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Sapkal Knowledge Hub

Kalyani Charitable Trust’s

Late G. N. Sapkal College of Engineering

Kalyani Hills, Anjaneri, Trimbak Road,Nashik - 422212.

Ph.: 02594 - 220171 /172 /173 /174 /175

Engineering PhysicsAcademic Year

20 - 20



Late G. N. Sapkal College of EngineeringKalyani Hills, Anjaneri, Trimbak Road, Nashik - 422212.

Ph.:02594 - 220171 /172 /173 /174 /175

CERTIFICATEThis to certify that

Mr/Miss...........................................................................................................................................................

..................................of Class...................................Div...................................Roll No.................................

University Examination Seat No....................................................has completed satisfactorily

Practical work in Engineering Physics as prescribed by University of Pune, during the academic

Year 20 - 20

Date: / / 20

Sub In-charge Head PRINCIPAL

Engineering Physics Dept. of Engineering Science


Engineering Physics 1

Late G. N. Sapkal College of Engineering

Kalyani Hills, Anjaneri, Trimbak Road, Nashik - 422212.

INDEXExpt. No. Title of Experiment Date Page No. Remark & Sign

1.To determine the radius of curvature of a Plano convex lens using Newton's rings

3

2. To determine wavelength of Red Color using Diffraction Grating. 10

3 To determine the divergence of a LASER beam. 14

4 To determine velocity of ultrasonic waves in a given Liquid. 17

5 To determine the forbidden energy gap in a semiconductor.( Ge Diode) 20

6To study characteristics of Solar cell and to find out the Voc,Isc, Fill Factor (FF) for given solar cell.

24

7 Half shade Polarimeter 28

8 Study of Hall Effect. 31

Date of Experiment: / / .

Signature of teacher with date


EXPERIMENT NO: 1TITLE: Newton’s Rings

Evaluation

Timely/late : /2Performance : /2Understanding : /3Lab Quiz : /2Neatness : /1---------------------------------------

Total marks : /10Aim: To determine the radius of curvature of a Plano convex lens using Newton’s rings.

Apparatus: A Traveling microscope with long focus objective, Newton’s rings apparatus,Reading lamp, Magnifying glass, Sodium lamp

Diagram:

Formula:

R = Slope / 4 λ

Where, R = Radius of curvature of Plano convex lens


λ = Wavelength of sodium light (5893 A0)

Theory:

When a Plano convex lens is placed over a glass plate, a thin wedge shaped film is formed between the two glass surfaces. When illuminated by a beam of monochromatic light, interference takes place between the rays reflected from the lower surface of the lens and the upper surface of the glass plate (i.e., top and bottom surfaces of the air film). The interference pattern is in the form of circular fringes. The fringes are the locus of points of constant thickness of the air film.

The surfaces of the lens and the glass plate should be perfectly clean. The glass plate on which the lens is placed should be optically flat and the curved surface of the Plano convex lens must be perfectly spherical, in order to be able to observe the circular rings. Otherwise the fringes are irregular in shape. Newton’s rings pattern when observed in the reflected light has a dark center. This is because at this position the lens and the plane glass plate touch each other and the thickness of the air film is zero. In the reflected light, the wave getting reflected at the surface of the denser medium has an additional phase shift of radians, and hence the path difference corresponding to λ/2

In case some dust particles are present on the surfaces, the two surfaces do not have a perfect contact at the center leading to a bright spot at the center of the pattern. If both the surfaces are cleaned and rearranged we get a dark-spot.

The concentric alternated dark and bright rings in the Newton’s rings pattern are numbered according to the following convention. The central dark spot in the pattern is called the ‘zero th dark ring.’ This is followed by the ‘first bright ring’, which in turn is followed by the ‘first dark ring’. The next ring is bright and is the ‘second bright ring’.

By theory, we know that the radii of dark rings are proportional to the square root of even natural numbers, whereas those of the bright rings are proportional to the square root of odd natural numbers. This fact is made use of in the present experiment.

Procedure:

1) Arrange the apparatus as shown in the Fig. Clean the lens and glass plate thoroughly and then place the Plano convex lens on the glass plate. With its convex surface touching the plate, place it below the microscope.

2) Allow the parallel light from sodium lamp to be incident on a glass plate, which is held at 45 degrees to the horizontal so that the reflected light falls upon the lens. Adjust the position of the microscope so that the point of contact is just below the microscope objective. Focus the microscope until the rings are distinctly visible and the center of ring system lies below the point of intersection at the cross wires. Rotate the eyepiece so that one of the cross wires is perpendicular to the horizontal scale.

3) Move the microscope across the field of vision and focus it on a distinct dark ring, say, the 20th dark ring from the center on the left. Read the position of the microscope with the cross


wire tangential to the ring. Focus on successive dark rings in steps of two and note the corresponding positions of the microscope. Continue this till you reach dark ring number 4. While focusing take care that every time the cross wire remains tangential to the ring.

4) Move the microscope across the center until it is focused at the other end of the diameter of the dark ring number 4.Read the position of the microscope and proceed onwards until you reach 20th dark ring on the right. From the set of readings, diameters Dn of the rings can be determined.

5) Calculate Dn2 and plot a graph of Dn2 Vs. ring number, n. Calculate the slope of the straight line and knowing the wavelength of sodium light, calculate the radius of curvature of the Plano convex lens using the given formula.

Observations:

L.C. of travelling microscope =..............

Observation Table:

Obs. No.

No. of dark rings (n)Microscope reading (cm) Diameter (cm)

Dn = x1 - x2 Dn2 (cm2)Left (x1) Right (x2)

1

2

3

4

5

6

7

8

9

10

Nature of the Graph:


Sample Calculation:

Slope of the graph =D2n - D 2m--------------- n-m

R =Slope

---------4λ

Result:

Radius of curvature of the Plano convex lens is . . . . . . . . . . . . . . . . . . . . . . . . .

Conclusion:

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .



EXPERIMENT NO: 2TITLE: DIFFRACTION GRATING

Evaluation

Timely/late : /2Performance : /2Understanding : /3Lab Quiz : /2Neatness : /1--------------------------------------

Total marks : /10

Aim: To determine wavelength of Red color using Diffraction Grating.

Apparatus: Spectrometer, mercury lamp, diffraction grating, magnifying glasses, etc.

Diagram:

Formula: We know that,(a + b) sin0 = n λ,


λ=(a + b)

nsin0

Where, (a + b) = grating elementa = width of a slitb = distance between two adjacent slitsn = order of the spectrum

% error =λexp _ λstd

______________

λstd

X 100

Where, λexp = experimental value of wavelengthλstd = standard value of wavelength

Theory: Spectroscopy is a powerful research technique in Physics. It is useful for production,

observation, analysis and interpretation of the spectra of given sources. The properties of a spectral source

can be studied by calculating parameters like intensity, line width and wavelengths of the spectral lines.

Spectroscopy has been found applicable in determination of atomic and molecular structures and

energy levels, study of interatomic and intermolecular interactions, identification of elements,

identification of chemical compositions of unknown substances and their physical properties, etc.

Verification of Bohr’s model of the atom has been made possible due to spectroscopic analysis.

Spectroscopy is also an important research tool in Astrophysics where it is used for studying the

compositions of planets and stars, their atmosphere, as well as their physical properties like pressure,

density, temperature, etc. The fact that the sun is composed of 65 % helium and 35 % hydrogen has been

discovered using spectroscopic analysis.

The aim of the experiment is one of the basic tasks in spectroscopy i.e. determination of

wavelength of spectral lines. The distribution of dispersed radiated energy arranged periodically according

to certain parameters like energy, frequency, wavelength, etc.is called the spectrum. The source can be

made to emit the radiation by suitably exciting it by direct or electrical heating. The excited atoms re-emit

the absorbed energy while returning to the ground state. This radiated energy can be converted into a

spectrum using the spectrometer which is an instrument used to produce, observe and analyze the

spectrum of a given source. It consists mainly of a collimator, a spectroscopic device, and a telescope.

The bending of any kind of wave when encountered by any obstacle of dimension comparable to

its wavelength is called diffraction. A diffraction grating consists of a large number of exactly parallel and

equidistant slits. The grating diffracts the light waves of various wavelengths emitted by the source


through different angles, thereby making it possible to determine the wavelengths. Gratings are often used

to measure wavelengths and to study the intensity of spectral lines.

Using mathematical analysis, it can be shown that the principal maximum of order ‘n’ for a given

grating lies at an angle ‘9’ which satisfies the following condition:

(a + b) sin0 = n λ

Where, (a + b) = grating element

a = width of a slit

b = distance between two adjacent slits

n = order of the spectrum

The source used in this experiment is a mercury vapour lamp, which is a source of white

light. It contains vapors of mercury in which electric discharge is produced by applying high voltage. The

mercury spectrum consists of a few lines in the ultraviolet region and a few lines in the visible region.

Procedure:

1) Adjust the telescope and collimator for parallel rays by Schuster’s method.2) mount the diffraction grating on the prism table such that its plane is perpendicular to the

prism table.3) arrange the face of the grating so that it lies normal to the collimator.4) Focus the vertical cross-wire of the eye-piece perfectly on the first (violet) line of the first

order spectrum of left side. This focusing must be perfect for accurate results.

5) Clamp the telescope and note its angular position using the angular scale through the observation window. Use a reading lamp and magnifying glass to note the reading.

6) Similarly, take readings for the other lines of spectrum. These readings are 01.

7) Turn the telescope to the right side and take the readings for the first order spectrum on this side. These readings are 02.

8) Calculate 20, 0, sin0 and therefore λ for each color by using the given formulae.

9) Compare your results (experimentally calculated wavelengths) with standard values and calculate how much the results deviated from standard ones by using the formula for percentage error.

Observations:


i) Grating element (a + b) =

2.54

------------

15000

= . . . . . . . . . . . . . . . . . cm

ii) Least count of spectrometer = . . . . . . . . . . . . . . . . .

Observation Table:

Obs.No.

Color of the

Spectral line

Angular Positions20=01-02 0=20/2 sin0 λ std. A0 λexp.A0

Left 01 Right 02

1. Violet 4387

2. Green 5460

3. Yellow 5890

4. Red 6390

Sample Calculation:

Result:

Wavelength of the Red color is . . . . . . . . . . . . . . . . . . . . . . . . .


Signature of Teacher with date


EXPERIMENT NO: 3TITLE: LASER

Evaluation


Total marks : /10Aim: To determine the divergence of a LASER beam.

Apparatus: He-Ne LASER, optical bench, screen, Wooden Scale etc.

Diagram:

Theory:-


Conventional light sources emit light in all directions around them. Therefore, the light emitted

by such sources is non-directional and its intensity rapidly decreases with increasing distance from the

source. It is assume that the source sends out spherical waves which have a very high divergence. Unlike

the conventional sources, a LASER emits light in a single direction because of which the intensity of

light is very large. It can be assume that LASER sends out plane wave which have zero versions. In

practice, LASER been diverges slightly due to diffraction effects. Therefore a LASER beam is

characterized by an extremely low divergence.

When a LASER beam is allowed to fall on a screen, circular bright spot is seen on the screen.

When the screen is moved near and far from source, the size of the spot decreases and increases

respectively, as illustrated in figure. If W1 the spot diameter of the LASER beam at some position and

W2 is spot diameter at the another position, then the diversions θO is given by the relation,

θ0 = [W 2

2 − W 12]

12

D

Where,

D = distance between two positions.

Procedure:

1. The LASER is mounted on an upright kept at one end of the optical bench.

2. A screen is mounted on another upright and it is held at about 50 cm from the LASER.

3. The LASER is switched on it forms a red spot on the screen.

4. The size of the spot W l on the screen is measured.

5. The screen is moved further say, to 100 cm. The size of the spot W2 on the screen in the second

position is measured.

6. The distance D between two positions of the screen is measured.

7. The measurements are repeated thrice in different positions of the screen.

8. They are recorded in the observation table.

9. The angular divergence is calculated from the above formula

Observation Table:


Obs.

No.

W1

(cm)

W2

(cm)

D

(cm)

θ0

(rad)

θ0

(degree)

θ0

( Mean)

1

2

3

Sample Calculation:

Result:

The angular divergence of the He - Ne LASER beam is found..................................................



EXPERIMENT NO: 4


TITLE: ULTRASONIC INTERFEROMETER

Evaluation


Total marks : /10

Aim: To determine the speed of ultrasonic waves in a given medium using ultrasonicInterferometer

Apparatus: Ultrasonic interferometer, liquid (distilled water), high frequency generator, Co-axial shielded cable to connect high frequency generator to the measuring cell of the interferometer.

Diagram:

Theory:

An ultrasonic interferometer offers a direct method of determining the speed of ultrasonic wave in

liquids. The measurement is based on accurate determination of the wavelength in the given medium.

Ultrasonic waves of known frequency are produced by a quartz crystal, which is fixed at the bottom of the


cell. These waves are reflected by a movable metallic plate, which is kept parallel to the quartz crystal. If the

separation between these two plates is exactly a whole multiple of the wavelength, standing waves are

formed in the medium. This creates an electrical reaction in the generator, which drives the quartz crystal.

As a result, the anode current of the generator become maximum. If the distance is increased or decreased

λ/2 or a multiple of it, the anode current becomes maximum. The velocity of ultrasonic wave can be

Calculate from the knowledge of the wavelength.

Formula:

v = f. λ

Where, v = speed of ultrasonic wavesf = frequency of waved = distance of the reflector plate from the quartz crystal plate

λ = wavelength of wave

n = no. of deflections in the ammeter when the reflector plate moves through the distance d

Procedure:1) Insert the cell in the square base socket and clamp it with the help of the screw provided on

one of its sides.2) Unscrew the curled cap of the cell and lift it away from the double wall. Pour the liquid in

the screw on the cap. The scale should be in front of observer.3) Connect the high frequency generator to the measuring cell using the cable provided.4) For the initial adjustment, two knobs are provided on the high frequency generator, one is

marked (Adj) and the other is marked (Gain).5) Switch on high frequency generator and wait for 15 sec to warm it up. Initially, the needle of

the ammeter goes to maximum, indicating that the ultrasonic waves are produced.6) Adjust the needle to the neutral position (at 50 A) using adjusting knob. Keep the sensitivity

(using the gain knob) such that the needle goes to 60 A.7) Keep the micrometer on exactly 10 mm.8) Keep the micrometer screw moving slowly so that the reflector plate is move up wards.

The current will peak and the ammeter will show maximum deflection of the needle. Immediately stop at this stage and note the reading on the scale. This is the first reading.

9) Continue rotating the micrometer screw. Every time the instrument satisfies the above condition, the ammeter needle will show maximum deflection. Continue rotating the screw while counting the number n of deflections. Stop immediately after the chosen value of n and note down the micrometer reading.

10) Calculate the speed of ultrasonic waves using the given formula.Observations:

1) Liquid used = distilled water2) Frequency applied = 2 x 106 Hz


3) Least count of micrometer = .........cm

Observation Table:

Obs. No.

No. of deflections (n)Micrometer

readingd1 (mm)

Micrometer readingd2 (mm)

d2-d1

(mm) 2dn = ------

n(mm)

v = f. λ(mm / s)

1

2

3

4

5

6

7

8

9

10

Calculation:

Result:

The speed of ultrasonic waves in the given liquid is........................ mm/s.



EXPERIMENT NO: 5


TITLE: FORBIDDEN ENERGY GAP

Evaluation


Total marks : /10

Aim: To determine the forbidden energy gap in a semiconductor (Ge. Diode)

Apparatus: DC Power supply, Electrical heater, 0-100 micro-ammeter, Thermometer, Germanium orSilicon diode etc.

Circuit Diagram:

Theory:

Solids are classified according to energy bands theory, it consists of two band namely conduction band and valance band. These two bands are separated by a gap known as forbidden energy gap (Eg).


The band due to which valance electronics, occupies highest energy at absolute zerotemperature is called valance band. It may partially or completely fill depending up on the structure ofthe solid. Another energy band in the solid is conduction band and it has lowest unfilled energy at zeroabsolute temperature. Energy band diagram of solid is as shown in Figure.

According to electrical properties of solid material and band theory of solids, the electrons

of valence band can be transferred to conduction band by providing them with energy equal to Eg

Therefore, on the basis of the forbidden energy gap energy, solids are divided into three groups,

(i) Conductors, (ii) Semiconductors (iii) Insulators.

In conductors, Eg = 0 eV, in semiconductors 0 < Eg < 1 eV and in insulators, Eg > 5 eV.

Germanium and silicon are two good examples of semiconductors. In silicon, the energy gap (Eg) is 1.1

eV, whereas that in germanium is 0.72 eV. At absolute zero in semiconductors the conduction band

found to be empty and the valence band is filled.With increase in temperature the valence band

electrons gain energy, and cross over the forbidden gap and transfer to conduction band. Thus, if the

number of electrons available for conduction with increases in temperature, the resistivity of the

semiconductor decreases. This is given by equation, (at temperature T > O ° K)

ρT = ρ 0 exp ( Eg2 kT )

………………………….. (1)

Where, ρT is resistivity at temperature T, po is resistivity at absolute zero, Eg is forbidden energy gap of

the semiconductor, and k is Boltzmann constant. But we know that resistance of a given specimen is

proportional to resistivity, therefore equation 1 becomes,

RT = R0 exp ( Eg

2 kT )…………………………….. (2)

Taking logarithm of both the side of the equation 2 we get,

Ln (RT) = ln (R0) +

Eg2 kT ……………………………. (3)


Mathematically equation (3) is in the form of equation of straight line (y = mx + c), in which

slope of graph of ln RT versus 1/T is equal to Eg/2k and intercept equal to In Ro. Using this relation Eg

can be computed.

Procedure:

1. Take cell / battery of 2 or 3 volts. Make the connections as shown in circuit diagram and get itchecked.

2. Note down room temperature and corresponding reverse leakage current through the diode (Reversebias)

3. Switch ON oven and wait until temperature is reached up to 90° C and switch Off the oven andimmediately switch ON the battery. The current observed in the micrometer decreases due tocooling down of the diode. Record the current in step of 5°C each (up to 40°C.)

4. Plot the graph loge l versus 1/T°K. Compute slope of the graph and using formula calculate band gapenergy (Eg) of the given semiconductor.

5. Compare the experimental result with standard value.

Observations:

Room temperature: ……….… ° C,

Voltage across diode: ………… V (volts)

Observation Table:

Sr.No.

Temp.t° C

Current I(A or 10 6A)

Loge(I)Temp in ° K

T= t + 273

1T°K

1

2

3

4

5

6

7

Graph: - Plot the Graph of log10 (I) Vs 1/T °K


Formula:

Band gap energy of given semiconductor

Eg = Slope of graph x k

Where,

k is Boltzmann’s constant= 0.8626 x 10-4 eV.

Eg = Slope x 8.617x10-5 eV

Sample Calculation:

Result:

The forbidden gap in germanium / silicon semiconductor is found to be……………………..

Eg=………………….. eV




EXPERIMENT NO: 7

TITLE: I-V CHARACTERISTICS OF SOLAR CELL

EvaluationTimely/late : /2Performance : /2Understanding : /3Lab Quiz : /2Neatness : /1--------------------------------------

Total marks : /10

Aim: To study characteristics of Solar cell and to find out the Fill Factor (FF) for given solar cell.

Apparatus: Solar Cell, Variable load, Digital Multimeter (DMM), 0 -50 A , etc.

Circuit Diagram:

Theory:


Solar cell is a p-n junction, which produces a voltage when visible light falls on it. Consider a

p-n junction made up of silicon (Si) having a forbidden band gap of 1.1 eV. If the light from sun or

some other visible radiation source is incident upon it, the incident photons having energies greater than

1.1 eV move the electrons from valence band to conduction band. This results in production of excess

electrons in the conduction band and excess holes in valence band. Suppose that the sunlight is inciden

ton the p-type region of the p-n junction. The incident light produces electron hole pairs in p-region. The

extra holes produced in p region are insignificant as compared to the large number of holes already

present in it. Some electrons immediately combine with holes and get neutralized while others shift to

the junction boundary.

In p-n junction there exists an electric field directed from n region to p region due

to the development of barrier potential. This electric field makes electrons move from p to n region.

Hence an extra negative charge builds up in n-region and extra positive charge in p-region. Due to this,

n-type region becomes like the negative terminal and p type region as positive terminal of a battery.

The voltage formed is about 0.6 V. If a load resistance RL is connected across the two terminals, the

electric current flows through RL. Such a current continues to flow as long as light is incident on the

junction. A similar process occurs when light is incident on n-region. Current produced depends upon the

number of charge carriers produced.

The voltage corresponding to RL = , is called open circuit voltage, Voc,. The current corresponding to RL =

0, is called short circuit current, Isc. The maximum power is delivered to a load resistance by the solar cell

when the product I x V maximum. The corresponding values of current and voltage (Im and Vm) can be

deduced by finding the rectangle of the largest area under the I - V characteristics. The ratio Im Vm / Isc is

called the fill factor (FF) and is the figure of merit for designing the solar cell.

Solar Cell

Nature of Graphs:


Observation Table:

Sr.

No.

VoltageV(mV)

Current

I ( µA)

Sr. No.

VoltageV(mV)

Current

I ( µA)

1 11

2 12

3 13

4 14

5 15

6 16

7 17

8 18

9 19

10 20

Calculations:


Short circuit current (Isc) = ………………………… µA

Open circuit voltage (V0C) = …………………………….volt

1) Fill Factor (FF) =

Im x V m

I sc x V oc

Where,

Im = current corresponding to maximum power point.

Vm= voltage corresponding to maximum power point.

Isc = short circuit current.

Voc = open circuit voltage.

Result:

Fill Factor of solar cell (FF) …………………………………………




EXPERIMENT NO: 7TITLE: HALF SHADE POLARIMETER

Evaluation


Total marks : /10

Aim: To determine the specific rotation of an optically active substance by Polarimeter.

Apparatus: A Polarimeter Optically active substance (known weight), a measuring cylinder, a beaker, a

scale Distilled water, Sodium light source

Diagram:

S – Slit A– Analyzing Nicol L – Collimating lens OE – Objective and eyepiece of the observing telescope P – Polarizing Nicol T2 – Tube housing of the observing telescope H – Half-shade plate T – Tube containing solution

Experimental set up to measure the specific rotations for an optically active solution by Polarimeter


Theory Let a plane polarized monochromatic light of wavelength pass through a column of solution

of length l centimeter and containing m gm of an active substance per cm3 at a given temperature. Then

the plane of polarization of the light is rotated through an angle given by

slm/10…. … (1)

Where, s is the specific rotation of the solution.

The quantity s is defined as the rotation produced by a column of solution of length 1

decimeter and containing 1 gm of the active substance per cm3 of the solution under the experimental

condition, i.e. at the temperature of the solution and for the wavelength used.

If 100 cm3 of the solution contains c gm s of the active substance, the strength of the solution is c

%, and

m c. Henceslc/1000 … (2)

s 1000lc. … (3)

Equation (2) shows that for a given length l, the plot of versus c will be a straight line.

This plot is known as the calibration curve of the Polarimeter for the active solution. This plot can be used

to find the concentration c of an unknown solution of the same solute by measuring the rotation

produced by it. From a set of values of c and , and by measuring the length of the tube containing the

solution, the specific rotation of the solution, s, can be determined from Eq. (3).

Procedure:-

1. Clean the Polarimeter tube, beaker, flask and measuring cylinder with water.2. Fill the Polarimeter tube with distilled water and note the angle of rotation by adjusting scale and observing uniform illumination through the eye-piece.3. Rotate the analyzer by approximately 1800 to record the other uniform illumination reading.4. Rotate the analyzer by another 1800 and note the readings for uniform illumination.5. Repeat the step 4.6. Prepare the solution of appropriate strength by dissolving a known amount of active substance in a known volume of distilled water.7. Fill the Polarimeter tube with the solution of known concentration. Note the rotation produced in by adjusting the scale and observing uniform illumination through the eye-piece.8. Repeat the steps 3, 4, & 5.9. Prepare solutions of different known concentrations by proper dilution of the parent solution.10. Repeat steps 7 & 8 for different concentrations.11. Now prepare an unknown concentration solution and repeat 7 & 8.12. Plot a graph of versus c.13. From the graph obtain the unknown concentrationObservations


Length of the Polarimeter tube: ……………..cm.

Wavelength of the light used: ………………….cm.

Room temperature: ………………………… 0C

Vernier constant (Least Count) of the Polarimeter: ……………………cm

Table 1Rotation of plane of polarization when tube contains distilled water

Sr. No.

1st Uniform illuminationposition

C.S.R V.S.R. Total Mean

(R0)

2nd Uniform illuminationposition

R0 + 1800

Rotation of plane of polarization when tube is filled

Sr. No.

Weight ofactivesubstance(gm)

Strengthof activesubstance(c %) 1st Uniform

illuminationposition

C.S.R V.S.R. Total Mean Θ=R0-R

(R0)

2nd Uniform illuminationposition

R0 + 1800

Graph:-Plot a graph of Vs C.


Results: - The specific rotation of an optically active substance by Polarimeter is………………




EXPERIMENT NO: 8TITLE: STUDY OF HALL EFFECT

Evaluation


Total marks : /10 Aim: To study Hall Effect in extrinsic semiconducting samples and determine the type and density of

majority charge carriers. This experiment demonstrates the effect of Lorentz force

Apparatus: Commercial setup with the following components: electromagnet with power supply, Hall

probe, Semiconductor sample, arrangement for pressure contact, current supply with meter, voltmeter etc.

Theory:

Consider a rectangular slab of semiconductor with thickness d kept in XY plane [see

Fig. 1] an electric field is applied in x-direction so that a current I flows through the

sample.If w is width of the sample and d is the thickness, the current density is given by

Jx=I/wd.

Fig. 1 Now a magnetic field B is applied along positive z axis (fig. 1). If the charge carriers are

positive (negative) and are moving with velocity v along positive (negative) x-axis then the direction of

force experienced by the charge carriers in presence of magnetic field is along negative y direction. This

results in accumulation of charge carriers towards bottom edge (fig1.). This sets up a transverse electric


field E y in the sample. The potential, thus developed, along y-axis is known as Hall voltage VH and this

effect is called Hall Effect.

Assuming Ey to be uniform the Hall voltage is given by

V H = E y w (1)

and the hall coefficient RH is given by

R H= E y / JxB = V H d / IB (2)

The majority carrier density n is related to the Hall coefficient by the relationRH =1/nq (3) Where q is the charge

From Equation (3), it is clear that the sign of charge carrier and density can be estimated from the sign

and value of Hall coefficient RH. RH can be obtained by studying variation of VH as a function of I for

given B.

Procedure1. Connect the leads from the sample to the "Hall effect Set-up" unit. Connect the electromagnet to

constant current generator.

2. Switch on the current through sample and measure the hall voltage without any magnetic field. There

may be some voltage due to misalignment of pressure contacts on the sample. This error must be

subtracted from the readings.

3. Switch on the electromagnet and set suitable magnetic field (<3 K gauss). You can measure this using

Hall probe. (Set magnetic field B=2 k G and B=3 k G for the experiment).

4. Insert the sample between the pole pieces of the electromagnet such that I, B and V are in proper

direction (Fig.1).

5. Record the hall voltage. Also record voltage by reversing both the current and magnetic field

simultaneously (Note down data for the first two columns with +B for all I’s and then reverse the field (-

B) to record data for the next two columns)

6. Keeping the magnitude of magnetic field constant, measure hall voltage as a function of I.

7. Repeat step 5 and 6 for various magnetic fields.

Plot VH as a function of I using the averaged data and find the value of Hall coefficient from the slope of

the graph. Hence determine charge carrier density and type of majority carrier in the given material.

Note down the sample number or details of the sample.


Observation Table:Sample number: Thickness of the sample: Magnetic

field: B= ………… Gauss

Magnetic field: B= ………………… Gauss

Graph:-

Plot two graphs and from the slope, calculate Hall coefficient RH. From the sign of the Hall voltage with the given current and field direction, determine the type of conductivity in the semiconductor material.


Sl.No.

I(mA) VH1(+I,

+B)VH2(-I,+B)

VH3(+I,-B)

VH4(-I,-B)

VH avg

1

2

3

4

5

6

7

Sl.No.

I(mA) VH1(+I,

+B)VH2(-I,+B)

VH3(+I,-B)

VH4(-I,-B)

VH avg

1

2

3

4

5

6

7

Sample Calculation:-

Calculate free carrier density (n) = 1/ (q RH) = …. m-3

Results:

For the given sample, RH = … ….. n = … ….Type of majority carrier = ……


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