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Laboratory Manual/Physics (EC) Department/IT/NU

Fundamental Units and Dimensions

Aim: To analyze fundamental units and dimensional formulas.

Introduction:

Physics is a science of exact measurement. In mechanics, the quantities mass, length and time are chosen as fundamental quantities because these quantities cannot be expressed in terms of any other more fundamental quantity.

A fundamental quantity is a quantity which cannot be expressed in terms of any other physical quantity. The units in which these quantities are measured are called fundamental units.

As there are mainly two types of physical quantities: Fundamental and Derived there are two types of units.

Fundamental Units: These types of units can be defined without any help of other units.

Derived units: These units are defined with the help of fundamental units and therefore are called derived units.

Fundamental Units: (IN SI System)

Sr. no.

Physical Quantity Unit Symbol

1. Length Meter M2. Mass Kilogram Kg3. Time Second S4. Temperature Kelvin K5. Quantity of Matter Mole Mol6. Electric Current Ampere A7. Luminous Intensity Candela Cd


Examples of Derived Units and Their Dimensional formulas:

Sr. No. Physical Quantity Formula Dimensional

formula1 Length -- [M0L1T0]2 Area Length x Length [M0L2T0]3 Volume Length x Length x Length [M0L3T0]4 Density Mass / Volume [M1L-3T0]

5 Relative Density or Specific Gravity

Density of Substance/ Density of water

Dimensionless Quantity

6 Velocity Distance / Time [M0L1T-1]7 Acceleration Velocity / Time [M0L1T-2]8 Force Mass x Acceleration [M1L1T-2]

9 Weight Mass x Acceleration due to gravity [M1L1T-2]

10 Tension Force [M1L1T-2]11 Thrust Normal Force [M1L1T-2]12 Pressure Force / Area [M1L-1T-2]13 Momentum Mass x Velocity [M1L1T-1]14 Impulse of Force Force x Time [M1L1T-1]

15 Moment of force or Torque Force x Distance form the point [M1L2T-2]

16 Work Force x displacement [M1L2T-2]17 Kinetic Energy (½) Mass x Velocity2 [M1L2T-2]18 Potential Energy M x g x h [M1L2T-2]

19 Angle or angular displacement Arc / Radius = L / L No Dimension

20 Angular Velocity Angular Displacement / Time [M0L0T-1]

21 Angular Acceleration Change in Angular Velocity / Time [M0L0T-2]

22 Angular Momentum Mass x Velocity x Distance [M1L2T-1]23 Angular Impulse Torque x Time [M1L2T-1]

24 Universal Gravitational Constant F x R2 / M1M2 [M-1L3T-2]

25 Intensity of Electric field Force / Charge [A-1M1L1T-3]

26 Electric Potential Work / Charge [A-1M1L2T-3]

27 Dielectric constant No Dimension

28 Electric Dipole Moment Charge x Distance [A1M0L1T1]

29 Electrical Resistance Voltage / current [A-2M1L2T-3]30 Potential energy of the (1/2) C x V2 [A0M1L2T-2]


Capacitor

Quiz

Sr. No. Physical Quantity Formula Dimensional

formula

1. Gravitational Potential

2. Moment of Inertia

3. Current Density

4. Capacitance

5. Specific resistance

6. Electrical Conductivity

7. Intensity of Magnetic field

8. Magnetic Flux

9. Surface Tension

10. Force Constant

11. Coefficient of Viscosity

12. Stress

13. Self Induction

14. Mutual Induction

15. Permeability of Vacuum

16.

17. Planck’s Constant

18. Charge


19. Potential Gradient

Date: Sign of Faculty:

Solar Cell CharacteristicsAim: Estimation of the solar energy in terms of solar power and V-I characteristics,

Power load characteristics of the solar cell.

Apparatus:

1. Solar cell mounted

2. Table lamp 100 W

3. Experimental set up board

Theory:

A solar cell is basically a P-N junction which converts solar energy into electrical

energy. Such conversion of optical energy into the electrical energy is known as

photovoltaic effect and therefore solar cell is also called as photovoltaic cell. Solar cells

are composed of various semi conducting materials. Over 95% of all the solar cells

produced worldwide are composed of the semiconductor material Silicon (Si). To

produce a solar cell, the semiconductor is "doped" with which one can obtain a surplus

of either positive charge carriers (p-conducting semiconductor layer) or negative charge

carriers (n-conducting semiconductor layer) from the semiconductor material. When the

solar cell is illuminated, the photons incident on the cell generate electron hole pairs. At

the junction, the barrier field separates the positive and negative charge carriers. Under

the effect of electric field, the electrons (minority carriers) from p side are swept in to n

region. A similar process occurs to holes in n region. This leads to an increase in the

number of holes in p- region and of electrons in n- region. Hence, such accumulation of

charges on the both sides of the junction produces an emf which is called a photo emf,

which is also known as open circuit voltage and is proportional to the illumination

(mW/cm2 or lumen/cm2) and the size of the illuminated area. When an external circuit is

connected across the solar cell terminals, the minority carriers return to their original

sides through the external circuit, causing the current to flow through circuit. Thus, the

solar cell is behaves like a battery with n- side as the negative terminal and p-side as a

positive terminal. The photo emf can be measured with a voltmeter.


The output (product of current and voltage) of a solar cell is temperature

dependent. Higher cell temperatures lead to lower output, and hence to lower efficiency.

The level of efficiency indicates how much of the radiated quantity of light is converted

into useable electrical energy. About 24% efficiency is achieved at the lab level with the

help of Monocrystalline Silicon. Following steps may be helpful for higher efficiency of

the solar cell panel:

Construction of the cell surface in a pyramid structure, so that incoming light hits

the surface several times. New material: for example, gallium arsenide (GaAs),

cadmium telluride (CdTe) or copper indium selenide (CuInSe²).

in order to be able to use a wide spectrum of radiation, different semiconductor

materials, which are suited for different spectral ranges, will be arranged one on

top of the other.

A higher light intensity will be focused on the solar cells by the use of mirror and

lens systems. This system tracks the sun, always using direct radiation.

1. Illumination Characteristics:

Procedure:

1.1 Make the circuit as shown in Fig. 1.

1.2 Allow the light from a 100 W bulb on the solar panel. Keep the intensity control

at its minimum and make the lamp on.


1.3 Keep the load at 0 Ohm and note down the short circuit current.

1.4 Vary the intensity of lamp in steps and note down the current at each setting. Note

down the readings in the given table 1.

1.5 Draw the graph between current versus intensity (in terms of lamp voltage) as

shown in the Fig. 2.

Observations: Table -1. Illumination Characteristics

RL=0 OhmSr. No. Intensity (DC Volts) Current (DC mA)

12

345

67

8

9

10

2. Current – Voltage Characteristics:

Procedure:

2.1 Keep the intensity of lamp at its minimum, say I1. Make the load resistance RL =0.


Note the open circuit voltage and short circuit current in the tabulation 2.

2.2 Now vary the load dial in steps of 100 Ohms up to 900 Ohms and note down the

current and voltage readings in the table 2.

2.3 Repeat the above step for high lamp intensity, say for I2 and furthermore

for I3 and note down the readings in table 2.

2.4 Plot the graph for Current versus voltage for maximum intensity I3 and minimum

intensity I1 as shown in Fig. 3.

Observations: Table -2 Current – Voltage Characteristics

Sr.

No.

RL

Ohms

Intensity I1=50 Intensity I2=100 Intensity I3=150

V

(Volts)

I

(mA)

V

(Volts)

I

(mA)

V

(Volts)

I

(mA)

1 0

2 100

3 200

4 300

5 400

6 500

7 600

8 700

9 800

10 900

11 ∞


Graph:

Result:

Sr. No. QuantityAt following different Intensities

I1 I2 I3

1 Open Circuit Voltage (RL=∞) ______ Volts ______ Volts ______ Volts

2 Short Circuit Current (RL=0) ______ mA ______ mA ______ mA

3. Power – Load characteristics:

Procedure:

3.1 Calculate the output power using the sets of readings of voltage and current of

table 2 for the intensities I1, I2 and I3 and enter into the table 3.


3.2 Plot the graph of power versus load as shown in the fig. 4 and determine the value

of optimum load that draws the maximum power from the panel.

Observations: Table -3 Power – Load Characteristics

Sr. No. RL Ohms

Power for

Intensity I1

(mW)

Power for

Intensity I2

(mW)

Power for

Intensity I3

(mW)

1 0

2 100

3 200

4 300

5 400

6 500

7 600

8 700

9 800

10 900

11 ∞

Result:

Sr. No. QuantityAt following different Intensities

I1 I2 I3

1 Max. Power Pmax

------mW ------mW ------mW


2 RL at Pmax ------Ohms -------Ohms -------Ohms

The Maximum Power Output is found to be ________ mW.

Graph:

Quiz:

1. What is solar cell? Give its applications.

2. Give the construction of solar cell penal with proper schematic diagram.

3. On which parameters, the efficiency of the solar cell penal depends?

4. What are the different ways to improve the efficiency of the solar cell?

5. Write a note on the contribution of Paul M Dirac in Physics.



e/m FOR AN ELECTRON BY THOMSON METHOD

Aim: Evaluating the charge to mass ratio for electron by applying perpendicular magnetic field on the electron beam in CRT

Apparatus: Magnetometer wit wooden bench, two bar magnets, cathode ray tube \ (CRT), Power supply.

Procedure:

The wooden frame & the wooden bench are coupled together as shown in figure .A magnetometer is kept on the wooden bench such that 90 -90 diameter of its circular scale is parallel to the length of the bench

The entire setup is oriented such that the pointer of the magnetometer coincides with 0-0 markings. This setting ensures tat when CRT is placed instead of the component of the earth’s magnetic field will be ineffective to influence the elec-tron beam. Once the position is adjusted. It is undisturbed till the experiment is completed.

The wooden bench is removed and the CRT is introduced in its place as shown in figure, the CRT is connected to the power supply. The power supply is switched on adjusting the intensity and focus controls a spot of sufficient brightness is ob-tained using X- and Y- shift controls it is brought to the center of screen.

Now apply a suitable deflecting voltage so that the lamination spot is deflected by about 0.5to 1.0 cm. Note the deflecting voltage V and position of the spot. Mea-sure the distance through which the spot has moved and let it be D.


Place the bar magnets symmetrically on either side of the cathode ray tube along the arms of the wooden stand on which the tube is fitted such that their opposite poles face each other and their common axis is exactly at right angles o the axis of the cathode ray tube adjust the polarity as well as the distance of the magnets so that the luminous spot comes back tots initial position. When the adjustment is perfect note the distance of the poles of the magnets on the side nearer to the cath-ode ray tube let the distance be r1 and r2.

Remove the bar magnets switch off the electric field off the electric field applied to the deflecting plates and again note the initial position of the luminous spot. Reverse the polarity of the potential difference applied to the luminous spot and calculate D.

Again place the bar magnets on the arms of the wooden stand as in the previous step and adjust their polarity as well as distance so that the spot comes back to its initial position. When the adjustment is perfect again note the distance of the poles of the magnets on the side nearer to the cathode ray tube let the distance be r1’ and r2’ switch off power supply.

To find the value of the magnetic field, carefully remove the magnets and the cathode ray tube from the wooden stand place the compass box such that its cen-ter lies exactly on the point where the common axis of the bar magnets and the axis of the cathode ray tube intersect. Rotate the compass box about its vertical axis so that the pointer lies along the 0-0 lines.

Place the magnets exactly in the same positions as in steps 5 an distance r 1 and r2. This produces a deflection in the magnetometer compass box and the two ends of the pointer deflection let the readings be θ1and θ2

Now place the magnets exactly in the same position as in step 6 at distance r 1 and r2 and note the deflection θ1’and θ2’ from the two ends of the pointer of the compass box. The mean of these four deflection θ1, θ2, θ1’and θ2’ gives the mean deflection θ if BH is the horizontal component of earth’s magnetic field then,

Where BH is the earth’s horizontal component of the field. It is different at the different place.

Take two more sets of observations by changing the values of B and hence that of electric field.

Observations:


1. Length of deflecting plates (l) = 0.045 meters

2. Separation between deflecting plates (d) = 0.0245 meters

3. Distance of the screen from the edges of the y-plate (L) = 0.125 meter

4. Horizontal component of earth’s magnetic field BH = 0.36×10-4 Tesla


Sr.

No.

Applied Voltage

V

Direct field Reversed Field

Position of Spot Magnetic pole and Distance Position of Spot Magnetic pole and

Distance

Initial Final Deflection D in meter Pole r1 Pole r2 Initial Final Deflection D

in meter Pole r'1 Pole r’2

1 V1 = D1 = D1 =

Observations Table:

For determination of B

Sr.

No.

Reading of two ends of pointer

when magnetic poles are at r1 & r2

Reading of two ends of pointer

when magnetic poles are at r1’ & r2’Mean θ

θ1 θ2 θ1’ θ2’

1

Laboratory Manual of Physics (PY101)

Calculation:

______________ C Kg-1

Quiz:

1. Write brief note on contribution in physics by Sir Thomas Alva Edison.

2. Draw diagram of CRT and discuss about its main parts.

3. Why CRT is to be highly evacuated?

4. Why do we need to orient entire set up in magnetic meridian before performing ex-periment?

5. In this experiment, the accuracy in the value of e/m largely depends upon which fac-tor? Why?

6. Can we use single bar magnet? Justify your answer.

7. The magnetic needle in the magnetometer aligns in which direction? Write answer in both polar and geographical terms.

Date: Sign of Faculty

15


emf GENERATION BY DYNAMIC MAGNETIC FIELD

Aim: To measure the electro motive force by dynamic magnetic field and verification of Faraday’s law

Theory:

The basic principle of generation of alternating e.m.f is electromagnetic induction was discovered by Michael Faraday. This phenomenon is the production of an induced e.m.f in a circuit caused by change of the magnetic flux linking the circuit. Faraday’s laws of induction tells us that the induced e.m.f is given by,

It represents the rate of change of flux linking the circuit. In MKS unit ε will be in volts, the flux will be in Weber and t in second.

APPARTUS:

In this experiment, we are using OMEGA TYPE ES_232 kit. It consists of a permanent magnet mounted on an arc of a semi-circle of radius 40 cm. The arc is a part of rigid frame of aluminum and I suspended at the center of the arc so that the whole frame can oscillate freely in its plane. (See figure 1).

Weights have been provided on the diagonal arm so that by alternating their positions the time period of the oscillations can be varied from about 1.5 to 3 seconds. Two coils of about

16

Laboratory Manual of Physics (PY101) 10,000 turns of copper wire loop the arc so that the magnet can pass freely through the coil. The two coils are independent and can be connected either in series or in parallel. The amplitude of the swing can be read from the graduated circle by pointer. On the panel of the measurement board one voltmeter with four ranges 0-2.5 V, 5 V, 10 V and 20 V and one ammeter with two ranges 1 mA, 2.5 mA are provided. Board also has four different valued condensers, five resistances and two diodes and one SPST switch for performing experiments.

Magnet N-S passes through a coil C (by using coil 1 or 2) with varying velocities (See figure 2). A Aluminum frame can be swing about a pivot O, it’s period adjustable sliding the loads A, A. if D is released from an angle from the equilibrium, the velocity Vmax with which the magnet passes the coil is given by,

P is the distance between the arc center and the weight. The magnetic flux through the coil changes as the magnet NS passes through it.

Procedure:1. Connect the circuit as shown in figure 3. Keep the switch S in off position. Here R1 is

the internal resistance of the coil and the forward resistances of the coil and the forward resistance of the diode. Capacitor is of 100 F. At each swing the diode permits the capacitor to gain a charge once. The charging time RC = 50ms, the capacitor reaches the value in a few swings.

2. When the Milli-ammeter shows no more kicks, it means the capacitor C has reached the potential E0. Measure this potential by changing switch on position and take reading on voltmeter.3. Vary Vmax by choosing different values and measure E0 each time. 4. Plot a graph of E0 versus Vmax and observe it is linear.5. The ideal graph given below shows the results of an experiment.

17


Observation Table:

Sr. No.

Angle Induced EMF Vmax cm/Sec

Quiz:

1. Define electromagnetic induction.

2. Write all three statement of Faraday’s law for electromagnetic induction.

3. Write integral and differential form of Faraday’s law.

4. How can you say that AC emf is induced in the coil?

5. What is the use of a capacitor and diode in this experiment?

6. What is the practical use of Faraday’s law? List out five of them.

7. Write a brief note on contribution in physics by Dr. Vikram Sarabhai.


STUDY OF TRANSFORMER

18


Aim: Energy Efficiency measurement of a power transformer.

Apparatus : OMEGA TYPE ETB-110.

Features of the Board:

The board consists of following built-in parts:

1. Main transformer with primary 230 V at 50 HZ and secondary out put at 0-10 V A.C. at 1 ampere.

2. AC/DC voltmeter, to measure the mains input to the transformer of 0-250 V.

3. AC/DC milli-ammeter, to measure the primary circuit current of 0-100 mA.

4. AC/DC voltmeter, to measure the secondary voltage of 0-15 V.

5. AC/DC Ammeter, to measure the secondary circuit current of 0-1Amp.

6. High Wattage rating Potentiometer for transformer’s primary voltage variations.

7. High Wattage rating Potentiometer to be used as a load for the secondary circuit.

Theory:

When an alternating current passes through a conductor, the magnetic field is produced across it. The direction of the field is shown in figure 1.

When a varying DC current passes through an inductor, the magnetic field lines are produced around it. It is shown in fig. 2.

19

Laboratory Manual of Physics (PY101) When an alternating current passes through an inductor, the magnetic flux is

produced around the inductor and the direction of the magnetic field changes after each half cycle of the applied signal. This is shown in fig. 3.

When an alternating current is flowing in one coil it induces voltage in the neighboring coil. The changing magnetic field lines in one coil cuts the turns of the other coil and induce a changing voltage in each of the turn in the neighboring coil. Transformer works on this principle. The transformer serves many functions. It enables the transform of electrical energy from one electrical circuit to another by using changing magnetic field lines as the linked between the two. In this way it behaves as coupling device. Also, it provides a mean by which an alternating voltage of given amount can be changed to higher or lower amount. Transformer is a electro-magnetic circuit consisting of two windings namely primary and secondary. They are the bobbins on which winding is done and then are laminated. A practical transformer wound on the tack of E and I shape strips of iron core are shown in fig. 4.

The alternating emf, which is to be transformed, is applied to the primary coil. The alternating current in the primary gives rise to flux change. This change of flux induces e.m.f in the secondary coil by mutual coupling and also produces self-induced e.m.f in the primary. Evidently, e.m.f in the secondary depends on the following factors.

1. The number of turns on the primary and secondary coil.

2. The e.m.f applied across the primary.

3. The co-efficient of mutual induction.

4. The co-efficient of self-induction of primary and secondary coils.

20

Laboratory Manual of Physics (PY101) 5. The Ohmic resistance of the coils & the frequency of the alternating current.

Voltage Transformer:

If Ep were the applied e.m.f across the primary of turns Np, the e.m.f induced in the secondary of turns Ns is given by,

Here equation 1 gives the transformer ratio. If Es › Ep. The transformer is called step up

transformer. If Ep › Es then the transformer is called as step down transformer. These are shown in figure 5.

Current Transformer:

We know that the impendence in an AC circuit of resistance Rs, self-inductance Ls and AC of angular frequency w is given by,

Let Rs and Ls stand for the secondary of the transformer. Now if the secondary is made of thick copper wires or strips Rs ‹‹ w Ls. A necessary condition for current transformer ratio,

Meaning the current transformer is inversely related to the number of turns.

21

Laboratory Manual of Physics (PY101) Effects of secondary:

It may be shown that the resistance of the primary increases because of the induced currents in the secondary. Thus the effective resistance of primary in the presence of the secondary becomes,

Similarly the effective inductance is given by,

Where Rp and Rs are the resistances of the primary and secondary coils. M and Ls are the co-efficient of mutual induction and the self-induction of secondary. The above-mentioned discussion is applied to AC circuits.

Efficiency of transformer:

The efficiency of a transformer is defined as the ratio of output energy from the secondary to the input in the primary,

(6)

Copper Loss:

When the current passes through the primary coil, a part of the electrical energy is converted into heat, this loss is due to the resistance of the copper coil. Therefore it is called copper loss. The secondary reflects an additional resistance in the primary; therefore the current in the secondary must be kept constant for copper loss determination.

Iron Loss:

Iron losses occur in transformer because of (i) Hysteresis and (ii) eddy currents. To minimize these losses a laminated core of suitable iron is used.

22


Procedure:

Connect the circuit as shown in fig.-6. For this, connect the respective terminal marked link in fig. 7.

1. Switch on the supply.2. Set the position of pots VR1 and VR2 anywhere.3. Measure the primary and secondary voltages and currents.4. Calculate the transformer ratio by equation 1.5. Calculate the efficiency using relation 6.6. Calculate the copper loss using this equation: Copper loss = Ep x Cp watts7. Change the position of VR1 and VR2 and repeat the step (4) above and calculate

transformation ratio, Efficiency and copper loss.

Observation Table:

Sr. No.

Primary Voltage

Ep

Primary Current

Cp

Secondary Voltage

Es

Secondary Current

Cs

Ep / Es=Np / Ns

Efficiency

1 200 V2 190 V3 180 V4 170 V5 160 V6 150 V7 140 V8 130 V

Quiz:

1. What is the principal of working of transformer?2. Can we use DC current in this experiment?3. Why do we use iron core? Which type of material it is?4. What is copper loss? How can we reduce the copper loss in the circuit?5. What is hysteresis loss? Define retaintivity and coercivity.6. Write a short note on uses of eddy current.7. What are the practical uses of transformer? List at least five of them. 8. Write brief note on Dr. APJ Abdul Kalam.


23


RESITIVITY MEASUREMENT BY FOUR POINT PROBE METHOD

Aim: To measure the resistivity of semiconductor by four point probe method at different temperature.

Apparatus: Omega type ES-246 (The instrument is absolutely self contained and requires no other apparatus.)

Theory:

The properties of the bulk materials used for the fabrication of transistors and other semiconductors devices are essential in determining the characteristics of the complete devices. Resistivity and lifetime measurements are generally made of germanium crystal to determine their suitability. The resistivity, in particular must be measured accurately since its value is critical in many devices. In solids, atoms are usually arranged in a regular way to achieve dense packing and thereby form a crystal. The arrangement has very desirable characteristics; i.e. the transport of holes and free electrons is very smooth in these electrons. As a result of interaction between the tremendous numbers of atoms in crystal, the energy level found in isolated atoms will be spilt and form bands of allowed energies which contain almost continuum levels. The unoccupied energy levels also split up and form another band called conduction band. The interaction between the unused shells is very large and they spread widely. Therefore, while there is a band gap, Eg between the valance and conduction band, splitting of higher orbit is so wide that they usually overlap.

According theory, an insulator must have a large band gap, so that at room temperature the conduction band is completely empty and valance band is practically filled and semiconductors must have narrow band gaps so that appreciable number of carriers is present in the valance and conduction band at room temperature.

Four Point Probe Method:

It permits the measurements of resistivity in samples having a wide variety of shapes, including the resistivity of small volumes within bigger pieces of semiconductor. In this manner the resistivity of both sides of p-n junction can be determined with good accuracy before the material is cut into bars for making devices.

Procedure:

Put the sample on the base plate of the four probe arrangement. Unscrew the pipe holding the four probes and let the four probes rest in the middle of the sample.

24

Laboratory Manual of Physics (PY101) Connect the outer pairs of probes (green leads to the constant current power supply) and the

inner pair (red labels) to the probe voltage terminals. Place the four probe arrangement in the oven and fix the thermometer in the oven through

the hole provided. Switch on the AC mains of four probe set up and put the digital panel meter in the current

measuring mode through the selector switch. In this position LED facing mA would glow. Adjust the current to a desired value (say 5 mA).

Now put the digital panel meter in voltage measuring mode. In this position LED facing mV of V would glow depending on position of switch and the meter would read the voltage between the probes.

Connect the oven power supply. Rate of heating may be selected wit the help of S switch- Low or High as desired. Switch on the power to the oven. Te glowing LED indicated power to the oven is ON.

Observations:

1. Current I = 5 mA (Constant)

2. Width of the chip (W) = 0.05 cm

3. Distance between two probes (S) = 0.2 cm

4. G7 (W / S ) = = 5.546

25

Figure 6.1: Measurement of resistivity using Four Point Probe Method

Laboratory Manual of Physics (PY101) Formulas:

1. To calculate ρ use the following equation:

Where,

2. The Energy Band Gap= =__________ eV

where, k= Boltzmann constant= 1.38×10-23 J/˚K loge ρ

1/T (K-1)

Observation Table:

Sr. No.

TemperatureVoltage

(Volt)

ρ0

(Ω cm)

ρ

(Ω cm)logeρ

1/T(K-1)

(ºC) (ºK)

1

2

3

4

5

6

7

8

26

Laboratory Manual of Physics (PY101) Quiz:

1. Write a working principle of four point probe method.

2. How will u distinguish between insulator, conductor and semiconductor?

3. Draw the band structure of insulator, conductor and semiconductor.

4. What are the requirements to measure resistivity accurately using this method?

5. Can we perform the experiment using only two probes only? Why?

6. What is physical significance of band Gap?

7. Write a microscopic form of Ohms’ law.

8. Which type of material has higher conductivity- P type or N-type? Justify your answer.

9. What are the advantages of four probe method over the conventional method used to measure the resistivity of semiconductor?

10. Write brief not on contribution in physics by Prof. C V Raman.


27


Band Gap Determination in Semiconductor diode

Aim: Determination of forbidden energy band gap in a semiconductor using a junction

diode.

Apparatus: An experimental board consists of digital micro-ammeter, oven, semiconductor

diode, and thermometer.

Theory:

In a material, the energy difference between its non-conductive state (valance band) and its

conductive state (conduction band) is referred to as energy band gap or forbidden gap. In

other words, the band gap is the energy it takes to move electrons from the valence band to

the conduction band. The existence of energy band gap is successfully explained by Band

theory of solids using Linear Combination of Atomic Orbital (LCAO).

Procedure:

1. Insert the thermometer in the hole of an oven and switch on the power supply.

2. Insert the two leads to the diode in the socket i.e. insert red plug in +ve socket and

black plug in – ve socket as shown in the fig 1.

3. Now turn the oven on and allow the temperature to rise up to 75˚ C only.

4. As soon as the temperature reaches to 75˚ C, switch off the oven and let stabilize the

temperature around 70˚ C.

5. Now start to take the readings of temperature by taking current as reference i.e. for

each step of 10 µA of current reduction, note down the readings of temperature

from the thermometer in table 1.

6. Plot the graph of log Is versus 103/T as shown in the fig. 2 and determine the slope

of the line.

28


Fig. 1 The practical circuit diagram

29


Observation Table-1.

Sr. No.

Reverse Saturation Current iS

(µA)

Temperature 1/T (˚K-1)

(×103)Log is˚C ˚K

12345678910

Graph:

Determine the slope of the line and find the slope of the line to calculate the band gap as

= --------- eV

From any point C on the line draw a perpendicular projection on the x-axis and y-axis. Slope of the graph is CD/CE.

Fig 2 Graph of log is versus 1/T

30


1. What is the criterion for a material for possessing semi conducting properties?

2. Give the fundamental Classification of a solid based on energy band gap.

3. What do you mean by energy band gap? What is the reason behind its existence?

4. Give the categories of semiconductor based on energy band gap. Define each.

5. What are the other methods to measure the energy band gap of semiconductor?

6. What are the advantages and disadvantages (if any) of energy band gap of a

semiconductor?

7. Write brief note on contribution in physics by Max Plank


31


MEASUREMENT OF THIN FILM AND WAVELENGTH OF MONOCHROMATIC

LIGHT

Aim: To measure the wavelength of light from sodium vapor lamp and find the thickness of thin film using Newton’s rings method.

Apparatus: Newton’s ring apparatus, wooden reflector, traveling microscope, sodium Light source, lens.

Theory:

The formation of Newton’s rings is due to the interference of light waves reflected from the upper and lower surface of the air film enclosed between the lens and the plan glass plate. The thickness of the air film at the point of the contact is zero and increases as we move from the point of contact towards the periphery of the lens. The thickness of the air film will be uniform for all points lying on the circle with centre at the point of contact. Thus using reflected light, the central point will be dark surrounded by bright circles separated by dark rings.

If R is the radius of curvature of the lens and t is the thickness of the air film at A or B, the necessary condition for bright fringes using reflected light is

Where, n = 0,1,2,……….. Since θ is very small cos θ=1 ( θ is angle of refraction).

Here for air μ=1.

And for dark fringes,

Now, from the geometry of figure,

(EB) (AE) = (EO) (ED)

Here, EB=AE= r = the radius of interference fringe & EO= BQ = t = thickness of the film

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On neglecting t2 in comparison to 2Rt,

Therefore, for bright fringes,

or

and for dark fringes,

When n=0, the radius of dark fringes is zero and the bright fringe is . Therefore, alternately dark and bright fringes are produced with dark centre. Suppose the diameter of n th

dark fringe be Dn then,

and the diameter of (n+p)th ring be Dn+p , then

M- MicroscopeG- Glass PlateL- Convex LensP -Plano convex LensE – Glass PlateS - Source

Here, R is the radius of curvature of Plano convex lens and p is the interval of fringes.

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Description:

The optical arrangement for Newton’s ring is shown in figure. Light from a monochromatic source (Sodium lamp) is allowed to fall on convex lens through a broad slit. Now it falls on a glass plate inclined at an angle 45° to the vertical, thus the parallel beam is reflected from the lower surface. Due to the air film formed by a glass plate and a Plano convex lens of large radius of curvature, interference pattern formed which are observed directly through the traveling microscope. The rings are concentric circles.

Formulae:

1. The wavelength

2. Thickness of the air film is , where, r = D/2

Observations:

1. The radius of curvature of Plano convex lens = R =115 cm

2. The interval between fringes = p = 5

Observation Table:

For Determination of (Dn+p2 -D n

2 )

No. of Rings

Micrometer Readings

DiameterD= a-b

cm

D2

(a-b)2

cm2

(Dn+p2-Dn

2)

cm2

MeanCm2

t(cm)

Left enda cm

Right end b cm

30252015100500

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Calculation and Results:

1. The mean wavelength of light from sodium vapor lamp = ……….. Ǻ

Quiz:

1. What is interference? Discuss its types.

2. What is coherent source?

3. What are the conditions for maximum and minimum intensity for the fringes?

4. What is the mechanism for formation of the fringes in this experiment?

5. Show when you will get bands of fringes and circular fringes due to interference?

6. Which type of source is used in the experiment? What will you get if with an

ordinary electric lamp?

7. What will be the change in diameter of the rings if a drop of a transparent liquid

is put between the lens and the plate?

8. What information you can extract from the experiment?

9. Write a brief note on contribution in physics by James Clarke Maxwell.

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Laboratory Manual of Physics (PY101) Date: Sign of Faculty

ULTRASONIC INTERFERROMETER

Aim: Determination of velocity of ultrasonic waves in liquid and its compressibility using ultrasonic interferometer.

Apparatus: Ultrasonic interferometer (include high frequency generator and measuring cell) and given liquid.

Formula: The wavelength of the ultrasonic wave is given by meter. d→ is the distance moved by the micrometer between successive Maxima or minima measured in meter. The velocity of the ultrasonic sound waves is given by f→ is the frequency of the ultrasonic waves in Hz( usually in MHz) λ→ is the wavelength of the ultrasonic waves in meter.

The compressibility of the given liquid is given by

ρ→ is the density of the given liquied in kgm-3.for water = 996.45 kgm-3)v→ is the velocity of the ultrasonic waves in ms-1.

Diagram:

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Observations:

Least count of the micrometer (LC)=______mm.

TR=PSR+HSR; where HSR= (HSC x LC)

Frequency of the ultrasonic waves f = 2MHz.

Observation Table:

Sr. NO Micrometer Reading corresponding to Maximum / Minimum (in mm)

Difference between Consecutive maxima / minima ( λ /2)

1

2

3

4

5

6

7

8

9

10

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Laboratory Manual of Physics (PY101) 11

12

13

14

15

16

17

18

19

20

Procedure: unscrew the knurled cap of cell and lift it away from double walled

construction of the cell. in the middle portion of it pour experimental liquid and screw the knurled cap. wipe out excess liquid overflowing from the cell.

insert the cell in the heavy base socket and clamp it with the help of a screw provided on its side.

Connect the high frequency generator to base co-axial cable provided with the instrument.

keep the "Adj" and "gain" knobs approx at the middle position and adjust such that current meter needle is approx . at center (50mua A).

Move micrometer slowly either clockwise or anticlockwise direction. micrometer needle will move either right or to let. Using both knobs, do adjustment such that the turning point come within the scale. move the micrometer in the same direction and do adjustment so that other turning point comes within the scale.

note the readings of micrometer corresponding to the maximum or minimum (Whichever is Sharper) in micrometer. take about 20 reading of consecutive maximum or minimum and tabulate them in table.

take average of all differences (λ/2). once the wavelenth (λ) is known, the velocity (υ) in the liquid can be

calculated with the help of the relation.

υ = λ×f Result:(i) The velocity of ultrasound waves is v= f ×λmean =………ms-1

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(ii) The compressibility of given liquid =………..m2 N-1.

Quiz:

1. What are the ultrasonic waves? Discuss their properties.

2. What are the methods for production of ultrasound?

3. How the capitation occurs in liquids by ultrasound? Give its use.

4. Give applications of ultrasound in different fields like medical, mechanical etc.

5. Write brief note on contributions in physics by the organizations like ISRO, PRL.


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OPTICAL PROPERTIES OF LIQUID BY LENS METHOD

Aim: To determine optical properties by lens methodApparatus: Glass convex lens, parallax pin, plane-mirror, stand, given liquid etc.

Formula:

Were Fc Focal lent of combination of the liquid and glass lenses. FG= Focal lent of te liquid and lass convex lens. FL=Focal length of liquid Plano-concave lens.

Where, μ=Refractive index of the liquid R2 =Radius of curvature of plane surface of the liquid lens= . R1= Radius of curvature of concave surface of liquid lens=-R.

Theory:

The refractive index (μ) of a medium is defined as the ratio of the velocity of light in the vacuum to the velocity of light in that medium. Also for a ray of light in going from a rarer to a denser medium the ratio of the sine of angle of incidence (i) to the sine of angle of refractive (r) gives the refractive index of the denser medium with respect to the rarer medium. If a liquid of refractive index μ be introduced between a convex lens and a plane mirror a liquid Plano-concave lens if formed between the mirror and the lens. Determination of the focal length of this combination and focal length of the convex lens enable us to determine the focal length of the liquid lens from the formula,

Where FC is the focal length of the combination, FG that of the convex lens and FL that of liquid lens. From the above relation we get,

Thus knowing R and FL , the refractive index μ of liquid is calculate from te lens-maker relation.

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the radius of the curvature (R2) for the plane surface is . So from the above relation we get,

Procedure:

Arrange the plane mirror, glass convex lens and pin as shown in the figure. Adjust the height of the pin so that there is no parallax between the pin itself and its image observed in the convex lens. Measure the height of the pin from center of the lens. This is the focal length of glass (FG). Place a few drops of liquid on the plane mirror and keep the convex lens over it. Remove parallax between the pin and its image seen in the common region formed by the combination of the lenses. This is FC calculate μ.Observations:

1. FG =_____________ cm.

2. FC =_____________ cm.

3. R2 = Calculation: 1. R1= _______cm.

2.

Result: The refractive index (μ) of the given liquid is ____________.

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1. Define focal length of lens.

2. What are nature of real and virtual images?

3. What do you mean by parallax?

4. Define refractive index of a material with respect to vacuum.

5. Why is R1 = - R?

6. What is the function of the plane-mirror?

7. Why is the distance between pin and lens equal to the focal lent when parallax is re-

moved?

8. What is the unit of μ?

9. What is use of information about the refractive index of the material?

10. Can we perform this experiment using concave lens?


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