IMPORTANT CONCEPTS

Branch of physics based on the wave concept of light is called ‘wave optics’ or ‘physical optics’.Wave front:-A wave front is defined as the continuous locus of all the particles of a medium, which are vibrating in the same phase.Depending upon the shape of source of light wave front can be of different types.i)Spherical wave front :- source of light is a point sourceeg:-

ii)Cylindrical wave front:- source of light is linear.

iii) Plane wave front :- When the point source or linear source is at very large distance, a small portion of wave front appears to be plane which is called as planewavefront.

HUYGEN’S PRINCIPLE:-1) Every point on the given wave front acts as a fresh source of new disturbance, called secondary wavelets which travel in all directions with velocity of light.2) A surface touching these secondary wavelets tangentially in the forward direction at any instant gives the new wave front at that instant. It is called secondary wave front.

a) LAW OF REFLECTION (HUYGEN’S PRINCIPLE)

From the figureIn ∆ADC &∆ AEC∟D=∟E= 90˚ ;AC=AC= Common sideAE=DC=CTBy R.H.S Congruency these two Δ΄are congruentSo CPCT are equal∟DAC=∟ACEI=rAngle of incidence = angle of refraction

b) Law of refraction: Huygen’s Principle

From the figure we haveAD=V2 t ; BC=V 1tAccording to definition of law of refraction,n2 / n1 =sin i/ sin rFrom the ∆ABC Sin i= BC/ACIn the ∆ADC Sin r=AD/ACn2 / n1 =sin i/ sin r = BC/AC * AC/AD= V 1t/ V2 t= V 1/ V2

n2 / n1 =sin i/ sin r= V 1/ V2

Huygen’s principle obey the law of refraction

RELATION BETWEEN PATH DIFFERENCE &PHASE DIFFERENCE

λ IS The path differenceThen phase difference is 2ΠIf x is path differencei.e. x is λφ is 2Πφ = 2Πx/ λx= φλ/2Π

COHERENT SOURCES:The sources of light which emit continuous light waves of the same wave length, same frequency and in same phase or having a constant phase difference are called coherent sourcesTwo independent sources of light cannot be coherent

INTERFERENCE:Two waves super imposed each other as a result the intensity of resulting wave increase or decrease is called “interference”Interference are two types. They are

1) Constructive interference: the crust and crust, trough and trough of two waves are superimposed; as a result maximum intensity wave is formed

1) Destructive interference: when two waves are super impose crust of one wave combined with the trough of another wave and the intensity of resulting wave decreases

When there is a modification in energy distribution because of super imposing of waves is known as “interference”(I)I is directly proportional to A2I=a1

2+ a2

2+ 2 a1 a 2 cosφ

Imax=( a1+ a2)2 φ=0Imin=( a1- a2)2 φ=180°I=I1+I2+2√ I1 I2 cos φImax/ Imin=( a1+ a2)2 /( a1- a2)2= (r+1) 2/(r-1) 2

Intensity of slit is directly proportional to width of the slit → I ά WI1/I2= a1

2/ a22=W1/W2

I=a12+ a2

2+ 2 a1 a 2 cosφ a1= a2 = a

I=2a2+2a2 cosφI=2a2 (1+ cosφ)I=2 a2 X 2 cosφ/2I=4 a2 cosφ/2

CONDITIONS FOR BRIGHT BAND:

It is formed during constructive interference.Path difference φ = 0°,2Π,4Π,6Π,……2nΠPhase difference =δ =0; λ; 2 λ; 3 λ;…; n λCONDITION FOR DARK BAND:It is formed due to destructive interference.Path difference φ = Π,3Π,5Π,…(2n+1)ΠPhase difference: δ = λ/2, 3λ/2… (2n+1) λ/2

YOUNG’S DOUBLE SLIT EXPERIMENT:

In young’s doubles slit experiment 2 narrow slits s1ands2 separated by distance‘d’act as coherent sources.screen is placed at a distance ‘D’ from slits. coherent waves come out from the two slits. These are involved interference and alternatively dark and bright band are formed on the screen. At P there is always bright band which is called as central band.on either sides alternative bands are formed.

1)position of bright fringes is given by x=n λD/d where n=1,2,3,for 1st ,2nd ,3rd …bright fringes.2) Position of dark fringes is given by x = (2n+1) λ/2 D/d whereN=1, 2, 3,…for 1st ,2nd ,3rd dark fringes.3) Width of each bright / dark fringe is β=λD/d4) When the entire apparatus is immersed in a transparent medium of refractive index μ, fringe width β'=λ'D/d=λD/μd=β/μ5)angular width of interference fringes Ө=β/D=λ/d

PRERVIOUS YEARS QUESTIONS:

1Q)Two slits in young's double slit experiments are illuminated by two different sodium lamps emitting light of same wave length.do you observe any interference pattern on the screen?

Ans: No, interference pattern is not obtained. This is because phase difference between the light waves emitted from two lamps will change continuously.2Q)Why is interference pattern not detected ,when the two coherent sources are far apart?

Ans: As fringe width β ά 1/dTherefore when d is so large the width may reduce beyond the visible region.hence the pattern will not be seen.

3Q)No interference pattern is detected when two coherent sources are infinitely close to each other why?Ans: when d is negligibly small, fringe width β which is proportional to 1/d may become too large.Even a single fringemay occupy the screen. hence the pattern cannot be detected.

4Q) Consider interference waves from two sources of interferences I and 4I.Find interference at points where phase difference is i) Π/2 ii) ΠAns: As we knowA

2 = a1

2+ a2

2+ 2 a1 a 2 cosφ

Ia=I1+I2+2√I1√I2

When φ= Π/2Ia=I+4I+2√4I2(cos Π/2)Ia=5IWhen φ= ΠIa=I+4I+2√4I2(cos Π)Ia=5I-4IIa=I

5Q)In YDE the intensity of central maximum is I,what will be the intensity at the same place if the slit is closed?Ans: When one slit is closed, amplitude becomes ½ and hence intensity becomes ¼ and there is no interference.

6Q)Widths of two slits in young’s experiment are in the ratio 4:1. what is the ratio of their amplitudes of light waves from them?Ans: W1/W2=I1/I2= a1

2/ a22=4/1

a1/a2=2/1a1:a2=2:1

7Q) what are coherent sources of light? why no interference pattern is observed when two coherent sources are i)too close ii)very far apartAns:The sources of light which emit light waves of same wave length, same frequency and in same phase are having a constant phase difference are called coherent sources .1) when the sources are too close, distance (d) between them tends to zero. Fringe width β=λD/d tends to ά. Hence no interference pattern will be observed.2)when the distance (d) between them becomes large fringe width β becomes too small and no longer be in the visible region. hence no interference pattern will be observed.

8Q) State the condition which must be satisfied for two light sources to be consent.1. Coherent sources of light should be obtained from a single source by some device.2. The two sources of should give mono chromatic light.3. The path difference between light waves from two sources should be small.

Q9) The two slits in YDE are separated by distance 0.03m. when light of wave length 5000A0 falls on the slits an interference pattern is observed on the screen1.5m away. Find the distance of fourth bright fringe from the central maximum?Ans: d=0.03m λ=5000 X 10-10

D=1.5mN=4x=n λD/d

= 4 X 5000X10-10X1500/30=106-10-10

=10-4=0.1mm

10Q)In young’s experiment, the width of the fringes obtain with light of wave length 6000A 0is 2.00 mm what will be the fringe width is the entire apparatus is immersed in a liquid of n=1.33?Ans: β= λD/d β'=λ'D/dβ'/β = λ'D/d X d/ λD= λ'/ λ=1/nβ'= β/n= 2.0/1.33= 1.5 mm

11Q)In Y.D.E. the slits are 0.03cm apart and these screen is placed 1.5 m away.the distance b/w the central fringe and 4th bright fringe is calculate the wave length of light is used?Ans: d=0.03cm = 3X10-4m D=1.5 mx=1cm=10-2m n=4 λ=?For bright fringes x=n λD/dλ=xd/nD = 10-2X3X10-4X102/4X1.5λ=0.5X10-6mλ=5X10-7m

12Q) In Y.D.E the width of the fringes obtained with the light of wavelength 6000A0 is 2.0mm what will be the fringe width of the entire apparatus is immersed in a liquid of n=4/3?Ans: β = λD/dWhen the entire apparatus is immersed in water.λ'= λ/nβ'= β/n = 2.00/4/3 = 1.5mm

PROBLEMS:

1Q) The width of one of the slits is a Y.D.E is double of the other slit. Assuming that the amplitude of the light coming from a slit is proportional to the slit width, find the ratio of maximum intensity in the interference pattern.?Ans: Let ‘A’ be the amplitude of light from one slit. . . Amplitude of light from 2ndslit is 2A.Imax/Imin =(a1+a2)2/(a1-a2)2 = (2A+A)2/(2A-A)2 = 9A2/A2 = 9:1

2Q) A double slit is illumated by light of λ=6000A0 the slits are 0.1cm apart and the screen is placed 1m away. Calculate.1)angular position of 10th maximum in radian.2)separation of two adjacent minima?Ans: λ=6000A0 = 6x10-7

d=0.1cm = 10-3 ;D=1m n=10x=n λD/dif Ө is angle of diffraction then sin Ө=x/D=nλ/d=10X6X10-7/10-3

=6X10-3

As sin Ө is very small Ө≈sin Ө = 6X10-3rad2) separation of two adjacent minimaβ=λD/d = 6x10-7x1/10-3 = 6x10-3

3Q) I n young’s double slit experiment , two coherent sources are 1.5m apart and and the fringes are obtained at a distance of 2.5m from them. If the sources produce light of wavelength 589.3 nm ,find the no. of fringes in the interference pattern which is 4.9x10-3 long?Ans: Here d=1.5 nm= 1.5x10-3 m.D=2.5m, λ=589.3nm=589.3x10-3

fringe width β= λd/D589.3x10-9x2.5/1.5x10-3

=982.17x10-6m

No of fringes X/β=4.9X10-3/982.17X10-6

=4.99 ≈5

Q4) The interference frings for sodium light (λ=5890 AO) in a double slit experiment have an angular width oof 0.2AO .For what wave length will the width be 10% greater?Ans: as angular directly proportional to λλ’/λ=ө’/өnow ө=0.20 = Ө’ = 0.2+10(0.2)/100 = 0.22λ = 5890A0

λ’ = ө’/ө.λ = 0.22x5890/0.2 = 6479A0

5Q)Two slits are made 1mm apart and the screen is placed 1m away. What is the fringe separation when blue green light of wavelength 500nm is used?Ans: d = 1mm = 10-3mD = 1m λ= 500nm = 500x10-9m = 5x10-7mFringe separation = fringe widthβ= λD/d = 5x10-7x1/10-3 = 5x10-4mβ = 0.5mm

6Q)In aY.D.E λ = 500mm and D = 1.0m the minimum distance from the central maximum for which the intensity is half of the maximum intensity?Ans: IA =4 I0cos2 ¢/2 = ½ (4I0)Cos2¢/2 = ½ cos¢/2 = 1/√2 ¢/2 = 450

¢ = 900

Minmum distance from central maximum corresponding to ¢ = 900 would be ¢/2π x βX = π/2/2π x λD/dX = ¼ x 500 x 10-9x 1/1 x 10-3 = 1.25x10-4m=

7Q) In YDE while using a source of light of wavelength 5000AO, the fringe width obtained is 0.6 cm.If the distance between the screen and the slit is reduced to half , what should be the wave length of the source to get fringes 0.003 wide?Ans: λ=5000 AO=5000 x 10-10 mP1 =0.6 cm=0.006 mD1=D (say)λ

2=?β 2=0.003 D2=D/2Let‘d’ be the distance between b/w the slit is the two casesβ 1= λ1D1/d1 β2= β 1D2/d2

or, β1/ β1 = λ1D1/ λ2D2

λ2= λ1D1 β2/ β 1O2

λ2 =5000X10-10XDXDX0.003/0.006XD/2λ=5X10-7m

8 Q) Find the maximum intensity in case of interference of n identical waves each of intensity IO if the interference is a)coherent b)incoherentIntensity of each of the ‘n’ waves = I0

When interference is due to coherent sourcesI=I1+I2+2√I1I2cos ¢Imax= I1+I2+2√I1I2 cos 0°

= (√ I1+√I2)2

For n identical waves each of intensity (I0)Imax=(√ I0+√ I0+… n times) I0

=(n√ I1 + √ I2)2 = n2 I0

When the interference is due to incoherent source ¢ varies randomly with time.(cos¢)av= oImax= I1+I2For n identical waves each of intensity IOImax= IO+IO+……… ..(n times) Imax =n IO9Q) In young’s experiment two slits are 0.2mm apart the interference fringes for light of wavelength 6000A0 are formed on a screen 80cm away

a) How far is the 2nd bright image from the central image?b) How far is the 2nd dark image from the central fringe?

Ans: Here d = 0.2mm = 2 x 10-4mλ = 6000A0 = 6 x 10-7mD = 80cm = 0.8ma) X = ? n = 2x = nλD/d = 2 x 6 x 10-7 x 0.8/2 x 10-4

= 4.8 x 10-3mb) X =? n = 2x = (2n+1) λD/2d= 5 x 6 x 10-7 x 0.8/2 x 2 x 10-4

= 6 x 10-3m.

10Q) Green light of wavelength 5100A0 from a narrow slit is incident on a double slit. If the overall separation of 10 fringes on a screen 200cm away is 2cm, find the slit separation?Ans: λ = 5100A0 = 5.1 x 10-7mD = 200cm = 2m, d =?As β = λD/dd = λD/β = 5.1 x 10-7 x 2/2x 10-3

= 5.1 x 10-4m

DIFFRACTION:-

Diffraction of light is the phenomenon of bending of light around corners of an obstacle or aperture in the path of light on account of this bending light penetrates in to the geometrical shadow of the obstacle.The light does deualts from it’s linear path.The diffraction pattern due to a single slit consists of a central bright band having alternate dark and week bright bands of decreasing intensity on both sides .Condition for diffraction maximum.a sinө = (2n+1) x λ/2 where n = 1, 2,3,…………………Condition for diffraction minimum.a sinө = n λ where n = 1,2,3,4,5,………………………width of the central maximum 2x = 2x λf/a = 2x λ D/a =2βAngular width = 2Ө 2 x λ /a= 2 x/f = 2x/DDiffraction is supposed to be due to interference of secondary wavelets from the exposed portion of wave front from the slit Where as interference, all bright fringes have same intensity, in diffraction bright bands are of decreasing order intensity

Differences between interference & diffraction

Interference Diffraction1)Two coherent waves superimposed each other as a result interference pattern is observed2)width of all dark bands are equal as well as bright bands 3)Intensity of all bright bands are equal

4) conditions for interference maximum is a ¢ =2nπ δ=nλ5)conditions for interference minimum ¢ = (2n+1)π δ= (2n+1)λ6)intensity distribution graph

1)Different wavelets of the same wave superimposed as a result diffraction is obtained2)width of the central band is maximum

3)intensity of central band is maximum &decreases with increasing order 4)conditions for diffraction maximum is a sin Ө=(2n+1)λ/25)diffraction minimum a sinӨ=nλ

6) intensity distribution graph

Fresnel distance : It is the minimum distance a beam of light has to travel before it’s deviation from straight line path becomes significant Where Zf = a2/ λ Zf → fresnel distance

POLARISATION:

Unpolarised light:

When light is come out from the source it will move in all possible direction with equal velocity i.e. particles vibrates in all directions is known as unpolarised light If light particles vibrate in unidirectional it is called as polarized light.

→ Unpolarised light can be converted into polarized by using polarisers When the polarizer is placed parallel in front of the incident unpolarised light it will allows the light particles which are parallel to the direction of slit of the analysers is placed parallel to the polarisers same pattern is observed. Intensity of light come out from the analyser equal to intensity of light incident on analyser If analyser is placed perpendicular to polariser nolight come out from the analyser

MALUS LAW:

Intensity of light come out from the analyser is directly propotional to the square of cos angle I is directly propotionat to cos2Ө I=I0 cos2 ӨWhere Ө is an angle between polariser and analyser I0→ intensity of light which is transmit through the polariser Intensity vs Ө graph

POLARISATION OF LIGHT BY REFLECTION:

From the figure we have Ip+r2p=90o

r2p=90o-ip

A/c to def of refractive index n = sin i / sin r Sin ip/sin (90-ip)= n n=sin ip/cos ip=tan ip

n= tan ip

Refractive index is directly proportional to tan of polarizing angle

DOPPLER EFFECT:

Apparent change in frequency due to relative motion between a source of light and observer When the star approaches to the earth frequency of light increases then the spectra shift towards violet color is called violet shift δ /٧ ٧=u/cif the star receding to the earth the frequency of light decreases. Its wavelength increases.The spectra shift towards red color is called red shift δ /٧ ٧= =∆ λ/ λ=U/CDopplers effect in light is used in measuring speed of a star or galaxy ,satellite ;submarines etc

USES OF POLAROIDS:

1.To avoid glare of light 2.to avoid dazzling light of a car approaching from the opposite side during night driving3.In three dimentionals motional pictures in halography.4. To improve colour contrast in old oil paintings.5. Used in a optical stress analysis.

PREVIOUS YEAR QUESTIONS

1. Name one device producing polarized light?Nicole prism or calcite prism.

2. What is the value of refractive index of a medium of polarizing angle 600?N=Tanip

Tan 600=Ö3 N=Ö3

3. What does red shift in the spectra indicates?The universe is expanding.

4. Diffraction is common in sound but not common in light waves. Why ?For diffraction of a wave, an obstacle or operture of the size of the wave length of wave is needed. As wave length of light is the order of 10-6 m and obstacle of this size is rare, therefore diffraction is not common in light waves. On the contrary, wave length of sound is the order of 1 meter and obstacle of the size is readily available. Therefore diffraction is common in sound.

1 In a single slit diffraction the width of the slit of the made double the original width. How does this effect the size and intensity of these central diffraction brand?Width of the central maximum is 2x=2Dl/a. when width of the slit a is doubled

central maximum width is halfed. Its area becomes ¼ th . hence intensity of central diffraction brand becomes four times.

1. What evidence is there to show that sound is not electromagnetic in nature?Sound waves are not electromagnetic in nature. This is evident from the fact that sound waves are require a material medium for propagation. For example, when a jar fitted with an electric bell is evacuated no sound is heard though hammer is seen to strike the bell. This is because without air sound can not travel.

2. Can a naked eye detect polarization of light? If not how is polarization of light detected?No. a naked eye can not detect polarization of light. This distinction can be made by using a crystal. A calcite crystal , quartz crystal , a Nicole prism can be used a polarizer as well as analyzer.

3. Two narrow slits are illuminated by a single monochromatic source. Name the pattern obtained on the screen. One of the slit is now completely covered , what is the nature of the pattern obtained now on the screen? Draw intensity pattern obtain in the two cases? Also write two differences between the patterns obtained in the above two cases.When two narrow slits are illuminated by a single monochromatic sourece, the pattern obtained on the screen is interference pattern consisiting of alternative bright and dark frings. When one of the cities covered completely no interference occurs. When we obtain is diffraction pattern due to single slit.Intensity pattern in the two cases are shown.Diagram.

Interference:1. Two coherent waves superimposed each other as a result interference pattern is observed2.width of all dark bands are equal as well as bright bands 3.Intensity of all bright bands are equal 4. conditions for interference maximum is a

¢ =2nπ δ=nλ 5. conditions for interference minimum ¢ = (2n+1)π δ= (2n+1)λ 6. intensity distribution graph

Diffraction:1) Different wavelets of the same wave superimposed as a result diffraction is obtained2) width of the central band is maximum3) intensity of central band is maximum &decreases with increasing order 4) conditions for diffraction maximum is a sin Ө=(2n+1)λ/25) diffraction minimum a sinӨ=nλ6) intensity distribution graph

4. The light of wave length 600 nm is incident on an aperture of size 2mm. calculate the distance upto which the ray of light can travel, such that its spread is less than size of aperture.Here l=600 nm = 600X10-9 mA=2 mm = 2X10-3m, ZF = ?We know ZF = a2/l = (2X10-3)2/600X10-9 = 6.67 m.

5. Light of wave length 600 nm is incident normally on slit of width 3 mm . calculate the linear width of central maximum on a screen kept 3 m away from the slit?Here l = 600 nm = 600X10-9 mSlit width a = 3 mm = 3 X10-3 mD = 3 m.Width of the central maximum 2x=2Dl/a2x = 2Dl/a=2X3X6X10-7/3 X10-3=1.2 mm

6. Determine the angular spread between central maximum and first order maximum of diffraction pattern due to single slit of width 0.25 mm, when light of wave length 5890 A0 is incident on it normally.

Half the angular spread q is given byA sin q = (2n +1)l/2If n = 1 , a sin q = 3 l/2When q is very small sin q = » qa q = 3 l/2q = 3 l/2a = 3X5890X10-10/2X0.25X10-3

q=3.534X 10-3 radTotal angular speed is =±3.534X 10-3 rad

7. The R.I of a medium id Ö3. what is the angle of refraction if the unpolarized light in incident on it at the polarizing angle of the medium?m = Ö3Tan ip = m = Ö3Ip = tan-1Ö3=600

As r = 900-ip=900-600

R = 300

13. The spectral line of l=65000 A in the light coming from a distant star is observed at 65250 . determine the velocity of the star relative to earth?l=65000= 6500X10-10 m, ∆l=25 A0

∆l/l=v/cV=∆l/lXc=25/6500X3X108

=1.15X106 ms-1

PROBLEMS:

1. for what distance is ray optics a good approximation when the aperture is 3 mm wide and wave length is 500nm?Here a= 3 mm = 3X10-3 ml = 500 nm = 5X10-7 mTresnel distance Zf = a2/l = (3X10-3)/5X10-7

= 18 m

2. The spectral line for a given element is the light received from a distant star is shifted towards longer wave length side by 0.25%. calculate the velocity of star in the line of sight?Here ∆l/l = 0.025% = 0.025/100∆v/c=∆l/l V=∆l/lxc =2.5x10-4x3x108

=7.5x104m/sAs shift is towards longer wave length side the star is moving away.

3. Two polarizing sheets are placed with their planes parallel, so that light intensity transmitted is miximum through what angle must either sheet be transmitted so that light intensity drops to half the maximum value?According to malus lawI=I0 Cos2qCos2q = I/I0= ½Cos q = ±1/2q = ±450 or ± 1350

The effect will be same when any of the two sheets turned through q in any direction.

4. Light reflected from surface from glass plate of refractive index 1.57 is linearly polarized calculate the angle of refraction in glass?N=1.57, r=?According to Brewster’s lawTan Ip = n = 1.57Ip = tan -1(1.57)= 57.50

R = 90-Ip = 32.50

5. A slit of 4 cm wide is irradiated with micro waves of wave length 2 cm. find the angular spread of central maximum assuming incidence normal to the plane of the slit?Here a = 4 cm = 4X10-2 m

l = 2 cm = 2 X10- 2 mAngular spread of central maximum 2q is calculated from Sinq=l/a= 2X10-2/4X10-2=1/2

q= 300 then 2q = 2X300 = 600

6. A screen is placed 2 m away from the away from the single slit. Calculate the slit width if the first minimum lies 5 mm on either side of the central maximum incident plane wave have a maximum wave length of 5000A0. Here the distance of the screen from the slit D = 2 m X =5 mm =- 5X10-3 m l= 5000A0

For the first secondary minima Sinq = l/a = x/DThen a = lD/x = 2X5000X10-10/5X10-3

A=2X10-4 m

1. two spectral lines of sodium D1 and D2 have wave length of approximately 5890A0 and 5896 A0. a sodium lamp sends incident plane wave on to a slit of width 2micrometer. A screen is located two meters from the slit find the spacing between the first maxima of two sodium lines as measured on the screen?l1 = 5890 A0 = 5890X10-10 m then a = 2X10-6 ml2 = 5896 A0 = 5896X10-10 mD = 2 mFor the secondary maxima Sinq= 3/2l1/a = x1/DX1 = 3l1D/2a and x2 =3l2 D/2a Spacing between the first secondary maxima of two sodium lines = x2-x1=3D/2a(l2-l1)= 3X2X6X10-10/2X2X10-6

9X10-4 m2. An unpolarized beam of light is incident on a group of four polarizing sheets which are arranged in

such a way that the characteristic direction of each polarizing sheet makes an angle of 300 with that of the preceding sheet what fraction of incident un polarized light incident?If I0 is intensity of unpolarized light , then intensity of light from first polarized sheet is I0/2Intensity of light from second polarized sheet I1 = I0/2(Cos 300)2 = I0/2(Ö3/2)2

=I0/2 X3/4 = 3I0/8Intensity of light from third polarized sheet I11=I1 Cos 300)2

I11=I0/2(3/4)X(Ö3/2)2

Intensity of light from fourth polarized sheet I111 = I11X (Cos 300)2

I111= I0/2(3/4)2X(Ö3/2)2

=I0X27/128Therefore I111/I0=27/128.

9. In diffraction of a single slit, a screen is placed 2m away from lens to obtain pattern. If the slit width is 0.2mm and the first minimum lies 5mmon either side of central maximum, find the wave length of light used. D= 2m, d= 0.2 mm = 2X10-4 mThen x1= 5 mm = 5X10-3 mAnd x = lD/dl=xd/D = 5X10-3X2X10-4/2 = 5X10-7 m

1. A plane wavefront of wave length 6X10-7 m falls on a slit 0.4 mm wide. A convex lens of focal length 0.8 m placed behind the slit focuses the light on a screen . what is the linear diameter of i) first minimum ii) second maximum?Here l=6X10-7 m ,a = 4X10-4 m and f = 0.8 m

a. linear diameter of first minimum 2x=2lf/a = 2X10-7X0.8/0.4X10-3

= 2.4X10-3 m = 2.4 mmb. Linear diameter of second maximum

2x1=2(2n+1)fl/2a = 2X5X6X0.8X10-7/8X10-4

=6 mm

11. A beam of light of wave length 600 nm from a distant source falls on a single slit 1mm wide and the resulting diffraction pattern is observed on a screen two meters away . what is the distance between first dark band on the either side of the central bright band?Here dSinq = nl , n=1dSinq = ld(x/D) = lx = lD/d= 600X10-9X2/10-3

=12X10-4 m

12. Unpolarised light of intensity 32 watt/m2 process through a 3 polarizers such that the transmission access of the last polarizer is crossed with the first . if the intensity of emerging light is 4w/m2, what is the angle between the transmission axis of first two polarizer’s? At what angle will the transmitted intensity be maximum?

IMPORTANT QUESTIONS:

1. Derive the lens maker’s formula in case of a double convex lens.n State the assumptions made and conventions of signs used. A converging lens has a focal length of 20 cm in air. It is made of a material of refractive index 1.6. If it is

immersed in a liquid of refractive index 1.3, what will be its new focal length2 Draw a labelled ray diagram of an astronomical telescope used in the normal adjustment position.

Two astronomical telescopes T1 and T2 have the same magnifying power. The ratio ofaperture of their objectives is 3:2. Which one of these two produces Image of greater intensity?3. (a) A convex lens and a concave lens of same focal length are combined in contact. What would be the resulting focal length? (b) What changes will occur in the focal length of a (i) concave mirror and (ii) convex lens when the incident

violet light is replaced with red light. Give reason. ( c) A beam of parallel rays of light is incident on a concave lens of refractive index n1 held in a medium of

refractive index n2. Trace the path of the refracted rays when (i) n2 > n1 (ii) n2 < n1 (iii) n2 = n1

4. (a) The lens combination of three lenses L1 and L2 and L3 shown in the fig. forms an image at 30 cm to the right of the lens L3 . The focal lengths of lenses L1 and L2 are 10 cm and -10 cm respectively. What is the focal length of lens L3?

(b) A ray of light passes through an equilateral prism (msuch that angle of incidence is equal to angle of

emergence and the latter is equal to 3/4th of angle of prism. Calculate the angle of deviation. 5. Two nearby narrow slits are illuminated by a single monochromatic source. Name the pattern obtained on the

screen. One of the slits is now completely covered. What is the name of the pattern now obtained on the screen. Write any one difference between the patterns obtained in the two cases. Also, draw intensity distribution curve in each case.

6. (a)What is the effect on the interference fringes in a Young’s double-slit experiment due to each of the following operations:

(i) the screen is moved away from the plane of the slits; (ii) the (monochromatic) source is replaced by another (monochromatic) source of shorter wavelength; (iii) the separation between the two slits is increased.

(b) how would the angular width of central maximum change if (i) monochromatic yellow light is replaced with red light (ii)distance between the slit and screen is increased.

7. Which of the following can be polarized : (i) X-rays and (ii) Sound waves. Two polaroids are used to study polarization , one of them is kept fixed and other is initially kept with its axis

parallel to the polarizer axis . The analyzer is then rotated through angles of 45,90 and 180 degrees in turn. Show graphically the intensity variation. Green light is incident at the polarizing angle on a certain glass plate. The angle of refraction is 32 0 .

What are:(i) the polarizing angle, and(ii) the index of refraction of glass.

UNIT- VII (DUAL NATURE OF MATTER AND RADIATION):

1. Photo electric effect- Definition of work function and threshold frequency , laws of photoelectric emission ( with necessary graphs)2. Einstein’s photo electric equation and its application to determine Plank’s constant and work function 3. De-broglie’s Hypothesis ( De-briglie’s wave length)4. Davission and Germer experiment

QUESTIONS:

01 The de Broglie wavelengths, associated with a proton and a neutron, are found to be equal. Which of the two has a higher value for kinetic energy?

02 Ultraviolet radiations of different frequencies ν1 and ν2 are incident on two photosensitive materials having work functions W1 and W2 (W1 > W2) respectively. The kinetic energy of the emitted electrons is same in both the cases. Which one of the two radiations will be of higher frequency?

03 In an experiment on photo electric effect, the following graphs were obtained between the photoelectric current (I) and anode potential (V). Name the characteristic of the incident radiation that was kept constant in this experiment.

I

O V04 Two metals A and B have work functions 4 eV and 10 eV respectively. Which metal has lower

threshold frequency?05 If the maximum kinetic energy of the electrons emitted by a photo cell is 4 eV, what is the

stopping potential?06 Two metals A and B have work functions 2 eV and 5 eV respectively. Which metal has lower

threshold wavelength?07 Name the device that converts the changes in intensity of illumination into changes in electric

current. Give three applications of this device.08 Calculate the ratio of de-Broglie wavelengths associated with a deuteron moving with a

velocity 2v and an alpha particle moving with a velocity v 09 Compare the energy of an electron of de – Broglie wavelength 1Å with that of an X –ray

photon of the same wavelength.10 On the basis of photon theory, obtain Einstein’s photoelectric equation. Use this relation to

show that there must be a threshold frequency for each photo sensitive surface. What is the relation between stopping potential and maximum kinetic energy of photoelectrons?

11 The given graphs show the variation of photo electric current (I) with the applied voltage (V) for two different materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but same intensity of incident radiations.

12 An electron, an alpha-particle and a proton have the same kinetic energy. Which one of these particles has the largest de-Broglie wave length ? 2

13 Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of incident radiations. 2

14

54V

50o

Name the experiment for which the above graph, showing the intensity of the scattered electrons with angle of scattering, was obtained. Also name the important hypothesis that was confirmed by this experiment.

15 Write the name given to the frequency νc, in the following graph , showing the variation of the stopping potential (Vo) with the frequency (ν) of the incident radiation, for a given photosensitive material. Also name the constant for the photo sensitive material obtained by multiplying νc with Planck’s constant.

Vo

O νc ν 2

16 de Broglie wave length associated with an electron accelerated through a potential difference V is λ. What will be its wavelength when the accelerating potential is increased to 4V?

17 Ultraviolet radiations are incident on two photosensitive materials having work functions W1

and W2 (W1 > W2) respectively. In which will the kinetic energy of the emitted electrons be greater? Why?

UNIT-VIII (ATOMS AND NUCLEI):

1. Radius and energy of electron in the nth orbit (hydrogen atom) using Bohr’s postulates , energy level diagram , excitation energy/potential and ionization energy/potential 2. Mass defect and binding energy – B.E / A curve and numerical for B.E /A3. Laws of radioactive decay ( alpha, beta and gamma decay), decay law /decay equation.4. Decay constant, half life and mean life- definition and relation between them5. concept about the variation of radius and density of nucleus with mass number

QUESTIONS:

01 Write the nuclear decay process for β – decay of 15P32

02 State two characteristic properties of nuclear force.03 Define the term ‘activity’ of a radio nuclide. Write its S I unit.04 Explain with an example, whether the neutron – proton ratio in a nucleus increases or

decreases due to β – decay. 105 The wavelengths of some of the spectral lines obtained in hydrogen spectrum are 9546 Ả,

6463Ả, and 1216Ả. Which one of these wavelengths belongs to Lyman series? 106 Some scientists have predicted that a global nuclear war on earth would be followed by a

‘nuclear winter’. What would cause the ‘nuclear winter’? 107 The total energy of an electron in the first excited state of hydrogen atom is about – 3.4 eV.

What are the values of potential energy and kinetic energy of the electron in this state? 108 Energy of an electron in the nth orbit of hydrogen atom is given by the equation En = - (13.6/n2) eV. How much energy is required to take an electron from the ground state to

the first excited state? 209 Write four characteristics of nuclear forces. 210 Calculate the energy released in MeV in the following nuclear reaction:

92 U 238→ 90 Th 234 + 2 He4 + Q [Mass of 92 U 238 = 238.05079 u, mass of

90 Th 234 = 234.043630 u, mass of 2 He4 = 4.002600 u, 1 u = 931 MeV/c2] 211 Draw a graph showing the variation of potential energy between a pair of nucleons as a

function of their separation. Indicate the regions in which the nuclear force is (i) attractive and (ii) repulsive. 2

12 You are given two nuclei – 3 X 7 and 3 Y 4. Are they isotopes of same element? State with reason, which one of the two nuclei is likely to be more stable. 2

13 The ground state energy of hydrogen atom is - 13.6 eV. (i) What does the negative sign signify?(ii) How much energy is required to take an electron from the ground state to the 1st

excited state, of this atom? 214 A radioactive nucleus a decays as follows.

+1e0 α A A1 A2

If the mass number and atomic number of A2 are 176 and 71 respectively, what are the mass numbers and atomic numbers of A and A1? Which of these three are isobars?

15 Prove that the radius of the nth Bohr orbit of an atom is directly proportional to n2, where n is the principal quantum number. 2

16 Calculate the half life of a radioactive substance if its activity drops to (1/16)th of its initial value in 30 years. 2

17 The sequence of stepwise decays of a 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 answers. 2

18 The sequence of stepwise decays of a radioactive nucleus is;0n1

α D D1 D2

If the mass number and atomic number for D2 are 176 and 71 respectively, find the mass number and atomic number of D. Amongst D, D1 and D2, do we have any isobars or isotopes?

UNIT- IX (ELECTRONICS):

1. Concept of p- type and n-type semiconductors & their energy level diagram on the basis of bands theory’2, P-N junction- Characteristic curve in forward and reverse biased - Depletion layer and barrier potential and their variation in forward and Reverse bias3. Zener diode as a voltage regulator ( including characteristic curve)4. Junction diode as a full wave rectifier 5. Transistor as an amplifier – Phase reversal , expression for current / voltage/ power gain and transconductance6. Transistor as an oscillator7. Logic gates

QUESTIONS:

01 Draw the voltage – current characteristics of a zener diode. 102 Name the type of biasing of a p – n junction diode so that the junction offers very high

resistance. 103 Name one impurity each, which when added to pure silicon, produces (i) n – type and (ii) p –

type semiconductor. 104 Draw energy band diagram of a p – type semiconductor. 105 State the reason why GaAs is most commonly used in making of solar cell. 106 State two reasons why a common emitter amplifier is preferred over a common base amplifier.

107 What will be the values of input A and B for the Boolean expression,

( A + B ) . ( A.B) = 1? 1

08 How is a sample of an n-type semiconductor electrically neutral though it has an excess of negative charge carriers? 1

09 How is the band gap, Eg, of a photo diode related to the maximum wavelength, λm , that can be detected by it? 1

10 Zener diodes have higher dopant densities as compared to ordinary p-n junction diodes. How does it affect the (i) Width of the depletion layer? (ii) Junction field?

111 An unknown input (A) and the input (B) shown below, are used as the two inputs in a NAND

gate. The output Y, has the form shown below. Identify the intervals over which the input ‘A’ must be ‘low’.

112 How does the width of the depletion region of a p – n junction vary, if the reverse bias applied

to it decreases. 113 Draw the voltage – current characteristics of a zener diode. 114 Write the full forms of the terms (i) MSI, (ii) SSI and (iii) LSI used for different types of

integrated circuits. 115 Explain how the width of the depletion layer in a p-n junction diode changes when the junction

is (i) forward biased and (ii) reverse biased. 216 Draw a labelled circuit diagram to show the use of zener diode as a voltage regulator.

217 The output of a 2-input NOR gate is fed to a NOT gate. Draw the logic circuit of this

combination of gates and write the truth table for the output of the combination for all inputs. (i) attractive, (ii) repulsive. 2

18 The output of an unregulated d. c. power supply needs to be regulated. Name the device that can be used for this purpose and draw the relevant circuit diagram. 2

19 Give the logic symbol for an OR gate. Draw the output waveform for the input waveforms A and B for this gate.

A

Inputs

B 220 Draw the energy level diagrams for (i) n – type & (ii) p – type semiconductors. Mark the donor

and acceptor levels in it. 2

21 Draw and explain the output waveform across the load resistor R, if the input waveform is as shown in the given figure.

+5V

D

0 R -5V 2

UNIT- X (COMMUNICATION SYSTEM): 1. Block diagram of basic communication system 2. Modulation and its need 3. Production and detection of A.M. waves (block diagram)4. Advantages /disadvantages of AM over FM5. Propagation of E.M Waves – Ground /Sky/Space wave propagation (Expression for LOS distance)6. Numerical problems

QUESTIONS:For questions refer to NCERT text book exercise