wavelength 6000 Å is coming from a star. Find the limit of resolution of a telescope whose objective has a diameter of 250 cm. (b) Two slits are made 1 mm apart and the screen is placed 1 m away. What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern? Q11. Write the expression for the resultant intensity at a point due to the superposition of two monochromatic waves

y1 = a cos ωt, y2 = a cos (ωt + Ф), where Ф is the phase difference between the two waves and a and ω denote the amplitude and angular frequency respectively. Q12. In Young’s double slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is k units. Find the intensity at a point on the screen where path difference is λ/4. Q13. (a) Give two reasons to explain why reflecting telescopes are preferred over refracting type. (b) Use mirror equation to show that convex mirror always produces a virtual image independent of the location of the object. Q14. (a) Write the necessary conditions to obtain sustained interference fringes. (b) In Young’s double slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What information can one obtain from the slope of the curve? (c) What is the effect on the fringe width if the distance between the slits is reduced keeping other parameters same? Q15. State clearly how an unpolarised light gets linearly polarised when passed through a polaroid.

(i) Unpolarised light of intensity I0 is incident on a polaroid P1 which is kept near another polaroid P2 whose pass axis is parallel to that of P1. How will the intensities of light, I1 and I2, transmitted by the polaroids P1 and P2 respectively, change on rotating P1 without disturbing P2?

(ii) Write the relation between the intensities I1 and I2. Q16. (a) The light from a clear blue portion of the sky shows a rise and fall of intensity when viewed through a polaroid which is rotated. Describe, with the help of a suitable diagram, the basic phenomenon/process which occurs to explain this observation.

(b) Show how light reflected from a transparent medium gets polarised. Hence deduce Brewster’s law. Q17. (a) In what way is diffraction from each slit related to the interference pattern in a double slit experiment? (b) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the centre of the shadow of the obstacle. Explain, why. (c) How does the resolving power of a microscope depend on (i) the wavelength of the light used and (ii) the medium used between the object and the objective lens? Q18. An object is placed 15 cm in front of a convex lens of focal length 10 cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image is formed at the position of the object itself? Q19. What does a polaroid consist of? Show using a simple polaroid, that light waves are transverse in nature. Intensity of light coming out of a polaroid does not change irrespective of the orientation of the pass axis of the polaroid. Explain why? Q20. (a) Assume that the light of wavelength 6000 Å is coming from a star. Find the limit of resolution of a telescope whose objective has a diameter of 250 cm. (b) Two slits are made 1 mm apart and the screen is placed 1 m away. What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern? Q21. Compare and explain three distinguishing features observed in Young’s double slit interference pattern with those seen for a coherently illuminated single slit producing diffraction pattern.

5 MARKS QUESTIONS:

Q1. (a) Using Huygens’s construction of secondary wavelets explain how a diffraction pattern is obtained on a screen due to a narrow slit on which a monochromatic beam of light is incident normally. (b) Show that the angular width of the first diffraction fringe is half that of the central fringe. (c) Explain why the maxima at = ( n + ½) λ/a become weaker and weaker with increasing. Ans.

We can regard the total contribution of the wavefront LN at some point P on the screen, as the resultant effect of the superposition of its wavelets like LM, MM2, M2N. These have to be superposed taking into account their proper phase differences .We therefore get maxima and minima ,i.e a diffraction pattern, on the screen. (b) Condition for first minimum on the screen a Sinθ = λ => θ = λ/a ∴ angular width of the central fringe on the screen (from figure)

Condition for first minimum on the screen a Sinθ = λ, => θ = λ/a ∴ angular width of the central fringe on the screen (from figure) ὼ = 2 θ = 2 λ/a Angular width of first diffraction fringe (From fig) = λ/a Hence angular width of central fringe is twice the angular width of first fringe. Maxima become weaker and weaker with increasing n. This is because the effective part of the wavefront, contributing to the maxima, becomes smaller and smaller, with increasing n.

Q2. (a) A point object ‘O’ is kept in a medium of refractive index n1 in front of a convex spherical surface of radius of curvature R which separates the second medium of refractive index n2 from the first one, as shown in the figure. Draw the ray diagram showing the image formation and deduce the relationship between the object distance and image distance in terms of n1, n2 and R. (b) When the image formed above acts as a virtual object for a concave spherical surface separating the medium n2 from n1 (n2 > n1), draw this ray diagram and write the similar (similar to (a)) relation. Hence obtain the expression for the lens maker’s formula. Ans. (a)

For small angles < NOM = tan NOM = MN/OM <NCM = tan NCM = MN/MC < NIM = tan NIM = MN/MI In Δ NOC , i = NOM + NCM i = MN/OM + MN/MC …(i) Similarly r = NCM + NIM r = MN/MC + MN/MI ……(II) Using Snell’s Law n1 sin i = n2 sin r For small angles n1 i = n2 r Substituting for i and r, we get n1/OM + n2/MI = n2-n1/MC Here, OM = -u , MI = + v, MC = +R Substituting these ,we get n2/v - n1/u = n(b)

Q3. (a) A concave mirror produces a real and magnified image of an object kept in front of it. Draw a ray diagram to show the image formation and use it to derive the mirror equation. (b) A beam of light converges at a point P. Now a lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (i) a convex lens of focal length 20 cm, (ii) a concave lens of focal length 16 cm ?

Ans.

In Δ ABP and Δ A’B’P AB/A’B’ = AP/A’P In Δ MPF and Δ B’A’F MP/A’B’ = FP/A’F But AB = MP ∴ AP/A’P = FP/A’P = FP/A’P-FP -u/-v = -f/-(v-f) uv-uf = vf Dividing by uvf 1/f – 1/v = 1/u 1/f = 1/v + 1/u b) Here the object is virtual and image is real u = +12 cm (i) 1/v-1/u = 1/f 1/v – 1/12 = 1/20, v = 7.5 cm from the lens (ii) 1/v -1/u = 1/f, 1/v – 1/12 = 1/-16 V = 48 cm from the lens

Q4. (a) A point-object is placed on the principal axis of a convex spherical surface of radius of curvature R, which separates the two media of refractive indices n1 and n2 (n2 > n1). Draw the ray diagram and deduce the relation between the distance of the object (u), distance of the image (v) and the radius of curvature (R) for refraction to take place at the convex spherical surface from rarer to denser medium. (b) Use the above relation to obtain the condition on the position of the object and the radius of curvature in terms of n1 and n2 when the real image is formed. Ans.

From the diagram : By Snell‟s law , ∠ 푖= ∠푁푂푀+ ∠ 푁퐶푀 ∠ 푟= ∠푁퐶푀− ∠ 푁퐼푀 푛1/푂푀+ 푛2/푀퐼= 푛2 −푛1/푀퐶 푛1sin푖 = 푛2sin푟 Substituting for i and r. and simplifying, we get Substituting values of OM , MI and MC 푛2 /휐−푛1 /푢= 푛2 −푛1 /푅 (b) Condition for real image: v is positive

∴푛2/푣 >0 n1/u > n2-n1/R ∴ 푢 > 푛1푅/푛2− 푛1

Q5. (a) Draw a labelled ray diagram showing the formation of image by a compound microscope in normal adjustment. Derive the expression for its magnifying power. (b) How does the resolving power of a microscope change when (i) the diameter of the objective lens is decreased, (ii) the wavelength of the incident light is increased ? Justify your answer in each case.

Ans.

Definition of limit of resolution The minimum linear or angular separation between two point objects at which they can be just separately seen or resolved by an optical instrument. It depends on (i) Wavelength of light used (ii) Medium between object and objective lens Resolving power of microscope is the reciprocal of its limit of resolution (b) Resolving power of compound microscope can be increased by (i) Decreasing wavelength (ii) Increasing refractive index of the medium between object and objective of the microscope. (c) A telescope produces an (angularly) magnified image of the far object and thereby enables us to resolve them. A microscope magnifies small objects which are near to our eye.

Q6. (a) Define a wavefront. (b) Using Huygens’ principle, draw the diagrams to show the nature of the wavefronts when an incident plane wavefront gets (i) reflected from a concave mirror, (ii) refracted from a convex lens. (c) Draw a diagram showing the propagation of a plane wavefront from denser to a rarer medium and verify Snell’s law of refraction.

Ans. a) Locus of all the points which are in same phase / surface of constant phase b) (i)

(ii)

(c)

Q7. (a) In Young’s double slit experiment, deduce the conditions for obtaining constructive and destructive interference fringes. Hence deduce the expression for the fringe width. (b) Show that the fringe pattern on the screen is actually a superposition of single slit diffraction from each slit. (c) What should be the width of each slit to obtain 10 maxima of the double slit pattern within the central maximum of the single slit pattern, for green light of wavelength 500 nm, if the separation between two slits is 1 mm ?

Ans.

(a) Path difference (Δ) = 푆2푃−푆1푃=푑 sin휃=푑푥/퐷, for constructive interference, Δ=푛휆 [푛=0,1,2..] Destructive interference, Δ= (2푛−1) 휆/2[푛=1,2..], for bright bands, Δ=푛휆=푥푛푑/퐷 표푟 푥푛=푛휆퐷/푑 For dark bands, Δ= (2푛−1) 휆/2= 푥푛푑퐷 표푟 푥푛= (2푛−1) 휆퐷/2푑, Fringe width 훽=푋푛−푋푛−1= 휆퐷/푑 (b)

(c) 10 훽= width of central maxima, 10퐷휆/푑 = 2퐷휆/푎 푎=푑/5=1/5푚푚=0.2 푚푚 Q8. (a) Two thin convex lenses L1 and L2 of focal lengths f1 and f2 respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens L1. Draw a ray diagram to show the image formation by the combination and hence derive the expression for the focal length of the combined system. (b) A ray PQ incident on the face AB of a prism ABC, as shown in the figure, emerges from the face AC such that AQ = AR. A Q R P B C Draw the ray diagram showing the passage of the ray through the prism. If the angle of the prism is 60o and refractive index of the material of the prism is 4/3, determine the values of angle of incidence and angle of deviation.

Ans. (a)

For First lens 1/푣1−1/푢=1/푓1 (푖) For Second lens 1/푣−1/푣1=1/푓2 (푖푖) By adding (i) and (ii) 1/푣−1/푢=1/푓1+1/푓2 Or 1/푓=1/푓1+1/푓2 b) Ray Diagram

∴푟=퐴/2=30° ∴ √3×sin30=sin푖 ⇒푖=60° ∴푒=60° 푖+푒=퐴+퐷

60+60=60+퐷 ∴퐷=60° Q9. (a) State the essential conditions for the phenomenon of total internal reflection to take place. (b) Draw a ray diagram to show how a right isosceles prism made of crown glass can be used to obtain the inverted image. (c) Explain briefly with the help of a necessary diagram, how the phenomenon of total internal reflection is used in optical fibres. Illustrate giving an example how optical fibres can be employed for transmission of optical signals.

Ans.

(a) Essential conditions

(1) The ray should pass from an optically denser medium into an optically rarer medium. (2) Angle of incidence should be greater than the critical angle for the given pair of media. (b)

(c)

When ray of light enters into an optical fibre through one of its ends , it undergoes repeated total internal reflections along the length of the optical fibre as the angle of incidence at every point inside optical fibre is greater than the critical angle. Example : Optical fibres are used for transmitting and receiving optical signals to facilitate visual examination of internal organs of human body / for long distance communication through optical fibre cables. Q10. (a) Draw a suitable diagram to demonstrate that given the shape of a wavefront at t = 0, its shape at a later time t1 can be obtained using Huygens’ geometrical construction. (b) Consider the propagation of a plane wavefront from a rarer to a denser medium and verify Snell’s law of refraction. Show that when a wave gets refracted into a denser medium, the wavelength and speed of propagation decreases but the frequency remains the same.

Ans. (a)

(b)

Q11. (a) Draw a ray diagram for the formation of image by a compound microscope in normal adjustment. (b) Obtain the expression for the minimum separation between the two points seen as distinct in a microscope. What is its relation with the resolving power? Mention the factors by which the resolving power of a microscope can be increased. Ans.

a)

(a) Definition of limit of resolution The minimum linear or angular separation between two point

objects at which they can be just separately seen or resolved by an optical instrument. It depends on

i) Wavelength of light used ii) Medium between object and objective lens

(b) Resolving power of microscope is the reciprocal of its limit of resolution b) Resolving power of compound microscope can be increased by

i) Decreasing wavelength

ii) Increasing refractive index of the medium between object and objective of the microscope.

(c) A telescope produces an (angularly) magnified image of the far object and thereby enables us to

resolve them. A microscope magnifies small objects which are near to our eye. CURRENT ELECTRICITY

QUESTION OF ONE MARK (VERY SHORT ANSWER)

Q1. Plot a graph showing variation of current versus voltage for a material.

Ans 1.

1

Q2. The graph shown in the figure represents a plot of current versus voltage for a given semiconductor. Identify the region, if any, over which the semiconductor has a negative resistance.

Ans2. Region BC shows negative resistance. 1

Q3. Plot a graph showing the variation of resistance of a conducting wire as a function of its radius, keeping the length of the wire and its temperature as constant.

Ans3. R = ρ , R α 1

Q4. Define the term ‘Mobility’ of charge carriers in a conductor. Write its S.I. unit.

Ans4 .Mobility is defined as the magnitude of the drift velocity acquired by it in a unit electric

field. μ = 퐈풗풅 퐈

푬 , S.I. unit m2 V-1 s-1 . 1

Q5. The V – I graph for two ohmic conductors are as shown in figure. What is the ratio of resistances of conductor 1 and conductor 2 in terms of θ1 and θ2.

Ans5. cot휃 / tan 휃

1 Q6. A copper wire of length 1m and radius 1mm is joined in series with an iron wire of length 2m and radius 3mm and a current is passed through the wires. What will be the ratio of the current density in the copper and iron wires?

Ans6: = ⁄ ⁄

= 9 : 1 1

Q7. A wire 50cm long and 1mm2 in cross-section carries a current of 4A when connected to a 2V battery. What is the resistivity of the wire? Ans7: ρ=R = = 1x10-6 Ω m 1

Q8. Show variation of resistivity of ‘Si’ with temperature in a graph. Ans8. Resistivity of Si decreases rapidly with increasing temperature. 1

Q9. Two wires one of maganin and the other of copper have equal length and equal resistance. Which one of these wires will be thicker? Ans9. R = ρ 1 Resistivity ‘ρ’ of maganin is much greater than that of copper, therefore to keep same resistance R for same length of wire, the maganin wire must be thicker.

Q10. The variation of potential difference V with length L in case of two potentiometer P & Q is as shown in fig. Which of these two will you prefer for comparing emfs of two primary cells?

Ans10. The potential gradient (V/ l) must be small. The slope V/l is smaller for a potentiometer Q, hence we shall prefer potentiometer Q for comparing the emfs of two cells. 1 Q11. Why do we prefer a potentiometer to measure e.m.f of a cell rather than a voltmeter?

Ans11. A potentiometer does not draw any current from the cell whose emf is to be determined, whereas a voltmeter always draws some current. Therefore, emf measured by voltmeter is slightly less than actual value of emf of the cell. 1

Q12.A potential difference V is applied to a copper wire of length ‘l’ and thickness ‘d. What will be the drift velocity when V is doubled?

Ans12: vd =

휏 = 휏( ) or vd ∝ V 1

The drift velocity will be doubled

Q13.A capacitor of 10µF has a potential difference of 40 volts across it. If it is discharged in 0.2 seconds, what is the average current during discharge? Ans13. Current I= = =2x10-3 A= 2mA 1 Q14.A charge of 2x10-2 C moves at 30 revolutions per second in a circle of diameter 80cm.What is the current linked with the circuit? Ans14: I= q/T = q 휈 = 2x10-2 x30=0.60A 1

Q15.A steady current is set up in a metallic wire of non-uniform cross-section .How is the rate of flow of electrons(R) related to the area of cross-section (A)?

Ans15: Rate of flow of electrons will depend upon drift velocity and drift velocity is inversely proportional to area of cross-section, i.e. vd ∝ . Therefore R ∝ 1 / A 1

Q16.In the circuit shown potential difference between X and Y is

Ans16: As the circuit is open, therefore no current flows through the circuit. Hence potential difference across X and Y= emf of battery =120V 1

Q17.The V-I graph for a good conductor makes an angle 400 with V-axis. Here V denotes voltage and I denote current. What will be the resistance of the conductor?

Ans17: Resistance R= = cot 400 1

Q18.A uniform wire of resistance 36Ω is bent in the form of a circle. What is the effective resistance across the points A and B

Ans18: R= × = 2.75Ω 1

Q19.The amount of charge passed in time ‘t’ through a cross-section of a wire is Q(t) = At2 + Bt + C. If the numerical values A,B and C are 5,3 and 1 respectively in SI units, find the value of the current at t=5s. Ans19. I = = 2At + B, substituting the values we get-

I = 53 A 1 Q20.An electron gun emits 2.0x1016 electrons per second .What electric current does this correspond to?

Ans20: I = ne /t = × × . × =3.2x10-3A 1

Q21.The given figure shows a network of currents. The magnitude of currents is shown in figure. What is the value of ‘I’ in the figure?

Ans21. Applying Kirchhoff’s first law

I=12+3+5= 20A 1

QUESTION CARRIES TWO MARKS

Q22. A battery of emf 2V and internal resistance 0.5Ω is connected across a resistance of 9.5Ω. How many electrons cross through a cross-section of the resistance in 1 second? Ans22: The current in the circuit is- i =

. Ω . Ω = 0.2A 1

Thus, a net transfer of 0.2 C per second takes place across any cross-section in the circuit. The number of electrons crossing the section in 1 second is .

. =1.25x1018 1

Q23.Masses of the three wires of same material are in the ratios of 1:2:3 and their lengths in the ratio of 3:2:1. In what ratio will the electrical resistance of these wires be? Ans23: Mass =volume × density = Al × d A=M/l d 1/2

Resistance R= ρl / A= 1/2 R∝ 푙 /푀 R1: R2 : R3 = 27 : 6 : 1 1

Q24. Why are alloys used for making standard resistance coils? Ans24. Alloys have (i) low value of temperature coefficient and the resistance of the coil do not vary much with rise in temperature. (ii)High resistivity, so that even a smaller length of the material is sufficient to design high standard resistance. Q25.A current through a wire depends on time as I=10+4t. What will be the charge that crosses through the section of the wire in 10 seconds? Ans25: I = =10+4t

dQ =(10+4t)dt , Integrating Q =10 t + 2t2 1

Putting t=10 sec Q=300C 1

Q26. Two wires of the same material but of different diameters carry the same current I. If the ratio of their diameters is 2:1, then what will be the ratio of their mean drift velocities?

Ans26: vd = = × i.e. vd ∝ 1

= = 1

Q27. A copper wire of cross-sectional area 2.0mm2, resistivity =1.7x10-8 Ω m, carries a current of 1A.What is the electric field in the copper wire? Ans27: E= = = 1 Substituting the values E = 8.5x10-3 V/m 1 Q28. The specific resistance of a wire is ρ, its volume is 3m3 and its resistance is 3 ohm, then what will be the length of such a wire?

Ans28: R= =

/ = 1

l = ( ) / = ( × ) = 3 / 휌 1

Q29. 2, 4 and 6 S are the conductances of three conductors. When they are joined in parallel, what will be the equivalent conductance of the combination? Ans 29: R1=1

2 Ω, R2= Ω, R3 = Ω 1 In series R= R1 + R2 +R3 = 11/12 Ω Equivalent conductance G=1/R = S 1 Q30.Two resistances are joined in parallel whose resistance is (3/5) Ω .One of the resistance wire is broken and the effective resistance becomes 3Ω. What is the resistance of the wire in ohm that got broken? Ans30. Rp =

=3/5 and R1=3Ω, substituting the values, so we get R2 = (3/4) Ω

Q31. Nine resistors each of resistance R are connected in the circuit as shown in the figure .What is the effective resistance between points A and B

Ans31. The equivalent circuit is shown in figure

1

Effective resistance is Reff = ( / )

= R 1

Q32. Two cells of emfs E1 = 6V and E2 =5V are joined in parallel, without any external load .If their internal resistances are r1 = 2 Ω, r2 = 3 Ω respectively. What is the terminal potential difference across the combination? Ans32.

I = ( )

= A 1

Potential drop across 3Ω = × 3=0.6 V

Terminal potential difference across 5V= 5+0.6 = 5.6V 1 Q33.A wire of resistance R is divided in 10 equal parts. These parts are connected in parallel, what will be the equivalent resistance of such a connection? Ans33. Resistance of each part of wire r = R/10 1

Total resistance of 10 parts of wire connected in parallel = r/10= (R/10)/10 =R/100 = 0.01R 1

Q34. Two wires of same metal have the same length but their cross sections are in the ratio 3:1. They are joined in series. The resistance of the thicker wire is 10Ω. What is the total resistance of the combination? Ans34. For the same length and material,

= = R2= 3R1 and R1=10Ω 1 The resistance of thin wire R2=30Ω

Total resistance = 10+30 = 40Ω 1

Q35. What is terminal potential difference of a cell? Can its value be greater than the e.m.f of a cell? Explain. Ans35. Terminal potential difference of a cell is defined as the potential difference between the two electrodes of a cell in a closed circuit. The value of terminal potential difference of a cell is less than emf of a cell, when current is drawn from the cell. During charging of a cell the value of terminal potential difference is greater than the emf of a cell. Q36. A voltage of 30V is applied across a carbon resistor with first, second and third rings of blue, black and yellow colour respectively. Find the value of current through the resistor. Ans36. Resistance = 60 × 104 Ω. 1 I = = 5 × 10-5 A

QUESTIONS CARRY THREE MARKS

Q37. To get a maximum current through a resistance of 2.5 Ω, one can use ‘m’ rows of cells each row having ‘n’ cells. The internal resistance of each cell is 0.5Ω. What are the values of ‘m’ and ‘n’ if the total number of cells is 20? Ans37. mn =20 , R = 2.5Ω , r = 0.5Ω

R = nr/m, Substituting the values 1 n=5m 1

m× 5푚=20 or m2 = 4 So m = 2, n = 10 1

Q38.Two identical cells connected in series send 1.0A current through a 5Ω resistor. When they are connected in parallel, they send 0.8A current through the same resistor. What is the internal resistance of the cell? Ans38: case (i) E + E = ( r + r + 5 ) 1.0

2 E = 2 r + 5 ---------- (i) 1

case (ii) E = ( × + 5 ) × 0.8

E = 0.4 r + 4.0 ------------- (ii) 1 Multiplying (ii) by 2 and equating with (i) and then on solving we get, r = 2.5 Ω Q39. Two identical cells send the same current in 3Ω resistance, whether connected in series or in parallel .What is the internal resistance of the cell Ans39. Let E,r be the e.m.f and internal resistance of each cell. In series combination of two cells the current through external resistance R will be

I = 1

In parallel combination of two cells the current through external resistance r will be,

I' = = 1

If I = I' then 2 r + R = r + 2 R R = r = 3 Ω 1

Q40.In figure shown, AB is 1 metre long uniform wire of 10Ω resistance. Other data are shown in the diagram. Calculate (i) potential gradient along AB, (ii) length AO, when galvanometer shows no deflection. 3

Ans40. (i) Potential gradient along AB

= ( ) = 0.008 V cm-1 1

(ii) Current through 0.3 Ω = .

. . = 1A 1

Potential difference across 0.3 Ω = 1 × 0.3 = 0.3 V

Across length AO, V = 0.3 V is to be balanced.

Length AO = ,

. = 37.5 cm 1

Q41. 1kg piece of copper is drawn into a wire 1mm thick and another piece into a wire of 2mm thick. What will the ratio of their resistances?

Ans4. Let‘d’ be the density of the material of copper wire. Let l1, l2 be the lengths of copper wires of diameter 1mm and 2mm respectively. As Mass= volume × density = (π D2/4) l d 1/2 From this we get l1 =4 l2 1 Now R=

/ 1/2

= × =16

So ratio is 16:1 1

Q42.Determine the current drawn from a 12V supply with internal resistance 0.5Ω by the following infinite network; each resistor has 1Ω resistance. 3

Ans42.

Let R be the equivalent resistance between A & B. Since the network is infinite, therefore resistance between C & D is same as between A & B, then equivalent resistance of R & 1 Ω in parallel is R1 = × 1/2 RAB = R1 +1 +1 R1 +1 +1 = R

+ 2 = R R2 - 2R – 2 = 0, on solving 1 R = (1 + √3) Ω = 2.732 Ω 1

Current drawn I = . .

= 3.7 A ½

Q43. Establish a relation between drift velocity and time of relaxation. Use this relation to deduce the expression for the electrical resistivity of the material. Ans43. See NCERT Book Article No. 3.5

Prove that vd = - 휏 1

& use the above relation to prove 휌 = 1

Q44. A potentiometer wire of length 1 m has a resistance of 5Ω. It is connected to a 8 V battery in series with a resistance of 15Ω .Determine the emf of the primary cell which gives a balance point at 60 cm. Ans44. Given l = 1m =100cm, R= 5Ω, emf of battery = E = 8 V , R1 = 15 Ω, x = 60 cm.

I = = = 0.4 A 1

Potential drop across potentiometer wire V = I R = 0.4 × 5 = 2.0 V 1 ∴ EMF of primary cell = × x = × 60 = 1.2 V 1 Q45. A battery of emf E volt & internal resistance ‘r’ ohm is joined in series with two resistances X & Y ohm in a closed circuit. A standard cell of emf 1.06 V and a galvanometer are joined in series & the combination is connected across X. The galvanometer shows no deflection when X = 60Ω & Y = 224 Ω or when X = 40 Ω & Y = 140 Ω. Calculate the values of E & r.

Ans45. ( )

× 60 = 1.06 1

( )

× 40 = 1.06 1

On solving, we get E = 5.51 V & r = 28 Ω 1 Q46. In the potentiometer circuit shown in figure, the balance point with R = 10Ω when switch S1 is closed & S2 is open is 50 cm, while that when S2 is closed & S1 is open is 60 cm. What is the value of ‘x’? What will you do if you fail to find a balance point with the given cell E’?

Ans46. If ‘k’ is the potential gradient & I is the current through resistances 10Ω & x Ω due to cell E' . Potential difference across 10 Ω = I × 10 volt Potential difference across x Ω = I × x volt When switch S1 is closed & S2 is open

I × 10 = k × 50 -------------------------------- (i) 1/2 When switch S2 is closed & S1 is open I (10 + x) = k × 60 ---------------------------------- (ii) 1/2 Solving equations (i) & (ii)

10 + 푥10 =

6050 =

65

x = 2Ω 1 If we fail to get balance point on the potentiometer wire, it shows that potential drop across (10 +x) is greater than potential difference across potentiometer wire due to cell E. To find the balance point 1

(i) We can increase the voltage of cell E by adding one or more cells in series with E or (ii) We can reduce the current in resistance (10+ x) & hence potential difference across it by

using a suitable resistance in series with cell E’.

Q47. A potential difference V is applied to a conductor of length L, diameter D. How are the electric field E, the drift velocity vd & the resistance R affected when (i) V is doubled (ii) L is doubled?

Ans47. E = , drift velocity vd = 휏 =

휏, R= ρ L / A

(i) When V is doubled, E & vd become double, R remains unchanged. 1

(ii) When L is doubled, E & vd become half, R becomes double. 1 Q48. The heating element of an electric toaster is of nichrome. When a very small current passes through it, at room temperature 27oC, its resistance is 75.3Ω. When toaster is connected to a 230V supply, the current settles after a few seconds to a steady value of 2.68 A. What is the steady temperature of nichrome element? The temperature coefficient of resistance of nichrome over averaged temperature range is 1.7 × 10-4 oC-1. Ans48. R1 = 75.3 Ω,훼 = 1.7 × 10-4 oC -1,V= 230 V , I = 2.68 A , t1 = 27o C R2=

.= 85.82Ω 1

(t2 – t1) = ×

1

Substituting the values, t2 – t1 = 849 oC 1 Q49 .(i) Calculate the value of R in the balance condition of the wheatstone bridge , if the carbon resistor connected across the arm CD has the colour sequence red , red & orange as shown in the figure .

(ii)If now the resistance of arms BC & CD are interchanged, to obtain the balance condition, another carbon resistor is connected in place of ‘R’ . What would now be sequence of colour

bands of the carbon resistor? Ans49. (i) Let a carbon resistor S is given to the bridge, Then

= ⟹ = 1 R = S = 22 × 103Ω 1

(ii)After interchanging the resistance

= ×

× ×=

X = 4R = 4× 22 × 10 = 88 kΩ 1 The sequence of colour bands is Grey, Grey, and Orange. 1

Q50. A few storage cells in series are to be charged from a 30 V DC supply. The emf of each cell is 1.35 V & internal resistance is 0.1 Ω. The charging current is 3.0 A. In this arrangement how many cells can be charged and what extra resistance is required to be connected in the circuit? Ans50. Let ‘n’ be the maximum number of cells in series, which can be charged & R is the extra resistance required. If ‘I’ is the current in the circuit, then (n r + R ) I = 30 – n E 1 E = 1.35 V, r = 0.1 Ω, I =3.0 A Substituting the values, we get

n + .

R = .

= 18 +..

1

∴ n = 18, R = 0.1 Ω 1

QUESTION CARRIES FIVE MARKS

Q51.State the principle of potentiometer .Draw a circuit diagram used to compare the emfs of two primary cells. Derive the formula used. How can the sensitivity of a potentiometer be increased? Ans51. See NCERT Book Article No. 3.16

Principle V = k L, k = 1

Prove = , using circuit diagram. 2 +1

The sensitivity of a potentiometer can be increased by decreasing its potential gradient. The same can be achieved (a) by increasing the length of potentiometer wire. (b) by reducing the current in the potentiometer wire . 1 Q52. (i) Using Kirchhoff’s rules obtain the balance condition in terms of the resistances of four arms of Wheat stone bridge.

(ii) First a set of n equal resistors of R each are connected in series to a battery of e.m.f ‘ԑ’ and internal resistance R. A current ‘I’ is observed to flow. Then the ‘n’ resistors are connected in parallel to the same battery. It is observed that the current is increased 10 times. What is ‘n’? Ans52. (i) See NCERT Book Article No. 3.14

Prove = , using Fig. 2+1

(ii). When ‘n’ resistors are in series

I = 1/2 When ‘n’ resistors are parallel Reff = R / n

I' = = ( )

1/2

According to question, I' = 10 I Substituting the values of I & I’ n = 10 Q53. (a) A resistance of R Ω draws current from a potentiometer as shown in figure. The potentiometer has a total resistance Ro Ω. A voltage V is supplied to the potentiometer. Derive an expression for the voltage across R when the sliding contact is in the middle of the potentiometer.

(b) Why are the connections between the resistors in a meter bridge made of thick copper strips? (c)Why is it generally preferred to obtain the balance point in the middle of the meter bridge wire? Ans53. (a) The equivalent resistance of the circuit

R1 = .

1

Req = + .

= ( )

( ) -------------- (i)

Current in the circuit I = = V ( )

( ) 1

V1 = I R1 Substituting the values of I & R1

We get V1 = ( )

1

(b) Resistivity of copper wire is very less. The area of thick copper strip is more. So it offers negligible resistance in the circuit. 1 (c) The error in the measured value of unknown resistance using meter-bridge will be minimum, when the null point is obtained at the middle of meter-bridge wire .In this situation, the end error of the bridge will become ineffective. 1 Q54. State Kirchhoff’s laws. Calculate the value of current I1, I2 & I3 in the following circuit.

Ans54. Statements of Kirchhoff’s laws 1 Applying Kirchhoff’s rule to the loop PRSP 4 I3 + 40 I2 =1 --------------- (i) 1/2 Loop PRQP 5 I3 + 15 I1 = 1 ------------- (ii) 1/2 From first rule I3 = I1 + I2 --------------- (iii) Solving the above equations I1 = 퐴 = mA 1

I2 = 퐴 = mA 1

I3 = 퐴 = mA 1

Q55.(a) Six lead-acid type of secondary cells each of e.m.f 2.0V and internal resistance 0.015Ω are joined in series to provide a supply to a resistance of 8.5Ω. What are the current drawn from the supply and its terminal voltage?

(b) A secondary cell after long use has an e.m.f of 1.9V and a large internal resistance of 380Ω.What maximum current can be drawn from the cell? Could the cell derive the starting motor of a car? (c) The emf of the driver cell in the potentiometer experiment should be greater than the emf of the cell to be determined. Why? (d) Why do we prefer potentiometer of longer length for accurate measurements?

Ans55. (a) I = 푛퐸푅+푛푟 = ( ×

. × . ) = 1.4 A 1

Terminal voltage, V = I R = 1.4 × 8.5 = 11.9 V 1/2 (b) Imax = = . = 0.005 A, 1 This amount of current cannot start a car because to start the motor, the current required is 100A for few seconds. 1/2 (c) If it is not so, there will be smaller fall of potential across the potentiometer wire than the emf of the cell to be determined and hence the balance point will not be obtained on the potentiometer wire. 1 (d) The value of potential gradient decreases. Therefore the measurement is more accurate. 1 Q56. (a)Draw the circuit diagram of a potentiometer which can be used to determine the internal resistance ‘r’ of a given cell of emf (E). Describe a method to find the internal resistance of a primary cell. (b) Two students ‘X’ & ‘Y’ perform an experiment on potentiometer separately using the circuit given. Keeping other parameters unchanged , how will the position of the null point be affected if (i) ‘X’ increases the value of resistance R in the set-up by keeping the key K1 closed and the key K2 open ? (ii) ‘Y’ decreases the value of resistance S in the set-up, while the key K2 remains open and the key K1 closed? Justify your answer in each case.

Ans56. (a). See NCERT Book Article No. 3.16 Circuit diagram 1 Prove r = R ( − 1 ) 2

(b) (i) By increasing resistance R the current through AB decreases , so potential gradient decreases . Hence a greater length of wire would be needed for balancing the same potential difference. So the null point would shift towards B.

(ii) By decreasing resistance S, the current through AB remains the same, potential gradient does not change. As K2 is open hence there is no effect of S on null point. 1

Q57. (a) Find the potential drops across the two resistors shown in figure. (b)A voltmeter of resistance 600Ω is used to measure the potential drop across the 300Ω resistor. What will be the measured potential drop?

(c) As the temperature of a conductor increases, its resistivity & conductivity change. What will be the effect on the ratio of resistivity to conductivity? Ans57.

(a) The current in the circuit = Ω Ω

=0.2A The potential drop across the 300Ω resistor is 300 × 0.2=60V 1 The drop across the 200Ω resistor is 40V. 1/2

(b)The equivalent resistance, when the voltmeter is connected across 300Ω R=200Ω + Ω Ω

Ω Ω =400Ω 1

Thus, the main current from the battery is I =

Ω = 0.25A 1/2

The potential drop across the 200Ω resistor is 200Ω x 0.25A=50V and that across 300Ω is also 50V.This is also the potential drop across the voltmeter and hence the reading of the voltmeter is 50V. 1 (c) Increases. 1 Q58. (a) Derive the formula for the equivalent EMF & internal resistance for the parallel combination of two cells with EMFs E1 & E2 and internal resistances r1 & r 2 respectively. (b).Each of the resistors shown in figure has a resistance of 10Ω and each of the batteries has an emf of 10V. Find the currents through the resistors a and b in the two circuits.

Ans58. (a). See NCERT Book Article No. 3.12

Cells in parallel combination

Prove Eeq = 2

req = 1

(b). Fig. (a) Current across ‘a’ =

Ω = 1A 1/2

Current across ‘b’ = 0, ∵ First cell, resistance ‘a’ & (series combination of ‘b’ and second cell) are parallel. So potential difference is same. 1/2 Fig. (b) Current across ‘a’ =

Ω = 1A 1/2

Current across ‘b’ is zero. 1/2

Q59.(a). The circuit in figure shows two cells connected in opposition to each other. Cell E1 is of e.m.f 6V and internal resistance 2Ω; the cell E2 is of e.m.f 4V and internal resistance 8Ω. Find the potential difference between the points A and B.

E1 A E2 B

(b) State, with the help of a circuit diagram, the working principle of a meter-bridge. Write

the expression used for determining the unknown resistance. What happens if the galvanometer & cell are interchanged at the balance point of the bridge.

Ans59. (a) . I = 6−42+8 = A 1

Potential difference across A & B is the terminal voltage of cell E1

= E1 – I r = 6 - × 2 = 5.6 V. 1

(b)See NCERT Book Article No.3.15 Circuit diagram 1

Working principle 1

Expression ½

No change in the position of the balance point of the bridge. 1/2

Q60.Find the equivalent resistances of the networks shown in figure between the points a and b.

Ans60. Fig. (a) R1 = r + r =2r, = +

R2 = 1

R3 = R2 + r = 5r / 3

= +

Rnet = (5r / 8) Ω 1

Fig. (b) = + +

R1 = r / 3 1

Req = R1 + r = + r = (4r / 3) 1/2

Fig. (c) It is balanced condition of wheat stone bridge. Therefore resistance in diagonal are neglected. 1/2

Equivalent resistance between ‘a’ & ‘b’

= + , Req = r Ω

DUAL NATURE OF MATTER AND RADIATION

One mark Questions

Q1.Two metals A and B have work functions 3eV and 5eV, respectively. Which metal has a lower threshold wavelength? Ans 1: Metal B having higher work function will have lower threshold wavelength. Q2. For a given photosensitive material and with a source of constant frequency of incident radiation, how does the photocurrent vary with the intensity of incident light? Ans 2: For a given photosensitive material and with a source of constant frequency ν (ν ˃ νo) of Incident radiation, the photocurrent increases with increase in intensity of incident light. Q3. Can you find the value of Planck’s constant from Vo- ν graph for photoelectric effect? Ans:2. Yes, we can because slope of V0-ν graph = h/e. Hence h = e x slope of Vo-ν graph Q4. The stopping potential in an experiment on an photoelectric effect is 2V. What is the maximum kinetic energy of the photoelectrons emitted? Ans: 4 As stopping potential = 2V, hence maximum kinetic energy of photoelectrons emitted = 2 eV. Q5. Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons be emitted when the surface is illuminated by (a) red light and (b) blue light? Ans:5 No electron will be emitted when illuminated by red light. Electron emission takes place with blue light. Q6. Ultraviolet light is incident on two photosensitive materials having work functions W1 and W2 (W1˃W2). In which case will the kinetic energy of the emitted electrons be greater? Why? Ans:6. Kinetic energy of emitted photoelectron K = E –W, where E is energy of incident light photon. As W1> W2, hence K1< K2. Thus, K.E of electrons will be more in second case. Q7. Draw a graph showing variation of de Broglie wavelength with the momentum of an electron. Ans:7.

Q8. Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of the incident radiation. Ans:8.

Q9 .Work function of sodium is 2.3 eV. Does sodium show photoelectric emission for light of wavelength 6800A0 Ans9. E = hc/λ = 1.83 eV, which is less than ϕ0. Photoelectric emission cannot take place. Q10. The de-Broglie wavelength of a particle of kinetic energy K is λ. What would be the wavelength of the particle, if its kinetic energy were K/4? Ans10. λ= h/√2mK When kinetic energy is K/4

λ1 = h/ = 2λ.

Q11.An electron, an α particle and a proton have the same kinetic energy. Which of these particles has the largest de-Broglie wavelength? Ans11. λ α 1/√m Electron will have largest de Broglie wavelength. Q12. Define the term ‘threshold frequency’ in relation to photoelectric effect. Ans.12. For a given photosensitive surface the minimum value of frequency of the incident radiation for which no photoelectric emission takes place is known as threshold frequency. Q13. De-Broglie wavelength associated with the electron accelerated through a potential difference of V is λ. What will be the wavelength when the accelerating potential is increased to 4V? λ=12.27/√V A0 Q14. If the energy of a photon is 30eV and work function of the material is 10eV, then find the value of stopping potential. Ans.14 V0= (E – ϕ0)/e = 20eV

Q15. An electron and alpha particle have the same de Broglie wavelength associated with them. How are their kinetic energies related to each other? Ans:15. Kinetic energy E = p2/2m λ = h /√2mE Ee >Eα, Kinetic energy of electron is greater than α particle. Q16. The two lines A and B shown in the graph plot the de- Broglie wavelength (λ) as a function of 1/√V( V is the accelerating potential) for two particles having the same charge. Which of the two particles having the same charge. Which of the two represents the particle of heavier mass? Ans.16

de Broglie wavelength λ= = √

Slope of λ versus 1/√V graph =

For the particles of same charge q, slope α√

As the slope of line A is smaller than that of line B, so the line A represents the heavier particle. Two mark Questions

Q17. Two monochromatic radiations, blue and violet, of the same intensity, are incident on a photosensitive surface and cause photoelectric emission. Would (a) the number of electrons emitted per second and (b) the maximum kinetic energy of the electrons, be equal in the two cases? Justify your answer. Ans17. (a) Intensity I of both monochromatic radiations of blue and violet colour are exactly the same, but frequency of violet radiations is more than that of blue radiations. Hence the number of electrons emitted per second are equal in two cases because number of photoelectrons depend on the number of incident radiation photons i.e. the intensity of incident radiation. (b) The maximum kinetic energy of photo electrons emitted by violet radiation is more because Kmax = hν-ϕ0 where ϕ0 is the work function of given photosensitive surface. Q18.Two metals X and Y when illuminated with appropriate radiations emit photoelectrons. The work function of X is higher than that of Y. Which metal will have higher value of threshold frequency and why?

(Φ0)X> (Φ0)Y The work function is related to threshold frequency ϕ0= hν0, (ν0) X > (ν0)Y

Q19. A source of light is placed at a distance of 50 cm from a photocell and the cut off potential is found to be V0. If the distance between the photocell and the light source is made 25 cm. What will be the new cut-off potential? Ans19. The stopping potential depends only on the frequency of the incident light and not its intensity. When we reduce the distance of the source from the photocell from 50 cm to 25 cm, only the intensity of incident light changes but not the frequency. Hence, this change in the distance of the source makes no difference to the stopping potential. Q20. The threshold frequency of a certain metal is 3.3 x 1014 Hz. If light of frequency is 8.2 x 1014 Hz. Ans20. Is incident on the metal, predict the cut off voltage for photoelectric emission. Given Planck’s constant, h = 6.62 x 10-34Js. eV0 = h(ν –ν0) V0 = 202.74V Q21. Find the ratio of the de Broglie wavelengths associated with (i) protons, accelerated through a Potential of 128V, and (ii) α-particles, accelerated through a potential of 64V. ANS: λ= h/ 2qmV, λp: λα =2:1

Q22. Show graphically, the variation of the de-Broglie wavelength (λ) with the potential (V) through which an electron is accelerated from rest.

Q23. For a photosensitive surface, threshold wavelength is λ0. Does photoemission occur if the wavelength λ of the incident light is (i) more than λ0 (ii) less than λ0? Justify your answer. ANS: For photoemission, the energy of the photoelectrons must be greater than the work function. The work function ϕ and threshold frequency are related as hν0 = ϕ

hc/λ0 = ϕ For photoemission

hν> hν0 hc/ λ >hc/λ0

λ < λ0 Hence photoemission is possible only if the wavelength of incident light is less than threshold wavelength. Q24. An electron is accelerated through a potential difference of 100 volts. What is the de-Broglie wavelength associated with it? To which part of the electromagnetic spectrum does the value of wavelength correspond? ANS: λ= h/√2meV = 1.227 x 10-10m, X-rays

Three Mark Questions

Q25. Red light, however bright it is, cannot produce the emission of electrons from clean zinc surface. But even weak ultraviolet radiation can do so. Why? ANS: X-rays of wavelength ‘λ ‘fall on a photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broglie wavelength of electrons emitted will be hλ/2mc. The frequency of ultraviolet radiations is more while that of red light is less than the threshold frequency for a zinc surface, so ultraviolet radiations can cause emission of electrons and red light cannot. As the work function of the metal can be neglected, so K.E. of emitted electron = Energy of X ray photon ½ mv2 = hν

p/2m = hc/λ

p=

de Broglie wavelength of emitted electrons, ‘λ’ = h/p , ‘λ’ =

Q26.(a) For what kinetic energy of a neutron will the associated de-Broglie wavelength be 1.4 x 10-10m? (b) Also find the de-Broglie wavelength of a neutron, in thermal equilibrium with matter, having an average kinetic energy of 3/2kT at 300K. ANS: λ= h/√2mE

E = 6.634 X 10-21J Ek = kT λ=1.45x 10-10m

27.Plot a graph showing the variation of stopping potential with frequency of incident radiation for two different photosensitive materials having work functions W1 ˃ W2 .On what function does the slope and (b) intercept of lines depend?

Slope of V0-ν = ΔV/Δν = h/e, Slope of V0-ν graph depends on h and e. It has same value for both materials. Intercept of potential axis = - W/e = - hν0/e, which depends on the work function of the material. Q28. The work function of caesium metal is 2.14 eV. When light of frequency 6 x 1014Hz is incident on the metal surface, photoemission of electrons occurs. What is the (a) maximum kinetic energy of the emitted electrons (b) stopping potential and (c) maximum speed of the emitted photoelectrons? ANS: W0 = 2.14 eV Ν = 6 x 1014Hz

Kmax = hν- W0 = 0.34eV. eV0= Kmax= 0.34eV

Stopping potential V0 = 0.34V, Kmax = ½ mv2 (max) = 0.34 eV and vmax= 345.8 km/s

Q29. Radiation of frequency 1015Hz is incident on three photosensitive surfaces A, B and C. Following observations are recorded: Surface A: No photo-emission occurs. Surface B: Photo-emission occurs but the photo-electrons have zero kinetic energy. Surface C: Photo-emission occurs and photo-electrons have some K.E. Based on Einstein’s photoelectric equation, explain the three observations. ANS: From the observations made (parts A and B) on the basis of Einstein’s photoelectric equation, we draw following conclusions: For surface A, the threshold frequency is more than 1015 Hz, hence no photoemission is possible. For surface B, the threshold frequency is equal to the frequency of given radiation. Thus, the photo-emission takes place but kinetic energy of photo-electrons is zero. For surface C, the threshold frequency is less than 1015 Hz. So photo-emission occurs and photo-electrons have some kinetic energy. Q30. If the frequency of the incident radiation on the cathode of a photo cell is doubled, how will the following change: (a) kinetic energy of the electrons (b) photoelectric current (c) stopping potential. Justify your answer. ANS: The K.E. of the photoelectron becomes more than double of its original energy.

As the work function of the metal is fixed, so incident photon of higher energy will impart more energy to the photoelectrons. The increase in frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one. Electron from the metal surface with the increase in frequency, the K.E of the photoelectron increases, hence stopping potential also increases.

Q31. In a plot of photoelectric current versus anode potential how does

(i) the saturation current vary with anode potential for incident radiations of different frequencies but same intensity?

(ii) the stopping potential vary for incident radiations of different intensities but same frequency?

(iii) Photoelectric current vary for different intensities but same frequency of incident radiations? Justify your answer in each case. ANS: (i) The saturation current for incident radiations of same intensity but different frequencies. does not depend on the anode potential because saturation current is a sole function of intensity of incident radiation.

(ii) The stopping potential V0 for incident radiations of different intensities but same frequency remains unchanged. The stopping potential, in accordance with Einstein’s equation h(ν-ν0) depends solely on the frequency of incident radiation and is independent of intensity of radiation.

(iii)The photoelectric current increases with increase in intensity of incident radiation of a given frequency provided that the frequency of radiation is equal to or more than the threshold frequency for the given metal. Greater intensity of incident radiation means greater number of incident photons, which in turn leads to more photoelectrons and hence higher value of photoelectric current. Q32. An electromagnetic wave of wavelength λ is incident on a photosensitive surface of negligible work function. If the photoelectrons emitted from this surface have the de Broglie wavelength λ1, Prove that, λ = [ ] λ1

2 ANS: Photoelectric equation, ½ mv2 = hν0 – W0 = hν - 0 As work function is negligible mv2 = 2hν = 2hc/λ

mv =

de Broglie wavelength of photoelectrons,

λ1= = , we get λ= λ12

Q33. An electron and a photon each have a wavelength of 1.5nm. Find (a) their momenta (b) the energy of the photon and (c) kinetic energy of the electron. ANS: p =h/λ = 4.42 x 10-25kgm/s, En=hc/λ= 1.326 x 10-16J and Ek = p2/2m= 1.0723 x 10-19J

Q34. Define the term ‘work function’ of a metal. The threshold frequency of a metal is f0. When the light of frequency 2f0 is incident on the metal plate, the maximum velocity of electrons is v1. when the frequency of the incident radiation is increased to 5f0, the maximum velocity of electrons emitted is v2. Find the ratio of v1 to v2. ANS: Work function is the minimum amount of energy required to pull an electron from a metal surface. According to Einstein’s photoelectric equation,

h(ν-ν0) = mv2max, Here, ν0 = f0, and when ν = 2f0, vmax = v1 h (2f0-f0) = 1/2mv12, ν = 5f0, vmax = v2,

h(5f0-f0) = 1/2mv22 ,

v1:v2 = 1:2.

Q35. Light of wavelength 2000 A0 falls on an aluminium surface. In aluminium 4.2 eV are required to remove an electron. What is the kinetic energy of (a) the fastest (b) the slowest emitted photoelectrons, (c) what is the stopping potential? (d) What is the cut off wavelength for aluminium? Planck’s constant h = 6.6 X 10-34Js and speed of light, c=3 x108m/s. ANS: Energy of incident photon E = hc/λ= 6.2eV

K.E. of the fastest electron, Ek = E – ϕ0 = 6.2 -4.2 = 2eV K.E of the slowest electron = zero.

As the emitted electrons have all possible energies from 0 to certain maximum value Ek If V0 is the stopping potential, then V0 = Ek/e = 2 eV/e = 2V. If λ0 is the cut off wavelength for aluminium. Then, ϕ0 = and λ0= 2946.4A0

ATOMS AND NUCLEI

01 MARK QUESTIONS

1. What is the nuclear radius of Fe125, if that of Al27 is 3.6 Fermi? ANS: = 1/3 =5/3, Rfe= 5/3, RAl = x 3.6 = 6.0 fermi

2. Four nuclei of an element fuse together to form a heavier nucleus. If the process is

accompanied by release of energy. Which of the two – the parent or the daughter nuclei would have a higher binding energy per nucleon?

ANS: The daughter nucleus would have a higher binding energy per nucleon because the release of energy is accompanied by an increase of binding energy per nucleon.

3. Two nuclei have mass numbers in the ratio 27:125. What is the ratio of their nuclear radii?

ANS: R α A1/3, R1:R2 = 271/3 : 1251/3 = 3:5

4. The radius of innermost electron orbit of a hydrogen atom is 5.3 x 10-11m.What is the radius of orbit in second excited state?

ANS: r α n2, r2 = 22/12x 5.3 x 10-11 = 2.12 x10-10m.

5. Tritium has a half-life 12.5 year undergoing beta decay. What fraction of a sample of pure Tritium will remain un-decayed after 25 years?

ANS: = (1/2) n, since n = 25/ 12.5, therefore N/N0 = ¼

6. What is the ratio of radii of the orbits corresponding to second excited state and ground

state in a hydrogen atom? ANS: = ( )2 = 32/12 = 9:1

7. The ground state energy of hydrogen atom is -13.6eV. What are the kinetic and potential

energies of electron in this state? ANS: Total energy E = -13.6 eV

Kinetic energy T = -E = + 13.6 eV Potential energy U = -2T = -2 x 13.6 = -27.2 eV.

8. A nucleus undergoes one β – decay. How does its (i) mass number (ii) atomic number change?

ANS: (i) Mass number remains same. (ii) Atomic number increases by one.

9. Which is more likely to be stable 3X7 or 3Y4? ANS: The number of protons is same in X and Y. But the number of neutrons is greater

in X. Therefore X is more likely to be stable.

10. The mean life of a radioactive sample is Tm. What is the time in which 50% of this sample would get destroyed?

ANS: Time needed = half life = 0.693 Tm 11. Plutonium decays with a half life of 24000 years. If plutonium is stored for 72000 years,

what fraction of it remains? ANS:

=

12. Two nuclei have mass numbers in the ratio 2:5. What is the ratio of their nuclear

densities? ANS: Nuclear density is independent of mass number. So, the ratio will be 1:1.

02 Marks Questions

13. Write any two characteristic properties of nuclear force. ANS: (i) Nuclear force is a short range force.

(ii) Nuclear forces show saturation effect.

14. Name the spectral series of hydrogen spectrum which (i) has least wavelength (ii) lies in the visible region of electromagnetic spectrum.

ANS: (i) Lyman series (ii) Balmer series

15. What is the significance of the negative energy of electron in the orbit?

ANS: It signifies electron is bound to the nucleus. Due to the electrostatic attraction between electron and nucleus, the P.E is negative and is greater than K.E of electron. Total energy of electron is negative. It cannot escape from the atom.

16. If nuclei, with low binding energy per nucleon, transform to nuclei with greater binding energy per nucleon, would the reaction be exothermic or endothermic? Justify your answer and write two examples in support of your answer.

ANS: Greater the binding energ, less is the total mass of a bounded system, such as nucleus. Consequently, if a nucleus with less binding energy per nucleon, transforms to

nuclei with greater binding energy per nucleon, there will be a net energy release. Reaction will be exothermic. For example, in case of fission, when a heavy

nucleus decays into two or more intermediate mass fragments or if fusion, when light nuclei fuse into a heavier single nucleus, energy is released.

17. A radioactive nucleus ‘A’ undergoes a series of decays according to the following scheme: α β α γ A→A1→A2→A3→A4 The mass number and atomic number of A are 180 and 72 respectively. What are these numbers for A4?

ANS: Mass number of A4 is 172 and its atomic number is 69.

18. The half life of radon is 3.8 days. After how many days 19/20 of the sample will decay? ANS: N = N0-19/20N0 N = 1/20 N0 = n, t = nT, t= 16.43 days

19. In a given sample, two radioisotopes, A and B, are initially present in the ratio 1:4. The half lives of A and B are respectively 100 years and 50 years. Find the time after which amounts of A and B become equal.

ANS: For sample A N = N0e-λ1

t

For sample B N = 4N0e-λ2

t Here, N0e-λ

1t = 4N0e-λ

2t

4 = e (λ2-λ1)t loge 4 = λ2t –λ1t 2 = t ( − ) t = 200 years

20. The sequence of the stepwise decay of radioactive nucleus is α β α α D-- D1--- D2-- D3-- D4 If the nucleon number and atomic number for D2 are respectively 176 and 71. What are the corresponding values for D and D4 nuclei? Justify your answer.

ANS: In each α- decay the nucleon number decreases by 4 and atomic number by 2. In β Decay the nucleon number remains unchanged and atomic number increases by 1. D --180, 72 and D4—168, 67.

21. A heavy nucleus X of mass number 240 and binding energy per nucleon 7.6 MeV is split

into two fragments Y and Z of mass numbers 110 and 130. The binding energy of nucleons in Y and Z is 8.5MeV per nucleon. Calculate the energy Q released per fission in MeV.

ANS: For nucleus X : A = 240 B.E per nucleon = 7.6 MeV Total B.E of X = 240 x 7.6 = 1824 MeV For nucleus Y : A1 = 110 B.E per nucleon = 8.5 MeV Total B.E of Y = 110 x 8.5 = 935 MeV For nucleus Z : A2 = 130 B.E per nucleon = 8.5 MeV Total B.E of Z = 130 x 8,5 MeV = 1105 MeV Energy released per fission Q = B.E of Y + B.E of Z – B.E of X = 935 + 1105 – 1824 = 216 MeV

22. The energy of an electron in the nth orbit is given by En = - 13.6 /n2 eV Calculate the energy required to excite an electron from ground state to second excited state?

ANS: Energy in the ground state (n =1), E1 = - 13.6 / 12 = -13.6 eV Energy in the second excited state (n = 3), E3 = -13.6 / 32 = -1.51 eV Required energy = E3 – E1 = -1.51-(- 13.6) = 12.09 eV.

23. 4 g of radioactive material of half life 10 years is kept in store for 15 years. How much material is disintegrated?

ANS: λ= 0.693 / 10, λt = 2.303 log , 15 x 0.693 / 10 = 2.303 log m = 2.59 g.

24. A radioactive substance decays to 1/32th of its initial activity in 25 days. Calculate its half life.

ANS: R = R0/16, R = R0 n , = R0 n , 5 = n , Or n =5 Half life period =

. = 25/5 = 5 days

25. The value of ground state energy of hydrogen atom is -13.6 eV.

(a) What does the negative sign signify? (b) How much energy is required to take an electron in this atom from the ground state to the

first excited state? ANS:

(a) The negative sign signifies that the electron is bounded to the nucleus. (b) The energy required to take an electron from ground state to the excited state

= Ef - Ei = -3.4 – (- 13.6) = + 10.2 eV.

26. Calculate the wave number and wavelength of first line of Balmer series of hydrogen. Given R = 1.097 x 107 m-1.

ANS: = 1.097 x 107 [1/22-1/32] = x 1.097 x 107, λ = 6.56 x 10-7m Hence wave number = 1/λ = 1.52 x106 m-1.

27. In a Geiger - Marsden experiment, calculate the distance of closest approach to the nucleus of Z= 80 when an – particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of α particle is doubled?

ANS: Here, Z = 80, K = 8 MeV = 8 x 1.6 x 10-13J

r0 =

= 28.2 x 10-15 m = 28.2 fm (Fermi)

The distance of closest approach is halved when the kinetic energy of the α particle is doubled.

28. If the nucleons bound in a nucleus are separated apart from each other, the sum of their masses is greater than the mass of the nucleus. Where does this mass difference come from? Explain briefly.

ANS: If the nucleons bound in a nucleus are separated apart from each other, we need to transfer a total energy equal to the binding energy of the nucleus to these particles. This energy is equivalent to the mass defect. Consequently the sum of their masses becomes greater than mass of nucleus according to Einstein’s mass- energy equivalence, E = Δm c2

29. After a certain time lapse, the fraction of radioactivity polonium left un-decayed is 12.5% of the

initial quantity. What is the duration of this time lapse if the half life of polonium is 139 days? ANS: Final number of nuclei = 12.5% of N0 = 12.5/100 N0 = N0/8

Let n be the number of half lives elapsed N = N0 (1/2) n = N0/8 (1/2)n = 1/8 8 = 2n, n=3 Time elapsed = 3 half lives = 3x 139 = 417 days

30. A radioactive isotope has a half life of 5 years. How long will it take the activity to reduce to 3.125 %? ANS: = n = t/T . = = 5, = 5, t = 5T = 5x 5 = 25 years. 03 Marks Questions

31. A radioactive sample contains 2.2 mg of pure 11C6 which has half life period of 1224 second. Calculate (i) the number of atoms present initially.(ii) the activity when 5 µg of the sample will be left.

ANS: (i) 11g contains 6.023 x 1023 atoms Number atoms in 2.2 mg N0 = (2.2 x 10-3 x 6.023 x10-23)/11 = 1.2046 x 1020 (ii) Number of atoms actually present in 5x 10-6 g of sample N = (5 x10-6 x 6.023 x1023)/11 λ= 0.6931/1224 Activity R = λN = 1.55 x1014 Bq

32. The energy levels of atom are shown below. Which of the shown transitions will result in the emission of a photon of wavelength 275 nm? Which of these transitions correspond to emission of radiation of (a) maximum and (b) minimum wavelength?

ANS: Energy of photon E = hc/λ= 6.63 x 10-34x3x108/275 x 10-9 = 4.5 eV So, transition B corresponds to the emission of a photon of wavelength 275 nm. Since E α 1/λ, hence

(a) Minimum energy corresponds to maximum wavelength i.e, transition A (b) Maximum energy corresponds to minimum wavelength i.e, transition D.

33. Which of the following radiations α-rays, β-rays and γ-rays (i) are similar to X- rays (ii) are easily absorbed by matter. (iii) travel with greatest speed. (iv) are similar in nature to cathode rays

(v) possess maximum ionising power (vi) possess least penetrating power.

ANS: (i) γ- rays (ii) α- rays (iii) γ- rays (iv) β rays (v) α- rays (vi) α- rays.

34. Define the term decay constant of radioactive nucleus. Two nuclei P & Q have equal number of atoms at t = 0.Their half-lives are 3 hours and 9 hours respectively. Compare their rates of disintegration after 18 hours from the start. ANS: According to law of radioactive decay, the instantaneous rate of disintegration of a radioactive element is directly proportional to the actual number of nuclides of that Element present at that instant. i.e, - = λN, where λ is a decay constant. ANS: Decay constant of a radioactive element is defined as the ratio of its instantaneous rate of disintegration to the number of nucleides present at that time. Its unit is s-1 (RP)0/ (RQ)0 = 9h / 3h =3

Rt = R0 ( ) t / T = x /

/ =

35. Define the activity of a radionuclide. Write the SI unit. Give a plot of the activity of a radioactive species versus time. ANS: Activity of a radionuclide is defined as the rate of disintegration of the radionuclide. It is also called count rate and is represented by A. The activity of the sample at any instant is directly proportional to the number of atoms left un-decayed in the sample at that time. Activity varies exponentially with time and is represented as

A = A0 e-λt, SI unit of activity is Becquerel.

36. How is the size of the nucleus determined experimentally? Write the relation between the radius and mass number of the nucleus. Show that the density of nucleus is independent of its mass number.

ANS: The size of nucleus is determined by the Rutherford experiments on alpha ray scattering. The distance of closest approach is approximately the size of the nucleus. Here, it is assumed that only coulombic repulsive force caused scattering. With the α- rays of 5.5 MeV the size of the nucleus was found to be less than 4 x 10-14 m. Scattering experiments performed firstly with the electrons instead of α- particles that were bombarded with the targets of different elements, the size of nuclei of various elements have been determined accurately.

The required relation is R = R0 A1/3 Where R0 = 1.2 x 10-15m Density = mass of nucleus/ Volume of nucleus ρ= mA/(4/3πR3) = 3m /4πR0

3 Which is independent of the mass number A or size of nucleus.

37. Calculate the energy of the reaction- 3Li7 + 1H1 - 4Be7 + 0n1 Given m(n) = 1.00867 amu, m(H)= 1.00783 amu, m(Li) = 7.01601 amu, m(Be)= 7.01693 amu.

ANS: Energy of reaction = [m(Be) + m(n) – m(Li) + m(H)] x 931 MeV = 1.65 MeV.

38. State the laws of radioactive decay. Plot a graph showing the number (N) of un-decayed nuclei as a function of time (t) for a given radioactive sample having half life T1/2. Depict in the plot the number of un-decayed nuclei at (i) t = 3T1/2 and (ii) t = 5T1/2

ANS: Number of nuclei undergoing decay per unit time is proportional to the number of nuclei present in the sample at that instant.

39. The decay constant for a radionuclide has a value of 1.386 day-1.After how much time will a given sample of this radionuclide get reduced to only 6.25% of its present number?

ANS: N = N0 e-λt . = N0 e-λt, eλt=16 Taking natural logarithms of both sides we get λt logee = loge16 λt = 2.303 log16 t = 2.303x 4x 0.3010 / 1.386 = 2 days

40. The electron in a given Bohr orbit has a total energy of -1.5eV.Calculate (i) its kinetic Energy (ii) potential energy and (iii) wavelength of light emitted, when the electron makes a transition to the ground state. Ground state energy is -13.6 eV. ANS: K.E = - U/2 = - E (i) K.E = -E = - (-1.5 eV) = 1.5 eV (ii) U = 3 eV

(iii) = Ei - Ef

휆 = 1.03 x 10-7 m 41. The half-life of 14C6 is 5700 years. What does it mean? Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-life is 1hr and 2hr respectively. Calculate the ratio of their rates of disintegration after 2 hours.

ANS: = = = 2

After 2 hours (2 half lives of X) Nx = ( ) = ) But for Y it is only one half life, hence Ny = (N0)y/2 = N0/2 R = λN Rx/Ry = 2 /

/ = 1.

42. Draw a plot of potential energy of a pair of nucleons as a function of their separation. Write two important conclusions which you can draw regarding the nature of nuclear forces.

ANS:

Two important conclusions:

(i) Nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This explains constancy of the binding energy per nucleon for large size nucleus.

(ii) Graph explains that force is attractive for distances larger than 0.8 fm and repulsive for distances less than 0.8 fm. 43. The energy levels of an element are given below: Identify using necessary calculations, the transition, which corresponds to the emission of a spectral line of wavelength 482nm.

ANS: hc/λ = E1-E2 For spectral line C, we have

E1-E2 = -0.85-(-3.4) = 2.55 eV λ=482nm. This confirms transition.

44. The ground state of hydrogen atom is -13.6eV. If an electron makes a transition from

an energy level -0.85eV to -3.4eV. Calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum, does this wavelength depend?

ANS: The ground state of hydrogen E1= -13.6 eV Energy Ei = - 0.85 eV corresponds to a level ni, where -0.85 = -13.6 / ni2, ni = 4 Energy Ef = -3.4 eV corresponds to level of nf such that -3.4 = -13.6 / nf

2, which leads to result nf = 2 As transition is taking place from ni=4 level to nf =2 level, the wavelength belongs to Balmer series of hydrogen spectrum.

45. Draw a diagram to show the variation of binding energy per nucleon with mass number of different atomic nuclei. Calculate binding energy/nucleon of 40Ca 20 nucleus. Given mass of Ca = 39.962589 u; mass of proton = 1.007825 u; mass of Neutron = 1.008665 u; and 1u = 931 MeV/c2. ANS:

Number of protons = 20 and Number of neutrons = 20

Expected mass of the nucleus m = 20 (mp+ me) = 20 (1.007825 + 1.008665) u = 40.3298 u Actual mass is m’ = 39.962589 u Mass defect Δm = m – m’ = 0.367211 u

Total binding energy = 0.367211 x 931 MeV = 341.873441 MeV B.E. per nucleon = 341.873441/40 = 8.547 MeV/nucleon.

46. Using the relevant Bohr’s postulates derive the expression for the radius of the nth orbit of electron in hydrogen atom. ANS: When an electron revolve in a stable orbit, the centripetal force is provided by Electrostatic force of attraction acting on it due to a proton present in the nucleus. Hence mvn

2/rn = 1/4πϵ0 e2/rn2 (i)

From Bohr’s quantum condition, we have mvnrn = nh/2π (ii)

Squaring equation (ii) and equating with (i) we get rn = ϵ0h2n2/πme2

47. The wavelength of the second line of the Balmer series in the hydrogen spectrum is 4861 A0. Calculate the wavelength of the first line of the series. ANS: 1/λ1 = R[ 1/22- 1/32] = 5/36 R

1/λ2 = R [1/22 – 1/42] = 3/16 R, λ1 =27/20 λ2 = 27/20 x 4861 = 6562.35 A0

48. The energy of an atom of an element X is as shown. A photon of wavelength 620 nm is emitted. This corresponds to which of the transitions: A, B, C, D or E?

ANS: As the energy of the photon is E = hc/λ On substituting the given values and solving Energy of photon = 2 eV

From the diagram this energy corresponds to the transition D as the energy emitted is 2 eV = [(-1eV)-(-3eV)].

49. (a) If both the number of protons and number of neutrons are conserved in each nuclear reaction, in what way is mass converted into energy or vice- versa in a nuclear reaction?

(b) The half life of a radioactive element A is same as mean life time of another radioactive element B. Initially, both have same number of atoms. B decays faster than A. Why?

ANS: (a) As neutron number and proton number are conserved in a nuclear reaction, the total rest mass of neutrons and protons is the same on either side of nuclear reaction. But the total B.E of nuclei on the left side need not be the same as that on the right side. The difference in the binding energies on two sides appears as energy released or absorbed in the nuclear reaction. As BE contributes to mass, we say that the difference in total mass of nuclei on the two sides gets converted into energy or vice versa. It is in this sense that a nuclear reaction is an example of mass energy inter-conversion.

(b)TA = τB = 1.44 TB Therefore, TA>TB λA < λB. Therefore, B decays faster than A.

50. What are limitations of Bohr’s theory of hydrogen atom?

ANS: (i) This theory is applicable only to the simplest hydrogen or hydrogen like atoms i.e., atoms having only one electron orbiting the nucleus. The theory fails to explain the radiation spectra of heavier elements.

(ii) The theory has no method for explaining why the orbits must be circular and not elliptical. (iii) This model failed to explain the fine structure of the spectral lines in the radiation spectrum of hydrogen atom.

(iv) This theory uses two conflicting concepts, i.e., the classical electrodynamics and quantum mechanics. Since both these are conflicting in nature, there is an inherent conflict in Bohr’s theory. (v) This model does not comment on the relative intensities of the various spectral lines.

SEMICONDUCTOR DEVICES One mark Questions Q1.At what temperature would an intrinsic semiconductor behave like a perfect insulator? Ans: At absolute zero temperature (0 K) Q2. Name two factors on which electrical conductivity of a pure semiconductor depends. Ans: The energy gap (band ga) between valence band and the conduction band. the temperature. Q3. Explain the difference in the temperature variation of resistivity of metals and intrinsic Semiconductors. Ans: Resistivity of metal increases and of semiconductor decreases when temperature increases.

Q4. Draw the I-V characteristics of a solar cell.

Q5. Calculate the current gain β of a transistor, if the current gain α= 0.9. ANS: β=

= 0.9/ 1- 0.9 = 9.

Q6. How does the width of depletion layer of a p-n junction change with decrease in reverse bias?

ANS: The width of depletion layer of a p-n junction decreases with the decrease in reverse bias. Q7. How does the d. c. current gain of a transistor change, if the width of the base region is Increased? ANS: Value of current gain decreases on increasing the width of base region.

Q8.Why mobility of holes is less than that of free electrons?

ANS: It is due to the smaller drift velocity acquired by the holes. Less drift velocity makes Mobility µ= v / E less for same electric field E. Q9.The a.c. current gain of a transistor is 100. What is the change in the collector current in the transistor whose base current changes by 100 µA? ANS: ΔIC = β x ΔIB = 100 x 100 µA = 10 mA. Q10. How does the forbidden gap of an intrinsic semiconductor vary with increase in temperature? ANS: The forbidden gap does not change with temperature. Q11. Frequency of input voltage to a half wave rectifier is 50 Hz. What will be the frequency of the output voltage? ANS: The frequency of output waveform will be 50 Hz.

Q12. Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. How Does it affect (a) width of the depletion layer. (b) junction field ? ANS: Width of depletion layer is small. Junction field is high. Q13.How does the collector current change in a junction transistor, if the base region has larger width? ANS: The collector current becomes smaller because of the increase in the rate of recombination of electrons and holes as they move across the emitter junction. Q14. Why is a common emitter amplifier preferred over a common base amplifier? ANS: Because current gain of a common emitter amplifier is more than that of a common base amplifier. Q15. How does the energy gap in an intrinsic semiconductor vary, when doped with a pentavalent impurity? ANS: The energy gap decreases when a semiconductor is doped with a pentavalent impurity due to the creation of a donor energy level just below the bottom of the conduction band.

Q16. Why is the conductivity of n-type semiconductor greater than that of p-type semiconductor even when both of these have same level of doping? ANS: This is because under a given electric field, free electrons have higher mobility than holes. Q17. In the given diagram is the diode D forward or reverse biased?

ANS: Reverse biased because n-end is at higher potential. Q18. A semiconductor has equal electron and hole concentration 6x 108 m-3. On doping with a certain impurity, electron concentration increases to 8 x 1012 m-3.Identify the type of semiconductor after doping. ANS: As the net concentration of electron has increased, it is n-type semiconductor. Q19. Can we take one slab of p-type semiconductor and physically join it to another n-type semiconductor to get p-n junction? ANS: The contact surface of a slab has a roughness much greater than inter-atomic crystal spacing and is therefore, not suitable for continuous flow of charge carriers. Q20. Pieces of copper and germanium are cooled from room temperature to 80K.what will be the effect on their resistances? ANS: Resistance of Ge will increase and that of Cu will decrease.

Q21. A specimen of silicon is to be made P-type semiconductor. For this one atom of indium, on an average is doped in 5 x 107 silicon atoms. If the number density of silicon is 5 x 1028 atoms/m3. ANS: Then find the number of acceptor atoms per cm3. Number density of silicon = 5 x 1028 atoms/ m3 = 5 x 1022 atoms/ cm3. No. of acceptor atom/cm3 = 5 x 1022 / (5 x 107) = 1015 /cm3. Q22. In a transistor, doping level in base is increased slightly. How will affect (a) collector current

and (b) base current? ANS: Collector current decreases, as more of the majority carriers going from emitter to Collector will get neutralised in base by electron hole combination and base current increases. Q23. Write the truth table for the combination of gates shown in diagram

ANS: A B Y’ =A.B Y 0 0 0 1 0 1 0 1 1 0 0 1 1 1 1 0 Q24. An unknown input (A) and the input (B) shown here are used as the two inputs in a NAND gate. The output Y, has the form shown in the figure. Identify the intervals over which the input ‘A’ must be ‘low’.

In intervals 0 to t1, t3 to t4, the input ‘A’ may be low.

Q25. In the working of a transistor, the emitter base [EB] junction is forward biased while collector- base [CB] junction is reverse biased. Why? ANS: Only forward biased EB junction can send the majority charge carriers from emitter to Base and only reverse biased collector can collect these majority charge carriers from the base region. If the emitter is reverse biased, no charge carriers will flow towards the collector and hence no current will flow through the transistor.

Two mark Questions Q26. Why is the base region of a transistor made very thin and lightly doped? ANS: When a transistor is biased, the charge carriers move from emitter to collector through base and if the base region is highly doped then charge carriers combine and hence they will not cross the base. This results as no flow of current through the transistor. Thus base is lightly doped.

Q27. How is forward biasing different from reverse biasing in a p-n junction? ANS: In forward biasing p-side is connected to the +ve terminal and n-side to –ve terminal of battery. In reverse biasing p-side is connected to the -ve terminal and n-side to +ve terminal of battery forward current is due to majority carriers. Reverse current is due to minority carriers. In forward biasing depletion layer becomes thin. In reverse biasing depletion layer becomes thicker. In forward biasing resistance across p=n junction decreases. In reverse biasing resistance across the p-n junction increases.

Q28.Why are Si and GaAs are preferred materials for solar cells? ANS: The energy for the maximum intensity of the solar radiation is nearly 1.5 eV. In order to have Photo-excitation the energy of radiation (hν) must be greater than energy band gap (Eg).Therefore the semiconductor with energy band gap about 1.5 eV or lower and with higher absorption coefficient, is likely to give better solar conversion efficiency. The energy band gap for Si is about 1.1eV, while for GaAs, it is about 1.53 eV. The GaAs is better inspite of its higher band gap than Si because it absorbs relatively more energy from the incident solar radiations being of relatively higher absorption coefficient.

Q29.In the following diagrams, indicate which of the diodes are forward biased and which are reverse biased.

ANS: Forward biased, because p side is at higher potential (+7V) than n-side (+5V)

Reverse biased, because p-side is at lower potential (0 V) than n-side (+2 V) Reverse biased, because p-side is at lower potential (-10 V) than n-side (0 V) Forward biased, because p-side is at higher potential (-5 V) than n-side (-12 V ) Q30. The current in the forward bias is known to be more (~mA) than the current in the reverse bias (~µA). What is the reason than to operate the photodiodes in reverse bias? ANS: In case of n-type semiconductor, let n be the majority carriers (electrons) density and p be the minority carrier (holes) density. When n ˃ p. On illumination of semiconductor, there will be production of equal number of electrons and holes. Let Δ n, Δ p be the increase in majority carrier density and minority carrier density due to illumination of semiconductor, where Δ n = Δ p. Hence, fractional change in majority carrier (= Δ n/n), fractional change in minority carrier ( = Δ p/ p).Since n >> p, so <

it means , due to photo-effects the fractional change due to minority carriers dominates. As a result of it, the fractional change in the reverse bias current is more easily measurable than the fractional change in the forward bias current. It is due to this reason, photodiodes are preferably used in the reverse bias condition for measuring light intensity. Q31. In a n-p-n transistor circuit, the collector current is 10 mA. If 90% of the electrons emitted reach the collector, find the base current and collector current. Ans: IC= 90%IE IE= 100IC/90 = = 11.11mA IB = 10% IE = 1.11 mA Q32. The output of an OR gate is connected to both the inputs of a NAND gate. Draw the logic circuit of this combination of gates and write its truth table. ANS: NOR gate is formed.

Q33. Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams.

ANS:

Q34. In Zener regulated power supply a Zener diode with VZ = 6.0 V is used for regulation. The load Current is to be 4.0 mA and the unregulated input is 10.0V. What should be the value of series resistance RS? Presume that Zener current is five times the load current. ANS: As load current IL= 4 mA and Zener current IZ = 5IL= 5x 4 = 20 mA. Hence, total current flowing through series resistance RS will be IL + IZ = 4 + 20 = 24 mA = 24 x 10-3 A. Moreover out of supply voltage of 10V, the voltage across load is only 6 V, hence fall in potential across the series resistance RS will be VZ = 10 - 6 = 4V RS = VS / IS = 4 V/ 24 mA = 167 Ω Three mark Questions Q35. Draw the transfer characteristics curve of a base biased transistor in CE configuration. Explain Clearly how the active region of the VO versus Vi curve in a transistor is used as an amplifier. ANS:

The transfer characteristics curve of a base biased transistor base biased transistor in CE configuration has been shown in figure. Q36. Name the type of biasing which results in very high resistance of a p-n junction diode. In the given circuit, a voltmeter ‘V’ is connected across a bulb ‘B’. What changes would occur in bulb ‘B’ and voltmeter ‘V’, if the resistor ‘R’ is increased in value? Give reason.

ANS: The reverse biasing results in very high resistance across the p-n junction. If the value of the resistance R is increased, the current in the forward biased input circuit decreases. The emitter current IE decreases and hence the collector current (IC = IE- IB) also decreases. The intensity of the bulb decreases. Due to decrease in IC, the potential drop across the bulb B decreases and hence voltmeter shows a lower voltage. Q37. In only one of the circuits given below, the lamp L lights. Which circuit is it? Give reason.

ANS: The lamp lights in the figure (b) because in this circuit, the input circuit is forward biased and the output circuit is reverse biased. Q38. Give the symbol and the truth table of each of the two logic gates obtained by using the two circuits shown in the figure.

ANS: The circuit (a) represents OR gate.

A B Y 0 0 0 0 1 1 1 0 1 1 1 1

The circuit (b) represents AND gate.

A B Y 0 0 0 0 1 0 1 0 0 1 1 1

Q39. Figure shows input waveform (A, B) and the output waveform (Y) of a gate. Identify the gate and write its truth table.

ANS: The output is low when both the inputs are high otherwise it is high. Hence it is a NAND gate.

A B Y 0 0 1 0 1 1 1 0 1 1 1 0

Q40. Identify the logic gates marked P and Q in the given logic circuit. Write down the output at X for the inputs (i) A=0, B=0 and (ii) A=1, B=1

ANS: The logic gate P neither is NOR gate and Q is AND gate.

Input Input Output of NOR gate Output of AND gate

A B Y X = B.Y

0 0 1 0

1 1 0 0 Q41. (i) Sketch the output waveform from an AND gate for the inputs A and B shown in the figure.

(ii)If the output of the above AND gate is fed to a NOT gate, name the gate of the combination so formed. ANS: The output of an AND gate is 1 when both the in outs are 1. Hence the output waveform from the AND gate for the inputs A and B is of type as shown below.

A NAND gate is obtained when the output of an AND gate is fed to a NOT gate. Q42. Draw the output waveform at X, using the given inputs A and B for the logic circuit shown below. Also, identify the logic operation performed by this circuit.

ANS: X = A + B Thus the given circuit performs OR operation. The output waveform obtained at X is of type shown below.

Q43.The input resistance of a silicon transistor is 665Ω .Its base current is changed by 15µA which Results in the change in collector current by 2mA. This transistor is used as a common emitter ANS: Amplifier with a load resistance of 5kΩ. Calculate (a) current gain βa.c (b) transconductance ‘gm’. (c)Voltage gain ‘A’ of the amplifier.

βa.c = = 2 mA / 15 µA = 133.

Rin =

Δ V = Rin x ΔI = 665 x 15 x 10-6 V g =

= 0.2 Ω-1 Av = βa.c x Ro / Ri = 1000.

Q44. Draw the circuit diagram of a common emitter amplifier, with appropriate biasing. What is the phase difference between the input and output signals? State two reasons why a common emitter amplifier is preferred to a common base amplifier.

ANS:

The phase difference between the input and output signals is π radian. Two main advantages of a common emitter amplifier are:

The voltage gain as well as power gain is comparatively higher. The output is constant for almost whole range of audio frequency signals.

Q45. For a common emitter amplifier, the audio signal voltage across the collector resistance of 2kΩ is 2V.The current amplification factor of the transistor is 100, find the input signal voltage and base Current, if the base resistance is 1kΩ. ANS: Voltage gain Av = Vo/ Vi = βro/ri = 100 x 2000 / 1000 = 200.

Vi = Vo / 200 = 2 / 200 = 0.01 V Base current I = 0.01 / 1000 = 10 µA

Power gain = Av x β = 200 x 100 = 2 x 104 Q46.The base current of a transistor is 105µA and collector current is 2.05 mA. Determine the value of β, Ie and α. ANS: A change of 27 µA in the base current produces a change of 0.65 mA in collector current. Find βa.c.

β= IC/IB = 2.05 mA / 105 µA = 19.5 IE= IB + IC= 105 µA + 2.05 mA = 2.155 mA.

α=IC/IB = 0.95 β a.c = ΔIC / ΔIB = 24.07

Q47. How is Zener diode fabricated so as to make it a special purpose diode? Draw I-V characteristics of Zener diode and explain the significance of breakdown voltage. ANS: A Zener diode is a special purpose diode which is operated under reverse bias in the break

Down region and used as voltage regulator. Zener diode is fabricated by heavily doping both p- and n-sides of junction so that depletion region formed is very thin (< 10-6m) and electric field of junction is extremely high (> 5 x 106V/m) even for a small reverse voltage of about

5V. As a result of this the breakdown voltage in reverse bias is small (about 5V) for a Zener diode. I-V characteristics of Zener diode is as shown.

When reverse bias voltage reaches breakdown voltage, a large current begins to flow. After breakdown (Zener) voltage, a large change in the current can be produced by almost

Insignificant change in the reverse bias voltage. Thus, the Zener voltage remains constant, Even though current through Zener diode varies over a wide range. This property of Zener

Diode plays a significant role in its use as a voltage rectifier.

Q48.Explain with the help of a circuit diagram how a zener diode works as a d.c. voltage regulator. A Zener diode is used to get a constant d.c. voltage from the unregulated filtered output of a full wave rectifier. The circuit diagram is as shown. ANS:

The unregulated d.c. voltage is connected to Zener diode through a series resistane Rs. Zener diode is connected in reverse bias arrangement. Output voltage is obtained across load resistor RL which is joined in parallel to Zener diode. If the Zener voltage increases, the current through RS and Zener diode increases. This increases the voltage drop across RS but the voltage drop across Zener diode (or load RL) remains constant.

This is because in the breakdown region, Zener voltage remains constant even though the current through the Zener diode changes. Again if the input d.c. voltage decreases, The voltage drop across RS decreases but voltage drop across Zener diode (or load RL) remains practically constant. Thus any increase / decrease in the input voltage Vi results in an increase / decrease of voltage drop across RS without any change in voltage across the Zener diode. Hence, the Zener diode acts as a voltage regulator, in which the output d.c voltage is constant at the Zener (breakdown) voltage VZ of the Zener diode. Q49.Draw a circuit diagram to study the input and output characteristics of a n-p-n transistor in its common –emitter configuration. Draw the typical input and output characteristics and explain how these graphs are used to calculate (i) input resistance, (ii) output resistance and (iii) current amplification factor of the transistor.

ANS: From the input characteristics, input resistance is the ratio of change in emitter base voltage to the resulting change in the base current at constant emitter voltage. Input resistance =

From the output characteristics, output resistance of transistor is the ratio of change in collector emitter voltage to the resulting change in collector current at constant base current. Output resistance =

Current amplification factor of a transistor in CE configuration is defined as the ratio of change In collector current to the change in base current at a constant collector emitter voltage when The transistor is in active state. Current amplification factor can be obtained from its output Characteristics.

Q50. Draw a labelled diagram of a full wave rectifier circuit. State its working principle. Show the input- output waveforms.

ANS: The forward bias resistance of a p-n junction is very low as compared to the reverse bias resistance. This property is used to rectify a.c signal i.e, to restrict the voltage variation of a.c to one direction only by the use of p-n junction diodes. The circuit arrangement of a full wave rectifier using two diodes D1 and D2 has been shown in the figure. We use a centre tap transformer having its secondary wound into two equal parts such that voltages at any instant at A (input of diode D1) and B (input of diode D2) are out of phase with each other. Let initially input voltage at A is positive then it will be negative voltage at B at that instant. Consequently diode D1, being in forward bias, conducts while D2, being in reverse bias, does not conduct. We get output current and hence output voltage across load resistance RL joined between X and Y due to D1. After half cycle of a.c input, voltage at A becomes negative and that at B positive. Now diode D1 does not conduct but D2 conducts and output voltage is again obtained across RL in the same direction. Thus, output is obtained throughout the input wave cycle. The Input and output waveforms have been shown.

COMMUNICATION SYSTEMS One mark Questions Q1. Why do we need carrier wave of very high frequency in the modulation of signals?

ANS: High frequency carrier waves are used to increase operating range, to reduce antenna length and convert the wide band signal into narrow band signal. Then the signal can be easily recovered and distinguished from other signals at the receiving station. Q2. What is sky wave propagation? ANS: Propagation of frequencies less than 40 MHz using the reflecting property of ionosphere is called sky wave propagation Q3.What is ground wave propagation? ANS Propagation along the surface of earth for signals of frequency < 10 MHz is called ground wave propagation. Q4. What is space wave propagation? ANS When the signal travels directly from the transmitting antenna to the receiving antenna with frequencies more than 40 MHz the propagation is called space wave propagation. Q5. What is the need for modulation? ANS: To reduce the size of antenna/ high power transmission/ avoiding mixing up of signals Q6. Name an appropriate communication channel, needed to communicate a signal of bandwidth 100 kHz over a distance of 8 km. ANS: Space wave communication using low wavelength microwaves. Q7. Which part of the electromagnetic spectrum is used in satellite communication? ANS: Microwaves. Q8. Explain the function of repeater in a communication system. ANS: Repeater extends the range of communication system. It picks up signals from the transmitter, amplifies and retransmits the signals to the receiver. Q9.Why is ground wave transmission of signals restricted to a frequency of 1500 kHz? ANS: In the ground wave propagation, attenuation increases with increase in frequency for frequencies above 1500 kHz the loss of power becomes extremely large. Hence, for ground wave propagation, the upper frequency limit is 1500 kHz. Q9. Audio signals, converted into an electromagnetic wave are not directly transmitted. Give reason. ANS: If a number of transmitters are transmitting base band signals simultaneously, then all these signals cover the same frequency range (20 Hz to 20 kHz) and will get mixed up. If the transmission is being done at a high frequency different users may use adequate bandwidth for that given frequency.

Q10.What should be the length of the dipole antenna for a carrier wave of frequency 6 x 108Hz? ANS: λ= c/ν = 3 x108 /6x 108 =0.5 m. Length of dipole antenna l = λ/4 = 0.125 m. Q11.The maximum amplitude of an AM wave is found to be 12V while its minimum amplitude Is found to be 3V. What is the modulation index? ANS: µ =

= = =

Q12.What is the significance of modulation index? ANS: Modulation Index determines the strength and quality of the transmitted signal. Higher modulation index ensures better quality and better strength. Two marks questions Q13. Why do we need a higher bandwidth for transmission of music compared to that for commercial telephonic communication? ANS: Frequency of speech signals range from 300 Hz to 3100 Hz and hence a bandwidth of 2800 Hz (or 2.8 kHz) is sufficient for commercial telephonic communication. As frequencies produced by musical instruments extend up to 20 kHz, hence a higher band width of 20kHz is needed for transmission of music programme. Q14. (i)Define modulation index.

(ii)Why is the amplitude of modulating signal kept less than the amplitude of carrier wave? ANS:The ratio of modulating signal to the amplitude of carrier waves is called modulation index. µ = Am/Ac Q15. In order to avoid distortion Am<Ac i.e, µ < 1. ANS: If Am> Ac then over-modulation takes place resulting in the distortion of signal. Q16. A device X can convert one form of energy into another. Another device Y can be regarded as a combination of a transmitter and a receiver. Name the device X and Y. ANS: X is transducer and Y is repeater. Q17.What is the range of frequencies used in satellite communication? What is common between these waves and light waves? ANS: Range of frequencies used in satellite communication is: uplink::5.925 to 6.425 GHz

and downlink: 3.7 to 4.2 GHz both, the waves used in satellite communication and light waves are electromagnetic waves and can travel through vacuum with the same speed.

Q18. What is the range of frequencies used for TV transmission? What is common between these waves and light waves? ANS: Range of frequency used for TV transmission is 54 MHz to 890 MHz (VHF and UHF). These waves and light waves are electromagnetic waves. The ionosphere is unable to reflect back these waves to earth. Q19. By what percentage will the transmission range of a TV tower be affected when the height of the tower is increased by 21%? ANS: Transmission range d = √2Rh

h‘= h + 21% of h = 1.21 h

d‘ = 2R(1.21h) = 1.1 √2Rh = 1.1 d

Percentage increase in transmission range = x 100% = 10% Q20. Sky waves are not used in transmitting TV signals. Why? State two factors by which the range of TV signals can be increased? ANS: TV signals have a frequency of 100 to 220 MHz which cannot be reflected by the ionosphere. So transmission of TV signals via sky wave is not possible. Range of TV transmission can be increased by using tall antenna and(b) geostationary satellites. Q21. Why is communication using line of sight mode limited to frequencies above 40 MHz? ANS: At frequencies above 40 MHz, communication is limited to line-of-sight communication.

At these frequencies, the sizes of transmitting and receiving antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. During line of sight communication, the waves coming directly from transmitting antenna towards receiving antenna get blocked at some point by the curvature of the earth.

Q22. A carrier wave of peak voltage 12V is used to transmit a message signal. What would be the peak voltage of the modulating signal in order to have a modulating index of 75%? ANS: µ = Am/Ac Am = µ / Ac = 75/100 x 12 = 9V Q23. What is sky wave communication? Why is this mode of propagation restricted to the frequencies only upto few MHz? ANS: Ionospheric reflection of radio waves back towards the earth is known as Sky wave communication. The ionospheric layers act as reflector for a certain range of frequencies (3-30 MHz)

Electromagnetic waves of frequencies higher than 30 MHz (or few MHz) penetrate the ionosphere and escape. Q24. A transmitter transmits a power of 25 kW, when modulation is 40%. Calculate the power of carrier waves. ANS: Pt= 25 kW, µ = 40/100 = 0.4

Now, Pc = Pt / (1 + µ2/2 ) =25 / 1+ (0.4)2/2 = 25 / 1.08 = 23.14 kW Q25. A message signal of 12 kHz and peak voltage 20V is used to modulate a carrier of frequency 12 MHz and peak voltage 30V. Calculate (i) modulation index (ii) side- band frequencies. ANS: Modulation index = Am/Ac = 20/30 = 0.66

Side bands fc + fm = 12 MHz + 0.012 MHz = 12.012 MHz and fc + fm = 12MHz + 0.012 MHz = 11.988 MHz Three marks Questions Q26. Name the three different modes of propagation of electromagnetic waves. Explain, using a proper diagram the mode of propagation used in the frequency range above 40 MHz. ANS:Three modes of propagation of electromagnetic waves are:

Ground waves Sky waves Space waves

In the frequency range above 40 MHz space wave propagation is used. Space waves travel in a straight line from transmitter to receiver. Space waves are used for line of sight (LOS) communication as well as satellite communication.

Q27. Give reasons for the following: Long distance radio broadcasts use short- wave bands. The small ozone layer on the top of the stratosphere is crucial for human survival. Satellites are used for long distance TV transmission. ANS: As radio waves from short-wave bands get reflected from ionosphere, we use them for long distance communication. It absorbs large portion of ultraviolet radiations in the solar radiation which are otherwise harmful for living organisms. Television signals of the frequency range from 100 MHz to 200 MHz neither follow the curvature of earth nor get reflected by ionosphere. Therefore, Satellites are used for long distance TV transmission. Q28. Consider an optical communication system operating at λ = 800 nm. Suppose, only 1% of the optical source frequency is the available channel bandwidth for optical communication. How many channels can be accommodated for transmitting (a) audio- signals requiring a bandwidth of 8kHz, and (b) Video TV signals requiring an approximate bandwidth of 4.5 MHz. ANS: Optical wavelength used λ= 800 nm = 8 x 10-7 m

Optical source frequency f = c/λ = 3 x 108/8x 10-7 = 3.75 x 1014Hz

Total bandwidth of channel = 1 % of 3.75 x 1014Hz = 3.75 x 1012 Hz

Number of channels =

Number of channels for audio signals = 3.75 x 1012 / 8 x 103 = 4.7 x 108 Number of channels for video TV signal = 3.75 x 1012 / 4.5 x 106 = 8.3 x 105

Q29. The TV transmission tower at a particular station has a height of 160 m. (a) What is the coverage range? (b) How much population is covered by the transmission, if the average density around the tower is 1200 km-2? (c) By how much should the height be increased to double its coverage range? Given radius of the earth =6400 km. ANS: Coverage range, d = √2hR = √2 X160 X 6400000 = 45255 m = 45.255 km

Population covered = ρ x πd2 = 1200 X X (45255)2 = 77.24 lakh

Q30. Draw a block diagram showing the important components in a communication system. What is the function of a transducer. ANS: Block diagram of communication system

ANS: Transducer is a device which converts one form of energy into another form. Q31. What is remote sensing? What type of satellite is required for it? Give its two applications. ANS: Remote sensing is the technique of obtaining information about an object, area or phenomenon acquired by a sensor that is not in direct contact with the target of investigation. Sun synchronous satellite is used. (i) Spying work for military purposes (ii) In ground water surveys. Q32. Which mode of propagation is used by short wave broadcast services having frequency range from a few MHz upto 30 MHz ? Explain diagrammatically how long distance communication can be achieved by this mode. Why is there an upper limit to frequency of waves used in this mode? ANS: Sky wave propagation / ionospheric reflection

As signals get reflected back to the earth (due to internal reflection) from 65 km to 400 km above its surface, range of transmission increases from ionosphere. Electromagnetic waves of these frequencies are reflected by the ionosphere towards the earth. E.M waves of frequencies higher than 30 MHz penetrate the ionosphere and escape.

Q33. A ground receiver station is receiving a signal at a 8MHz, transmitted from a ground transmitter at a height of 200m located at a distance of 100 km. Identify whether it is coming

via space wave or sky wave propagation or satellite transponder Radius of earth = 6.4 x 106m. Maximum number density of electrons in ionosphere = 1012 m-3 ANS: For space wave propagation, maximum distance covered is given by d= √2Rh = 2 (6400000)289 = 60.83 X 103 = 61 km Since the distance between transmitter and receiver is 100 km hence for the given frequency Signal of 8 MHz, the propagation is not possible via space wave propagation. For sky wave propagation, the critical frequency ν= 9 (Nmax)1/2 = 9 (1012)1/2 = 9 MHz Since the frequency of signal 8MHz is less than νc. Hence its propagation via sky wave is possible. Q34. (i) Why is communication using line of sight mode limited to frequencies above 40 MHz?

(ii) A transmitting antenna at the top of a tower has a height 36 m and that of the receiving antenna is 49 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of earth is 6.4 x 106m.

ANS: EM waves having frequencies above 40 MHz cannot be reflected by the ionosphere. The sky propagation is not possible. Moreover, such high frequency waves are heavily absorbed during their propagation over the surface of the earth. Their ground wave propagation is not possible. Such waves can only propagate by line of sight mode.

Here, hT = 36 m, hR = 49 m R = 6400 km = 64 x 105m Maximum distance for satisfactory communication in LOS mode

dM = √2Rh + √2Rh = 8 x 6 x 102 x √20 + 8 x 7 x 102 x √20 = 46510.2 m = 46.5 km. Q35. Draw a block diagram of amplitude modulation. Explain briefly how it is achieved.

ANS: To achieve amplitude modulation, modulating signals are added to carrier signal

x(t) = Am sin ωmt + Ac sin ωmt Signals thus, produced, x(t), is then passed through a square law device. Square law device converts the signal in a non –linear form.

y(t) = B x(t) + C x2(t)

This signal is passed through a band pass filter which rejects low and high frequencies allows a band of frequencies to pass through. Output, thus obtained, is AM wave. Q36. What is the role of band pass filter in amplitude modulation? Draw a block diagram of a detector of A.M signal and briefly explain how the original signal is obtained from the modulated wave. ANS: A band pass filter rejects low and high frequencies and allows a band of frequencies to pass through. It rejects d.c. and the sinusoids of frequencies ωm, 2ωm and 2ωc and retains the frequencies ωc, ωc-ωm and ωc+ωm

In order to retain original signal from the modulated wave it is passed through a rectifier first. Rectified wave then passed through an envelope detector which filters out rectified wave and signal is retrieved. Q37. What is meant by detection of a signal in a communication system? With the help of a block diagram explain the detection of A.M. signal.

ANS: Detection or de- modulation is the process used to recover the original signal from the received signal which is a higher frequency signal. STEPS: (i) It rectifies the modulated carrier wave. (ii)It filters out the components, the carrier and the useful audio signal. The waveforms at different stages can be shown below:

Q 38. Draw a schematic diagram showing the (i) ground wave (ii) sky wave and (iii) space wave Propagation modes for electromagnetic waves. Write the frequency range for each of the following; Standard AM broadcast, Television, Satellite communication. ANS:

(a) Ground wave Standard AM broadcast 530 kHz to 1710 kHz

(b) Sky wave Television 1710 kHz to 40 MHz

(c)Space wave Satellite communication upto 1710 kHz

A modulating signal is a sqaue wave as shown.

Q39. The carrier wave is given by c(t) = 2 sin (8πt) volts. Sketch the amplitude modulated waveform. What is the modulation index? ANS: The modulating signal wave is a square wave of maximum amplitude Am= 1V and frequency 1 Hz. The carrier wave is c( t) = 2 sin (8πt) volts i.e, it is a sinusoidal wave of amplitude AC =2 V and angular frequency ω = 8π or frequency 4Hz As superposition of two waves we get amplitude modulated waveform as shown. The maximum and minimum values of AM wave are 2+1 = 3V and 2-1= 1V respectively.

Frequency of AM wave is 4 Hz. Am = 1V and AC = 2V Modulation Indexµ = Am / AC = 1V/2V = 0.5

Q40. Name any three types of transmission media that are commonly used for transmission of signals. Write the range of frequencies of signals for which these transmission media are used. ANS: The commonly used transmission media are (i) wire (cable) (ii) free space and (iii) optic fibre cable. Co axial cable is a widely used wire medium. These cables are normally operated below 18 GHz and offer a bandwidth of approximately 750 MHz. Communication through free space using radiowaves takes place over a wide range of frequency from a few hundred of kHz to a few GHz. Communication through optical fiber cable is performed in the frequency range of 1 THz to 1000 THz .An optical fibre can offer a transmission bandwidth of 100 GHz or even more. VALUE BASED QUESTIONS (04 marks)

Q1. Mahesh was doing social work during vacation. He visited a village where there was no electricity. He decided to help the villagers in purchasing solar panels. For this he awared them about the technology and advantages. Villagers applied for solar panels and got them from government at nominal charges. What type of a person is Mahesh? What is the principle of a solar cell? How does it work?

ANS: Mahesh is a very social person He wants to help people with modern technology. He has scientific temper and is a caring person. Solar cell works on the principle of photovoltaic effect. When photons of appropriate frequency fall on the p-n junction of solar cell, they generate electron-hole pairs. These are separated out by electric field at the junction .this way e.m.f is developed across the solar cell. Q2. In an experiment on photoelectric effect, Summit observed that when monochromatic yellow coloured lift beam is incident on a given photosensitive surface, photoelectrons are not ejected, while the same surface gives photoelectrons when exposed to green coloured monochromatic beam. Sumit consulted his testbooks but could not understand the reason behind it. He told his problem to his Physics teacher who explained that yellow light has a frequency below the threshold frequency of the given photosensitive surface, so no photoemission occurs with the yellow light. (a). What are the values being displayed by Sumit? (b). What will happen if the same photo-sensitive surface is exposed to (i) violet light and (ii) Red light? Give reason. Ans:(a). Power of observation, determination and critical thinking. (b)(i). Violet light ejects photoelectrons because its frequency is greater than that of green light. (ii). Red light will not eject photoelectrons because its frequency is less than that of yellow light. Q3.A function was arranged in the school auditorium. The auditorium has the capacity of 500 students. When entry, started. Student entered in groups and counting became a great problem. Then, science students took the responsibility at the gate. All the students entered the hall one by one. This helped them to maintain discipline and counting became easy with the help of a devise used by these students. (a.).What values are displayed by science students? (b). Name the devise which is based on application of photoelectric effect. Ans.(a). Knowledge and sense of responsibility. (b). Photo-cell connected to a digital counter. Q4. Mrs. Sharma’s family was fast asleep during night. They has no clue that their living room has caught fire due to a short circuit. Suddenly they heard sound of an alarm and woke up. They were surprised to see that the sound was coming from the model of fire alarm prepared by their son. They were all happy that a small science model had saved their life. (a). Give the values displayed by the parents and the son. (b). Name the devise used in the model. Ans. (a). Knowledge and scientific thinking.

(b). A photo-cell connected to an electric bell. When the fire breaks out, light radiations fall on the photo-cell. Photo-electric current begins to flow and electric bell starts operating as warning signal. Q5.Anju’s mother was having a constant headache and was diagnosed with tumour. She was avoiding treatment because of financial constraints. When Anuj learn about it, he cancelled his plans to go abroad and decided to use that money for the treatment and care of his mother. Answer the following questions. (a). What, according to you, are the values displayed by Anuj? (b). Which type of radiation do you think could be used for the treatment? (c). When are γ-rays emitted by a nucleus? Ans. (a). Affection, care and concern. (b). γ-rays. (c). After losing an α- or β-particle, a daughter nucleus is left in the excited state. It comes to its ground state by emitting one or more γ-ray photons. Q6.Ajita’s uncle was advised by his doctor to get an ECG for his heart. Her uncle did not know much about the details and significance of this test. She told her uncle that an ECT (Electrocardiograph) would enable the doctors to know the conditions of his heart without causing any harm to him. Her uncle was convinced and got the required ECG done. The resulting information greatly helped the doctors to treat him well. (a). What were the values displayed by Ajita? (b).What is an ECT? What is its importance? Ans. (a). Empathy, helping and caring. (b). An electrocardiograph (ECG) is a graph of the electrical activity of the heart. Recordings are made from electrodes fastened over the heart and usually on both arms and a leg. Changes in the normal pattern of an ECG indicate heart irregularities or disease. Q7. A semiconductor devise is used as a rectifier that allows the voltage to flow in positive direction and a very small value in the reverse direction. Now a day, there is a problem of supply of less voltage that damages the household appliances. You are asked to give the technique to save the appliances in use. (a). What can you think to solve the situation ? (b). Can a diode be fabricated in terms of doping and choice of material to control the input voltage to save your appliances from damage? Ans. (a).Critical thinking.

(b). Yes, it is Zener diode which is specially designed to operate in the reverse breakdown region continuously and can be used as a voltage regulator. Q8. A child uses a semiconductor device in listening radio and seeing pictures on TV. He was asked to suggest the techniques as the cost of LPG/CNG is going up, to cope with future situations. (a) What are the values developed by the child? (b) What may be the suitable semiconductor material used for utilization of maximum solar energy ? Ans. (a). Awareness of social problems and generating new idea with fluency. (b).The suitable semiconductors for the fabrication of solar cells are Si, GaAs, CdTe, etc., because of their suitable bandgap (1.0 to 1.45 eV) and high optical absorption. Q9. Sabby and Sonali wanted to purchase a new TV set. They visited electronic shops to look for sine branded TVs. The dealer showed them both LCD and LED TVs Now they may confused which set to buy. Finally after discussing with friends, reading relevant literature and searching the internet, they decided to purchase an LED. (a). Which value is being highlighted by Sabby and Sonali ? (b). What is difference between LED and LCD? Ans. (a). Interpretation skill. (b). LED is a forward biased junction diode which spontaneously converts the biasing electrical energy into optical energy like visible light. Liquid crystals are liquids which show some degree of molecular order. Liquid crystal displays (LCDs) make use of light-polarizing properties of such crystals. When no voltage is appled to the crystal, it twists the light polarization through 900 and polarized light is reflected by a mirror. When voltage is applied, there is no twisting, the light does not pass through second polarizer and is not reflected. Q10. Nidhi has to take admission in some professional college. It was last date of admission and Nidhi left her birth certificate at her home. College was very far from the home. She called her brother and he faxed the birth certificate. She got the admission and thanked her brother. (a). What value was displayed by Nidhi ? (b). What value was displayed by her brother? (c). What is the function of a Fax? Ans: (a).Awareness. (b). Understanding. (c). It is used to produce an exact copy of a document or picture at a distant place.

COMMUNICATION SYSTEMS

01 Mark Questions

1. Why do we need carrier wave of very high frequency in the modulation of signals? ANS: High frequency carrier waves are used to increase operating range, to reduce antenna length and convert the wide band signal into narrow band signal. Then the signal can be easily recovered and distinguished from other signals at the receiving station.

2. What is sky wave propagation? ANS: Propagation of frequencies less than 40 MHz using the reflecting property of ionosphere are called sky wave propagation

3. What is ground wave propagation?

ANS: Propagation along the surface of earth for signals of frequency < 10 MHz is called ground wave propagation.

4. What is space wave propagation? ANS: When the signal travels directly from the transmitting antenna to the receiving antenna with frequencies more than 40 MHz the propagation is called space wave propagation.

5. What is the need for modulation? ANS: To reduce the size of antenna/ high power transmission/ avoiding mixing up of signals

6. Name an appropriate communication channel, needed to communicate a signal of

bandwidth100 kHz over a distance of 8 km. ANS: Space wave communication using low wavelength microwaves.

7. Which part of the electromagnetic spectrum is used in satellite communication? ANS: Microwaves.

8. Explain the function of repeater in a communication system. ANS: Repeater extends the range of communication system. It picks up signals from the transmitter, amplifies and retransmits the signals to the Receiver.

9. Why is ground wave transmission of signals restricted to a frequency of 1500 kHz?

ANS: In the ground wave propagation, attenuation increases with increase in frequency. For frequencies above 1500 kHz the loss of power becomes extremely large. Hence, for ground wave propagation, the upper frequency limit is 1500 kHz.

10. Audio signals, converted into an electromagnetic wave are not directly transmitted.

Give reason. ANS: If a number of transmitters are transmitting base band signals simultaneously, then all these signals cover the same frequency range (20 Hz to 20 kHz) and will get mixed up. If the transmission is being done at high frequency different users may use adequate bandwidth for that given frequency.

11. What should be the length of the dipole antenna for a carrier wave of frequency 6 x 108Hz?

ANS: λ= c/ν = 3 x108 /6x 108 =0.5 m. Length of dipole antenna l = λ/4 = 0.125 m.

12. The maximum amplitude of an AM wave is found to be 12V while its minimum amplitude is found to be 3V. What is the modulation index?

ANS: µ =

= = =

13. What is the significance of modulation index? ANS: Modulation Index determines the strength and quality of the transmitted signal. Higher Modulation Index ensures better quality and better strength.

02 Marks Questions

14. Why do we need a higher bandwidth for transmission of music compared to that for commercial telephonic communication?

ANS: Frequency of speech signals range from 300 Hz to 3100 Hz and hence a bandwidth of 2800 Hz (2.8 kHz) is sufficient for commercial telephonic communication. As frequencies produced by musical instruments extend up to 20 kHz, hence a higher band width of 20 kHz is needed for transmission of music programme.

15. (i) Define modulation index.

(ii) Why is the amplitude of modulating signal kept less than the amplitude of carrier wave? ANS:(i) The ratio of modulating signal to the amplitude of carrier waves is called Modulation index.

µ = Am/Ac (ii) In order to avoid distortion Am < Ac i.e, µ < 1.

If Am > Ac then over-modulation takes place resulting in the distortion of signal.

16. A device X can convert one form of energy into another. Another device Y can be regarded as a combination of a transmitter and a receiver. Name the device X and Y.

ANS: X is transducer Y is repeater

17. What is the range of frequencies used in satellite communication? What is common between these waves and light waves?

ANS: Range of frequencies used in satellite communication is: uplink: 5.925 to 6.425 GHz and downlink: 3.7 to 4.2 GHz. Both the waves used in satellite communication and light waves are electromagnetic waves and can travel through vacuum with the same speed.

18. What is the range of frequencies used for TV transmission? What is common between

these waves and light waves?

ANS: Range of frequency used for TV transmission is 54 MHz to 890 MHz (VHF and UHF). These waves and light waves are electromagnetic waves. The ionosphere is unable to reflect back these waves to earth.

19. By what percentage will the transmission range of a TV tower be affected when the

height of the tower is increased by 21%? ANS: Transmission range d = √2푅ℎ

h’ = h + 21% of h = 1.21 h d’ = 2푅(1.21ℎ) = 1.1 √2푅ℎ = 1.1 d Percentage increase in transmission range = x 100% = 10%

20. Sky waves are not used in transmitting TV signals. Why? State two factors by which the range of TV signals can be increased?

ANS: TV signals have a frequency of 100 to 220 MHz which cannot be reflected by the ionosphere. So transmission of TV signals via sky wave is not possible. Range of TV transmission can be increased by using (a) Tall antenna and (b) geostationary satellites.

21. Why is communication using line of sight mode limited to frequencies above 40 MHz? ANS: At frequencies above 40 MHz, communication is limited to line-of-sight communication. At these frequencies, the sizes of transmitting and receiving antennas are relatively smaller and can be placed at heights of many wavelengths above the ground. During line of sight communication, the waves coming directly from transmitting antenna towards receiving antenna get blocked at some point by the curvature of the earth.

22. A carrier wave of peak voltage 12V is used to transmit a message signal. What would be

the peak voltage of the modulating signal in order to have a modulating index of 75%? ANS: µ = Am/Ac

Am = µ / Ac = 75/100 x 12 = 9V

23. What is sky wave communication? Why this mode of propagation is restricted to the frequencies only upto few MHz?

ANS: Ionospheric reflection of radio waves back towards the earth is known as Sky wave communication. The ionospheric layers act as reflector for a certain range of frequencies (3-30 MHz). Electromagnetic waves of frequencies higher than 30 MHz (or few MHz) penetrate the ionosphere and escape.

24. A transmitter transmits a power of 25 kW, when modulation is 40%. Calculate the power

of carrier waves. ANS: Pt= 25 kW, µ = 40/100 = 0.4

Now, Pc = Pt / (1 + µ2/2) =25 / 1+ (0.4)2/2 = 25 / 1.08 = 23.14 kW

25. A message signal of 12 kHz and peak voltage 20V is used to modulate a carrier of frequency 12 MHz and peak voltage 30V. Calculate (i) modulation index (ii) side- band frequencies.

ANS: (i) Modulation index = Am/Ac = 20/30 = 0.66 (ii)Side bands fc + fm = 12 MHz + 0.012 MHz = 12.012 MHz and fc + fm = 12MHz + 0.012 MHz = 11.988 MHz

03 Marks Questions

26. Name the three different modes of propagation of electromagnetic waves. Explain, using

a proper diagram the mode of propagation used in the frequency range above 40 MHz. ANS: Three modes of propagation of electromagnetic waves are:

(a) Ground waves (b) Sky waves (c) Space waves

In space wave propagation the frequency above 40 MHz is used. Space waves travel in a straight line from transmitter to receiver. Space waves are used for line of sight communication as well as satellite communication.

27. Give reasons for the following:

(i) Long distance radio broadcasts use short- wave bands (ii) The small ozone layer on the top of the stratosphere is crucial for human

survival. (iii) Satellites are used for long distance TV transmission.

ANS: (i) As radio waves from short-wave bands get reflected from ionosphere, we use them for long distance communication.

(ii) It absorbs large portion of ultraviolet radiations in the solar radiation which are otherwise harmful for living organisms.

(iii)Television signals of the frequency range from 100 MHz to 200 MHz neither follow the curvature of earth nor get reflected by ionosphere. Therefore, Satellites are used for long distance TV transmission.

28. Consider an optical communication system operating at λ = 800 nm. Suppose, only 1% of

the Optical source frequency is the available as channel bandwidth for optical communication. How many channels can be accommodated for transmitting (a) audio- signals requiring a bandwidth of 8kHz and (b) Video TV signals requiring an approximate bandwidth of 4.5 MHz.

ANS: Optical wavelength used λ= 800 nm = 8 x 10-7 m Optical source frequency f = c/λ = 3 x 108/8x 10-7 = 3.75 x 1014Hz Total bandwidth of channel = 1 % of 3.75 x 1014 Hz = 3.75 x 1012 Hz

Number of channels =

(a) Number of channels for audio signals

= 3.75 x 1012 / 8 x 103 = 4.7 x 108 (a) Number of channels for video TV signal

= 3.75 x 1012 / 4.5 x 106 = 8.3 x 105

29. The TV transmission tower at a particular station has a height of 160 m. (a) What is the coverage range? (b) How much population is covered by the transmission, if the average density around the tower is 1200 km-2? (c) By how much should the height be increased to double its coverage range? Given radius of the earth =6400 km.

ANS: (a) Coverage range, d = √2ℎ푅 = √2 푋160 푋 6400000 = 45255 m = 45.255 km

(b)Population covered = ρ x πd2 = 1200 X X (45255)2 = 77.24 lakh

30. Draw a block diagram showing the important components in a communication system.

What is the function of a transducer?

ANS:

Transducer is a device which converts one form of energy into another form.

31. What is remote sensing? What type of satellite is required for it? Give its two applications.

ANS: Remote sensing is the technique of obtaining information about an object, area or phenomenon acquired by a sensor that is not in direct contact with the target of investigation. Sun synchronous satellite is used. (i) Spying work for military purposes (ii) In ground water surveys.

32. Which mode of propagation is used by short wave broadcast services having frequency

range from a few MHz upto 30 MHz ? Explain diagrammatically how long distance communication can be achieved by this mode. Why is there an upper limit to frequency of waves used in this mode?

ANS:

As signal gets reflected back to the earth (due to internal reflection) from 65 km to 400 km above its surface, range of transmission increases from ionosphere. Electromagnetic waves of these frequencies are reflected by the ionosphere towards the earth. E.M waves of frequencies higher than 30 MHz penetrate the ionosphere and escape.

33. A ground receiver station is receiving a signal at a 8MHz, transmitted from a ground transmitter at a height of 200m located at a distance of 100 km. Identify whether it is coming via space wave or sky wave propagation or satellite transponder Radius of earth = 6.4 x 106m. Maximum number density of electrons in ionosphere = 1012 m-3

ANS: For space wave propagation, maximum distance covered is given by d = √2푅ℎ = 2 (6400000)289 = 60.83 X 103 = 61 km Since the distance between transmitter and receiver is 100 km hence for the given frequency Signal of 8 MHz, the propagation is not possible via space wave propagation. For sky wave propagation, the critical frequency ν= 9 (Nmax)1/2 = 9 (1012)1/2 = 9 MHz

Since the frequency of signal 8MHz is less than νc. Hence its propagation via sky waves is possible.

34. (i) Why is communication using line of sight mode limited to frequencies above 40 MHz? (ii) A transmitting antenna at the top of a tower has a height 36 m and that of the receiving

antenna is 49 m. What is the maximum distance between them for satisfactory communication in line of sight mode? Given radius of earth is 6.4 x 106m.

ANS: (i) EM waves having frequencies above 40 MHz do not get reflected by the ionosphere hence sky wave propagation is not possible. Moreover, such high frequency waves are highly absorbed during their propagation along the surface of the earth. Ground wave propagation is also not possible because such waves can only propagate by line of sight mode. (ii) Here, hT = 36 m, hR = 49 m R = 6400 km = 64 x 105m Maximum distance for satisfactory communication in LOS mode dmax = √2푅h + √2푅h = 8 x 6 x 102 x √20 + 8 x 7 x 102 x √20 = 46510.2 m = 46.5 km.

35. Draw a block diagram of amplitude modulation. Explain briefly how it is achieved. ANS:

To achieve amplitude modulation, modulating signals are added to carrier signal x(t) = Am sin ωmt + Ac sin ωmt

Signals thus, produced, x(t), is then passed through a square law device. Square law device converts the signal in a non –linear form.

y(t) = B x(t) + C x2(t) This signal is passed through a band pass filter which rejects low and high frequencies allows a band of frequencies to pass through. Output, thus obtained, is AM wave.

36. What is the role of band pass filter in amplitude modulation? Draw a block diagram of a detector

of A.M signal and briefly explain how the original Signal is obtained from the modulated wave. ANS: A band pass filter rejects low and high frequencies and allows a band of frequencies to pass through. It rejects d.c. and the sinusoids of frequencies ωm, 2ωm and 2ωc and retains the frequencies ωc, ωc-ωm and ωc+ωm

In order to retain original signal from the modulated wave it is passed through a rectifier first. Rectified wave then passed through an envelope detector which filters out rectified wave and signal is retrieved.

37. What is meant by detection of a signal in a communication system? With the help of a block diagram explain the detection of AM signal.

ANS: Detection or de-modulation is the process used to recover the original signal from the received signal which is a higher frequency signal.

STEPS: (i) It rectifies the modulated carrier wave. (ii) It filters out the components, the carrier and the useful audio signal. The waveforms at different stages can be shown below:

38. Draw a schematic diagram showing the (i) ground wave (ii) sky wave and (iii) space wave. ANS:

39. Mode of propagations for electromagnetic waves.

ANS:(a) Standard AM broadcast (b) Television (c) Satellite communication.

40. Write the frequency range for each of the following; (a) Ground waves: (b) Sky waves:

(c) Space waves; ANS: (a) Ground waves: Standard AM broadcast 530 kHz to 1710 kHz (b) Sky waves: Television 1710 kHz to 40 MHz (c)Space waves; Satellite communication up-to 1710 kHz

41.A modulating signal is a square wave as shown. The carrier wave is given by c(t) = 2 sin (8πt) volts. (a) Sketch the amplitude modulated waveform. (b) What is the modulation index?

ANS: (a)

(b) Comparing with the standard sine wave equation

c(t) = a sin ωt ω = 8π while ω=

T = s and amplitude = 2

40. Name any three types of transmission media that are commonly used for transmission of signals. Write the range of frequencies of signals for which these transmission media are used.

ANS: The commonly used transmission media are (i) wire (cable) (ii) free space and (iii) optic fibre cable.

(i) Co axial cable is a widely used wire medium. These cables are normally operated Below 18 GHz and offer a bandwidth of approximately 750 MHz.

(ii) Communication through free space using radio waves takes place over a wide range of frequency from a few hundred of kHz to a few GHz.

(iii) Communication through optical fibre cable is performed in the frequency range of 1THz to1000 THz. An optical fibre can offer a transmission bandwidth of 100 GHz or even more.