DEPARTMENT OF PHYSICS R.V.COLLEGE OF ENGINEERING · PDF fileto enable engineering graduates to understand, leverage and appreciate the role of physics for the development of sustainable

DEPARTMENT OF PHYSICS R.V.COLLEGE OF ENGINEERING

(AUTONOMOUS INSTITUTION AFFILIATED TO VTU, BELAGAVI)

ENGINEERING PHYSICS

LABORATORY MANUAL

COURSE CODE: 16PH12/22

2017-18

For the First / Second Semester B.E

Name of the student

Section, Batch

Branch

Roll No. / USN

Faculty In-charge 1. 2.

DEPARTMENT OF PHYSICS

VISION

TO ENABLE ENGINEERING GRADUATES TO UNDERSTAND, LEVERAGE

AND APPRECIATE THE ROLE OF PHYSICS FOR THE DEVELOPMENT OF

SUSTAINABLE AND INCLUSIVE TECHNOLOGY.

MISSION

• EDUCATES THE STUDENTS WITH A PROGRAM CHARACTERIZED BY

ART OF TEACHING WITH EXPERIMENTING SKILLS, PROJECT

WORK/SEMINAR, SELF STUDY, EFFECTIVE COUNSELING AND AN

ACTIVE INVOLVEMENT OF STUDENTS IN THEIR EXPERIENTIAL

LEARNING.

• IMBIBE INQUISITIVENESS IN STUDENTS TO USE PHYSICS FOR

ENGINEERING INNOVATION.

• EMPOWER THE FACULTY AND STUDENTS TO INVOLVE IN RESEARCH

AND TO DEVELOP THE DEPARTMENT AS A KEY FACILITATOR FOR

R&D TO ALL ENGINEERING PROGRAMS.

INDEX

Name : Sec/Batch : USN/Roll No :

S.NO EXPERIMENTS PAGE NO

TEACHING FACULTY COURSE OUTCOMES 01 GENERAL INSTRUCTIONS TO STUDENTS 02 MEASUREMENTS 04 1. Characteristics of laser beam 06 2. Verification of Stefan’s law 10 3. Wavelength of LED’s 12 4. Torsional Pendulum 14 5. Series LCR Circuit 18 6. Fermi energy of copper 20 7. Hall Coefficient measurement 22 8. Energy band gap of a Thermistor 26 9. Dielectric constant 28 10. Thermal conductivity of a poor conductor by Lee’s and

Charlton’s method 32

11. Electrical resistivity by Four Probe method 36 12. Thermal conductivity of a good conductor by Searle’s method 38 13. Numerical Aperture and Loss in Optical Fiber 44

Sample viva questions 48

TEACHING FACULTY

Sl. No.

Name

Designation

Initials

01 Dr. T Bhuvaneswara Babu Prof. and HoD TBB

02 Dr. D N Avadhani Associate Professor DNA

03 Dr. T K Subramanyam Associate Professor TKS

04 Dr. Sudha Kamath Associate Professor SK

05 Dr. G Shireesha Assistant Professor GS

06 Dr. S Shubha Assistant Professor SS

07 Dr. Tribikram Gupta Assistant Professor TG

08 Dr. BM Rajesh Assistant Professor BMR

09 Dr. P Ramya Assistant Professor RP

10 Dr. Karthik Shastry Assistant Professor KS

Course Outcomes: After completing the course, the students will be bale to CO1 Understand the fundamental concepts of Optical Physics, Quantum mechanics, wave

theory and conductivities CO2 Apply the concepts of Optical Physics, Quantum mechanics, wave theory and

conductivities in Engineering domain. CO3 Analyze the theoretical concepts and investigate in the laboratory. CO4 Demonstrate team work and effective reporting

CO mapping for Engineering Physics (16PH12/22) lab experiments

S.No Experiments CO1 CO2 CO3 CO4

1. Characteristics of laser beam

2. Verification of Stefan’s law

3. Wavelength of LED’s

4. Torsional Pendulum

5. Series LCR Circuit

6. Fermi energy of copper

7. Hall Coefficient measurement

8. Energy band gap of a Thermistor

9. Dielectric constant

10. Thermal conductivity of a poor conductor by Lee’s and Charlton’s method

11. Electrical resistivity by Four Probe method

12. Thermal conductivity of a good conductor by Searle’s method

13. Numerical Aperture and Loss in Optical Fiber

Department of Physics, RVCE 2

GENERAL INSTRUCTIONS TO STUDENTS

1. Lab batches will be allotted at the beginning of the semester. Students will have to perform two experiments in one lab.

2. Every student has to perform the experiment whichever is allotted to him /her, no change of experiments will be entertained.

3. While attending every laboratory session the student must bring the data sheets pertaining to the experiments.

4. Submission of data sheets and practical record with observations and results in every class for evaluation is mandatory.

5. The data sheet must contain entries like aim of the experiment, apparatus required, circuit diagram or the diagram of the experimental setup, tabular columns, the necessary formulae of the experiments as given in the left hand side pages in the lab manual.

6. Separate data sheets should be prepared for each experiment. The procedure and principle of the experiment must be read by the student before coming to the laboratory and it should not be written on the data sheets.

7. All calculations pertaining to the two experiments should be completed in the laboratory. The results must be shown to the batch teacher and must obtain the exit signature from batch teacher before he or she leaves the laboratory.

8. Entries of observations should be made in data sheets only with pen.

9. Substitutions and calculations should be shown explicitly in the data sheet and the practical record.

10. In the event the student is unable to complete the calculations in the regular lab session, with the permission of the lab in-charge, the student should complete calculations, write in the practical record and submit for the evaluation in the next lab session.(In case of any difficulty in calculation the student can consult the batch teacher within two working days).

11. Mobile phones are not allowed to the lab. The student should bring his/her own

calculator, pen, pencil, eraser, etc.


12. The experiments are to be performed by the students in the given cyclic order. This will be made clear to the student in the instructions class. If for some reason a student is absent for a practical class then the student must move on to the next set in the subsequent class. The experiment, that he or she has missed, will have to be performed by him or her in the repetition class.

13. Please remember that practical record is evaluated during regular lab session. Therefore it is imperative that each student takes care to see that the experiments are well conducted, recorded and submitted for valuation regularly.

14. There will be a continuous internal evaluation (CIE) in the laboratory. An internal test will be conducted at the end of the semester. The internal assessment marks are for a maximum of 50 marks.

15. The semester end examination (SEE) of the lab will be conducted for 50 marks.

All students are strictly adhere to the Do’s and Don’ts in the laboratory:

Do’s Come prepared to the lab. Wear lab coat in the lab. Maintain discipline in the lab. Handle the apparatus with care. Confine to your table while doing the

experiment. Return the apparatus after completing

the experiment. Switch off the power supply after

completing the experiment Switch off the electrical circuit

breaker if there is burning of insulation.

Utilize the First Aid box in emergency situation.

Keep the lab clean and neat.

Don’ts Come late to the lab and leave the lab

early. Carry mobile phones to the lab Touch un-insulated electrical wires. Use switch if broken. Overload the electrical meters. Talk with other students in the lab. Make the circuit connection when the

power supply is on.


MEASUREMENTS

To conduct various experiments in the Physics Laboratory, we need to learn measurement of dimensions and other physical quantities using instruments. Measurements of various dimensions of object using Vernier Calipers, Screw gauge, Multi meter etc are discussed here.

Vernier Calipers

It is used to measure the length and breadth of some small objects accurately.

Least count of vernier calipers The vernier caliper has two scales – main scale and vernier scale. The main scale is graduated in cm while the vernier scale has no units. The vernier scale is not marked with numerals, what is shown above is for the clarity only. Least count (LC) of the vernier calipers is the ratio of the value of 1 main scale division (MSD) to the total number of vernier scale divisions (VSD). Example:

Value of 10 main scale divisions (MSDs) = 1cm Value of 1 MSD = 0.1cm Total number of VSD = 10 Therefore LC = 0.1cm/10 =0.01cm

To take readings using the calipers (1) First see if the 0 of the vernier scale coincides with a main scale reading. If it

coincides then take it as the main scale reading (MSR). (2) If the 0 of the vernier scale does not coincide with any main scale division then

the division just behind the zero of the vernier is the main scale reading (MSR). (3) Then see which vernier division coincides with a main scale reading. This division

of the vernier scale is noted as the coinciding vernier scale division (CVD). (4) The total reading is given by TR = MSR + VSR, TR = MSR + (CVD x LC)

For example, if MSR = 1cm, CVD = 6, TR = MSR + (CVD x LC) = 1cm + (6 x 0.01)cm = 1.06cm Screw gauge: Unlike the above instruments, screw gauge has a pitch scale and a head scale. The pitch scale is graduated in mm while the head scale has no units. The least count for this type of instruments is given by divisions scale head of No.

pitchCount Least =

Main Scale

Vernier Scale


The pitch of the screw gauge is the distance moved on the pitch scale for one complete

rotation of the head. To find pitch give some known number of rotations to the pitch scale

and note the distance moved by the head scale.

Pitch = distance moved on the pitch scale / No. of rotations given to head scale

Usually the pitch is 1mm, the head scale is divided into 100 divisions.

Therefore LC = pitch /No. of head scale divisions = 0.01mm

In the screw gauge the head is rotated until the plane faces (metal plug and screw head) that

hold the object touch each other. If the pitch scale reading is zero and the zero of the head

scale coincides with the pitch line then there is no zero error, otherwise there is a zero error

(ZE). Determination of zero error is shown in the following figure.

The total reading is calculated using the formula: TR=PSR+(HSR - ZE)×LC mm Multi meter: A multi meter is an instrument with ammeters, voltmeters (both AC and DC), ohmmeters

etc., of various ranges built into it. By conveniently switching the rotatable knob of the multi

meter, we can choose the electrical meter required for a particular measurement.

+ Ve

- Ve

ZE = 0 ZE = +2 ZE = -4


CHARACTERSTICS OF LASER BEAM

OBSERVATIONS:

Part A: Determination of the wavelength of LASER

Diagram:

Formula:

Wavelength of Laser source = nC sinθλn ……. m

Where C is the grating constant, n is the order of spectrum, θ is the angle of diffraction

Grating Constant:2

51 inch 2.54 10C 5.08 10 mNo. of lines (N) per inch 500

m−−×

= = = ×

Table:

Diffraction order (n)

Distance 2Xn(cm)

Distance Xn(cm)

Diffraction angle (θn) θn = tan-1 Xn

d

Wavelength λ (nm)

n= nC sinθλ

1. 2. 3. 4. 5. 6.

Result: The wavelength of laser light is found to be…………… nm

d


CHARACTERSTICS OF LASER BEAM

Experiment No: 01 Date:

Aim: Part A: To determine the wavelength of a given laser beam Part B: To measure the divergence angle of the given laser beam

Apparatus and other required materials: Laser source, Grating, optical bench with accessories and metre scale etc.,

Principle: The laser is a device, which gives a strong beam of coherent photons by stimulated emissions. The laser beam is highly monochromatic, coherent, directional and intense. The directionality of a laser beam is expressed in terms of full angle beam divergence. Divergence of a laser beam is defined as its spread of laser beam with distance.

Part A: Determine the wavelength of a laser

Formula:

Wavelength of Laser source nC sinθλ=n ……. m

Where C is the grating constant: Distance between successive lines, n is the order of spectrum, θ is the angle of diffraction, N = No. of lines per inch

Grating Constant:2

51 inch 2.54 10C 5.08 10 mNo. of lines (N) per inch 500

m−−×

= = = ×

Procedure: • Mount the laser on an upright and fix the upright at one end of the optical bench. Mount

a screen on another upright and fix it at the other end of the optical bench. • Mark four quadrants on a graph with ‘O’ as the origin and fix the graph sheet on the

screen using pins. Switch on the laser source and adjust the position of the graph sheet, so that the centre of the laser spot coincides with the origin.

• Mount the grating on the grating stand such that the length of the grating is on the grating stand and move the stand closer to the laser source. Adjust the grating plane such that the diffraction pattern is along the horizontal on the screen with the central maximum is at the origin. Note down the distance ‘d’ between grating and the screen.

• Mark the centres of the central maximum and secondary maxima on the graph sheet using pencil and remove the graph sheet from the stand. Measure the distance between the first order maxima on either side of the central maximum as 2X1, for the 2nd order maxima measure the distance as 2X2, continue this up to 6th order maxima.

• By using the grating constant C and the angle of diffraction θn, calculate the wavelength of laser light for all the orders. Finally find the average value of wavelength.


PART B: Determination of divergence angle of a laser beam: Diagram: Table :

Spot No

Distance ‘D’ cm

Horizontal diameter wh (cm)

Vertical diameter wv (cm)

Mean Diameter (cm)

2ww

w vh +=

-1 w = tanD

φ

I

II

III

Average ɸ= --------degree

Average ɸ= ---------rad

Result: The angle of divergence of given laser beam is found to be ɸ = ......................rad.

LASER

D3 D2

D1

I-spot II-spot

III-spot

ɸ


PART B: Determination of divergence of laser beam.

Formula: The angle of divergence ɸ of the laser beam is given by

-1 w= tanD

φ

Where, w is the mean diameter of the laser spot D is the distance of the screen from the source.

Procedure: • Arrange the laser source on a laser stand at one end of the optical bench. • Place the screen with graph sheet in front of the laser source at a distance “D1”. (Measure

this distance with metre scale.) • Draw the boundary of bright laser beam spot on the graph sheet using a pencil. • Measure the diameter of the spot along the horizontal (WH) and along the vertical (WV). • Repeat the above procedure for two more distances D2 and D3 • Calculate the average diameter (W) of the spot for each distance D. • Calculate angle of divergence ɸ by substituting the values of W and D.

Result:

1. The wavelength of laser light (λ) is ………… nm

2. The angle of divergence (ɸ) of given laser beam is = ......................rad.

Precautions: 1. Laser is harmful to eye. Hence don’t look at the laser source directly. 2. The laser source should be switched on only while taking the observation and switched

off thereafter. 3. The laser tube axis should be horizontal.


STEFAN'S LAW OBSERVATIONS:

Diagram Model Graph:

Table:

Trial No. Voltage, V

(Volt) Current, I

(A) R = V/I

(Ω) P = VI

(W) 𝒍𝒐𝒈𝟏𝟎𝑷 𝒍𝒐𝒈𝟏𝟎𝑹

1 2 3 4 5 6 7 8 9

Note: Take readings only up to 12V.

Electrical power given to bulb = energy radiated by bulb per second • P = VI = eσAT4 where A is the surface area of the filament, σ is Stefan's constant, T is the

absolute temperature of the filament and e is the emissivity of the filament. • Emissivity of the black body is 1 • Let P = e σ A Tn where it is to be proved that n = 4

Log P = Log (σ) + Log (A) + n Log (T)+loge But T α R (R is the resistance of the filament) Log P = Log (σ) + Log (e A) + n Log (R) Log P= n log(R) + Log (σ e A) Log P= n log(R) + K , Where log(σ e A) = K (a constant)

• This equation is of the form y = mx + c where y = Log P, m = n and c = K Therefore a plot of Log P versus Log R is a straight line with the slope = n If n= 4 Stefan’s law is verified

RESULT: Slope = _______. , Hence Stefan’s law is verified.

Bulb

RPS

V + -

+ -

A

+ -

Log R

Log

P A

B C

n=AB/BC

Slope=AB×Scale on y-axisBC ×Scale on x-axis


STEFAN'S LAW

Experiment No.:02 Date:

Aim: Verification of Stefan's law of radiation.

Apparatus and other required materials: Regulated power supply (RPS), electric bulb (12 V, 10 W) and DC Ammeter, Voltmeter. Principle: Stefan’s law states that the energy dissipated per unit area per unit time by a

perfect black body is proportional to the fourth power of its absolute temperature.

Stefan’s law E α T4 or E = σ T4

The filament is not a perfect black body, hence the Stefan’s law for the filament is E = σe T4

Where, E is the energy dissipated per second per unit area of a black body

e is the emissivity of the filament material

σ is the Stefan's constant (5.67 x 10-8 Wm-2K-4)

T the temperature of the black body in Kelvin

Procedure:

• Make the electrical connections as shown in the diagram.

• Keep the display in the regulated power supply (RPS) in volts mode.

• Keep the current knob at the maximum position, keep the course volts and fine volts

knobs at the minimum position before switching ON the RPS.

• Switch ON the RPS and vary the voltage knobs slowly till a faint glow is seen in the

filament.

• Vary the voltage in convenient steps and note down the current I in the ammeter and the

voltage V in multi meter set in voltage mode (0 – 20 V DC).

• Calculate the power (P) dissipated in the filament and resistance (R) of the filament

material.

• Plot a graph of 𝑙𝑜𝑔10𝑃 versus 𝑙𝑜𝑔10𝑅 and calculate the slope ‘n’ of the graph.

• If the slope ‘n’ is 4, the Stefan’s law is verified

Result: Slope n = _______. Hence Stefan’s law is verified.


WAVELENGTH OF LED’S OBSERVATIONS:

Diagram: Model Graph:

Formula: Energy of the PhotonE = hυ = hcλ

= eV

The wavelength of LED is K

hcλ= n meV

Where, h is Planck’s constant = 6.63 x 10 -34 Js c is Speed of Light = 3 x 108 m s-1

e is charge on electron = 1.602 x 10 -19 C VK is the Knee voltage in V of the LED, (to be measured).

Table: Sl No. LED 1 LED 2 LED 3

Colour: Colour: Colour: Voltage

V Current

mA Voltage

V Current

mA Voltage

V Current

mA 1. 2. 3. 4. 5. 6. 7. 8. 9. 10.

Result: The wavelengths of LED’s

Colour of LED Wavelength (nm)

Ba +

+ -

+

-

V

mA

+

- - LED

1

LED

2

-

+

R R

_

+

Knee Voltage Voltage V

Cur

rent

mA


WAVELENGTH OF LIGHT EMITTING DIODES

Experiment No:03 Date: Aim: To determine the wavelengths of the given Light Emitting Diodes (LED s).

Apparatus and other required materials: Power supply, LED’s, multi meter. milli ammeter, patch cords etc.,

Principle: Light emitting diode is special type of semiconductor diode. It consists of heavily doped P type and N type direct band gap semiconductor. The LED absorbs electrical energy and converts it into light energy. When the PN junction is in the forward biased the electrons from the N region migrate into P region and combine with holes. This recombination of electrons and holes results in the emission of photons.

Formula: Energy of the photons emitted by LED =E = hυ = hcλ

= eVK

The wavelength of LED is K

hcλ= n meV

Where, h is Planck’s constant = 6.63 x 10 -34 Js; c is Speed of Light = 3 x 108 m s-1

e is charge on electron = 1.602 x 10 -19 C; VK is the Knee voltage of the LED. Procedure: • Make the connections as shown in the circuit diagram. • Keep the display in the regulated power supply (RPS) volts mode. • Keep the current knob at the maximum position, keep the course volts and fine volts

knobs at the minimum position before switching ON the RPS. • Include the LED1 in the circuit. • Vary the voltage in convenient steps and note down the current I in the milli ammeter

and the voltage V in multi meter set in voltage mode (0 – 20 V DC). (Note down the few voltmeter reading for current below 0.5 mA).

• Repeat the above procedure for the second and third LED’s • Plot a graph of current versus voltage for each LED starting with true origin on x and y

axis. • Draw a tangent at the knee of the graph and extend the tangent to intercept the voltage

axis. The point of intersection is knee voltage VK. • Note down the knee voltage for each LED from the graph. • Calculate the wavelength of each LED using the relevant formula.

Result: The wavelength of LED’s are found to be

Sl. No Colour of LED Wavelength (nm) 1 2 3



TORSION PENDULUM

Experiment No:04 Date:

Aim: To determine the moment of inertia of the given irregular body

Apparatus and other materials required: Rectangular, circular and irregularly shaped

plates, steel or brass wire, chuck nuts, stop clock, pointer, metre scale, weight box.

Principle: Torsion pendulum is an illustration of angular harmonic oscillation. Moment of Inertia of a

body is the reluctance to change its state of rest or uniform circular motion. The moment of

inertia (MI) of a regular body about any axis is calculated by knowing its mass and

dimensions. A body whose moments of inertia about different axes are known, is made to

oscillate about the same axes, corresponding periods are noted. If the moment of inertia is I

and the time period is T then for all axes chosen for all the bodies the ratio 2TI

is a constant

as long as the dimensions of the suspension wire remains the same.

IT=2πC

, C is the couple per unit angular strain (twist) of the wire. It is equivalent to the

spring constant of the spring for linear oscillation, and is dependent on dimensions of the wire

only. Therefore if we change the pendulum geometry, T I∝ and thus squaring both sides

2TI

is constant.

Procedure:

• Measure the dimensions of the given circular and rectangular discs.

• Clamp one end of the wire through the chuck nut to a regular disc and other end to the

top end of the retard stand.

• Twist the wire through a small angle and then let free so that the body executes torsional

oscillations (The oscillations should be in a horizontal plane. Arrest the side ward

movement or wobbling if any).

• For each configuration of the pendulum, note down the time taken for 10 oscillations and

repeat the process thrice. Tabulate this in table 1.


To determine the period of irregular body

Axis through CG Time for 10 sec. (s) Mean time(t) for 10 oscillations in s

Period T=t/10 s

1. 2. 3.

Tα =

1. 2. 3.

Tβ =

(a) Moment of inertia of irregular body about the axis through the CG and perpendicular to its plane, I'

2α α2

mean

II = × T T

= ___________ Kgm2

(b) Moment of inertia of the irregular body about the axis through the CG and parallel to its plane

2β β2

mean

II = T T

×

= ____________ Kgm2

Result: 1. The moment of inertia of irregular body about an axis perpendicular to the plane Iα =______Kgm2. 2. The moment of inertia of irregular body about an axis parallel to the plane Iβ =______Kgm2.


• Calculate mean time (t) for 10 oscillations and hence the time period T then find 2I

T for

each axes.

• Follow the same procedure for two different axes of the irregular body and tabulate in

table 2.

• Find out the moment of inertia of irregular body using given formula.

Result: 1. The moment of inertia of irregular body about an axis perpendicular to the plane Iα =______Kgm2. 2. The moment of inertia of irregular body about an axis parallel to the plane Iβ =______Kgm2.


SERIES RESONANCE – OBSERVATIONS: Diagram: Model Graph :

L C R

Capacitance of capacitor in the circuit C = ________ µF, Resistance of the resistor in the circuit R = ________ Ω Table:

Frequency (Hz)

Current (mA)

Frequency (Hz)

Current (mA)

Formula: At resonance XL=XC

=π Lf2 0 Cf21

0π ⇒ Cf4π1=L 2

02

Where f0 is the resonant frequency (Hz), L is the inductance of the coil (H) and C is the value of capacitance (F). Band width = f2-f1 (Hz),

Where f1& f2 are the lower and upper cut off frequency

Quality factor = 0

2 1

Resonant frequency fQ =Band width f -f

=

Result: 1. Resonant frequency of the circuit = ______Hz 2. Self-inductance of the given coil = ______H 3. Quality factor from graph = ______ 4. Band width = ______Hz

L - Inductor , C- capacitor, R - variable resistor mA - ac milli-ammeter, SG - signal generator/oscillator

L C R

AC mA

S G


SERIES RESONANCE IN L-C-R CIRCUIT

Experiment No.05 Date:

Aim: To construct a series L-C-R circuit and to study the frequency response of the circuit. From the frequency response to calculate, a) the self-inductance of the given coil, b) quality factor of the circuit (Q- value) and c) the band-width of the circuit. Apparatus and other materials required: Signal generator, Inductor, Resistance box, Capacitor and AC milli-ammeter. Principle: L-C-R circuit fed by an alternating emf is similar to forced harmonic oscillator. The circuit contains an inductor L, capacitor C and resistor R in series. In this series LCR circuit, the inductive reactance XL increases with increase in frequency and the capacitive reactance XC decreases with increasing frequency. At resonant frequency f0 the inductive reactance and capacitive reactance are equal and the reactance is zero hence the total impedance of the circuit is minimum and there by the current is maximum. Therefore, at resonance XL = XC,

0 2 20 0

1 12πf L= L=2πf C 4π f C

⇒ Where f0 is the resonant frequency, L is the inductance of the coil and

C is the value of capacitance. After the resonance with increase in frequency the inductive reactance is more than the

capacitive reactance and hence the reactance and impedance will increase and the current decreases. The property of a reactive circuit to store energy is expressed in terms of quality factor or ‘Q’ factor. It is a figure of merit that enables us to compare different coils, it is given as follows

0ω ×energy storedQ=average power dissipated

where ω0 = 2πf0

Procedure: • Make the circuit connection as shown in the figure. • Unplug a small resistance say 1Ω or 2Ω from the resistance box. • Keep the frequency knob to the minimum position, DC offset in off position and set level

knob (ac voltage knob) to 25% of the maximum position before switch on the oscillator. • Select 1k Hz frequency knob. • Vary the frequency from the minimum value in convenient steps till the current in the circuit

reaches maximum and then decreases to minimum. Note down the current for increasing frequency.

• Take more number of readings of current for small variations in frequency around the maximum current.

• Plot a graph of frequency versus current and note down the maximum current and corresponding frequency (f0).

• Draw a horizontal line at IRMS=Imax/ 2 from the current axis such that it cuts the graph at two points A and B and note down the frequencies corresponding to A and B i.e., lower cut off frequency f1 and upper cut off frequency f2respectively.

• Calculate the self-inductance of the given coil, quality factor and band width using the relevant formulae.

Results: 1. Resonant frequency of the circuit= ___Hz; 2. Self-inductance of the given coil = ___H

3. Quality factor from graph= ______; 4. Band width= ______Hz


FERMI ENERGY OBSERVATIONS:

Diagram:

Model Graph:

Formula:

-15F

ρ A mE =1.36×10 Jl

-15

F 19

ρ A m1.36×10 JE = ............

1.6 10l eVC− =

×

EF is the Fermi energy (eV) T is the reference temperature (K), A is area of cross section of the given copper wire (m2) l is the length of the copper wire (m) Charge of the electron, e = 1.602 x 10-19 C. m is the slope of the straight line. ρ is the density of copper =8960 Kg/m3

Table:Sl. No.

Temp 0C

Temp K

V (mV)

I (mA)

R=V/I (Ω)

1. 90 2. 3. 4. 5. 6. 7. 8. 9. 10.

Result: Result: The Fermi energy of copper is EF =___________ J, ____________eV

mA

V +

-

- +

Thermometer

Copper coil

R

-

-

+

Res

ista

nce

in Ω

Temperature in K

A

B C


FERMI ENERGY OF COPPER


Aim: To determine the Fermi energy of copper Apparatus and other materials required: RPS, Multi meter, Milli ammeter, Beaker, Thermometer and copper wire. Theory: In a conductor, the electrons fill the available energy states starting from the lowest energy level. Therefore all the levels with an energy E less than a certain value EF(0) will be filled with electrons, whereas the levels with E greater than EF0 will remain vacant. The energy EF0 is known as Fermi energy at absolute zero and corresponding energy level is known as Fermi level. For temperature greater than zero Kelvin, Fermi energy is the average energy of the electrons participating in electrical conductivity. By measuring the resistance of the copper wire at different temperature, Fermi energy is calculated.

-15F

ρ A mE =1.36×10 Jl

EF is the Fermi energy T is the reference temperature (K),

A is area of cross section of the given copper wire (m2) l is the length of the copper wire (m) Charge of the electron,e = 1.602 x 10-19 C. ρ is the density of copper =8960 Kg/m3 m is the slope of the straight line obtained by plotting resistance of the metal against absolute temperature of the metal.

Procedure: • Make the circuit connections as shown in the diagram. • Set the multi meter to 200 or 2000 mV DC mode according to the requirement. • Set the current to about 6 to 8 mA by varying the power supply voltage and note down the

resistance of the coil at the ambient temperature. • Immerse the copper coil in a beaker containing water at about 90°C. • Note down the voltage in multi meter and current in milli ammeter for every decrement of

2°C to about 60°C. • Plot a graph of resistance along y-axis and temperature along x-axis and calculate the

value of slope m. • Calculate the Fermi energy of the material by using the relevant formula.

Result: Result: The Fermi energy of copper is EF =___________ J, ____________eV


HALL EFFECT OBSERVATIONS: BLOCK DIAGRAM OF EXPERIMENTAL SETUP

Figure 1.

Block diagram

Table 1: Sl. No Current

I (A) Magnetic field ‘B’ (Gauss)

1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

Magnetic Field B

Cur

rent

I

Calibration Curve

Voltage knobs of Hall effect setup

V I

Current knobs of

Hall effect

d A

B


HALL EFFECT

Experiment No: 07 Date: Aim: To study the Hall Effect in semiconductors / metals, to calculate the Hall Coefficient and to determine the concentration of charge carriers Apparatus and other materials required: Hall Effect setup, Hall Probe (Ge Crystal n or p type / Metal), Electromagnet, Constant Current Power Supply, Digital Gauss meter etc.,

Principle: When a metal or a semiconductor carrying current is placed in a transverse magnetic field B, a potential difference VH is produced in a direction normal to both the magnetic field and current direction. This phenomenon is known as Hall Effect. The Hall Effect helps to determine

1. The nature of charge carries e.g. whether semiconductor is an n-type or p-type 2. The majority charge carrier concentration 3. The mobility of majority charge carriers 4. Metallic or semiconducting nature of materials

Experimental Setup: The experimental setup for the measurement of Hall voltage and determination of Hall coefficient is shown in the figure 1. A thin rectangular germanium wafer is mounted on an insulating strip and two pairs of electrical contacts are provided on opposite sides of the wafers. One pair of contacts is connected to a constant current source and other pair is connected to a sensitive voltmeter. This arrangement is mounted between two pole pieces of an electromagnet, such that the magnetic field acts perpendicular to the lateral faces of the semiconductor wafer.

Procedure: Part A: Calibration of the magnetic field of the electro magnet • Adjust the gap between the pole pieces of the electromagnet such that the specimen kept

between the pole pieces is not in contact with them. Maintain the same gap throughout the experiment.

• Connect the electromagnet to the constant current supply source and keep the current varying knob at zero position.

• Place the Gauss probe between the pole pieces and switch on the digital gauss meter. Adjust the knob in it for zero reading.

• Switch on the constant current supply unit. Slowly increase the current from a minimum of 0.25 A, in convenient steps upto 4A and note down the corresponding magnetic field from the gauss meter.

• Note down the reading of current and magnetic field and enter the readings in the table 1. • Plot a graph of the current in the electromagnet and the magnetic field, this is called as

calibration graph. • Remove the gauss probe and switch off the digital gauss meter. • Reduce the current in the constant current supply to zero.


Model Graph: For n or p type

Table 2: Current through the crystal IC =……..mA Sl. No

Current I (A)

Magnetic field from calibration

curve (B)

Hall Voltage

VH (mV)

B (Gauss)

B (Tesla)

1. 2. 3. 4. 5. 6. 7. 8. 9.

10.

1 gauss = 10-4 Tesla

Formula:

i. Hall Coefficient: HH

C C

V t m tR =B I I

=

= …………….. Ωm/Tesla

Where VH = Hall Voltage in V IC = Current through the crystal in mA t = thickness of specimen in meters (t = 0.5 X 10-3m) B = Magnetic flux density in Tesla

ii. Carrier Concentration: 3

H

1n =qR

m−= − − − − −

Where q = Charge of electrons/holes in C, RH= Hall Coefficient in Ωm/Tesla Result:

Hall Coefficient (RH) of the material = …………… Ω-m/tesla Carrier Concentration (n) of the material = …………… /m3

Hall

Volta

ge V

H (m

V)

Magnetic Field B (Tesla)

B C

Hall Voltage V/s Magnetic

A

Slope=

AB×Scale on y-axisBC ×Scale on x-axis


Part B: Measurement of Hall voltage • Insert the Hall probe between the pole pieces in the electromagnet such that the crystal in

the Hall Probe is facing the north pole of the electromagnet. • The wires connected to the length of the crystal (Black and Red) are connected to the

current source, the wires connected to the breadth of the crystal (Green and Yellow) are connected to the voltage source in the Hall Effect setup.

• In the Hall Effect setup turn the selector knob (Toggle switch) to the current and set the crystal current (IC) to a small value say 1 mA by varying the current knob and note down the current IC. Maintain the same current IC throughout the experiment.

• Turn the selector knob to the voltage to measure Hall voltage VH and set the voltage to zero using offset knob.

• Vary the current in the electromagnet in random steps from 0.5A to 4.0 A with the help of constant current source. Note down the current ( I ) and the Hall voltage (VH) and enter the values in Table 2.

• For the currents ( I ) in the previous step, note down the magnetic field from the calibration curve and enter the values in the Table 2.

• Plot a graph of magnetic field B (Tesla) and a Hall voltage VH in V and find the slope. • Calculate the Hall co efficient and carrier concentration using relevant formulae. Result:

Hall Coefficient (RH) of the material = …………… Ω-m/tesla Carrier Concentration (n) of the material = …………… /m3


THERMISTOR – OBSERVATIONS: Diagram: Model Graph:

Formula: gE = 4.606 k m Joules× ×

g 19

4.606 k mE = eV1.6 10−

× ××

Where Eg = Energy gap of a given thermistor in eV k = Boltzmann constant = 1.381 x 10-23 J/K m = Slope of the graph Table:

Temp tC Temp T(K) R (Ω)

Log R 1/T

85

Result: The energy gap ( band gap) of the given thermistor is __________eV.

Ω

Thermometer

Thermistor

Multimeter

A

B C

Slope=AB×Scale on y-axisBC ×Scale on x-axis

Slope=AB/BC

1/T L

og R


BANDGAP OF A THERMISTOR Experiment No: 08 Date: Aim: To determine the energy gap (Eg) of a Thermistor. Apparatus and other materials required: Glass beaker, Thermistor, Multi meter, Thermometer.

Principle: A thermistor is a thermally sensitive resistor. Thermistor’s are made of

semiconducting materials such as oxides of Nickel, Cobalt, Manganese and Zinc. They are

available in the form of beads, rods and discs.

The variation of resistance of thermistor is given by Tb

eaR = where ‘a’ and ‘b’ are

constants for a given thermistor. The resistance of thermistor decreases exponentially with

rise in temperature. At absolute zero all the electrons in the thermistor are in valence band

and conduction band is empty. As the temperature increases electrons jump to conduction

band and the conductivity increases and hence resistance decreases. By measuring the

resistance of thermistor at different temperatures the energy gap is determined.

Formula : 194.606

1.6 10gk mE eV−

× ×=

×

Where Eg = Energy gap of a given thermistor in eV.

k = Boltzmann constant = 1.381 x 10-23 J/K .

m = Slope of the graph.

Procedure: • Make the circuit connection as shown in the figure.

• Keep the multi meter in resistance mode (200 Ω range).

• Insert the thermometer in a beaker containing thermistor and note down the resistance at

room temperature.

• Immerse the thermistor in hot water at about 90C.

• Note down the resistance of the thermistor for every decrement of 2°C upto 60C.

• Plot the graph of log R versus 1/T and calculate the slope ‘m’.

• Calculate the energy gap of a given thermistor using relevant formula.

Result: The energy gap of the given thermistor is __________eV.


DIELECTRIC CONSTANT OBSERVATIONS:

Diagram:

Model Graph 1: Charging Model Graph 2: Discharging

_____s ; C =Rτ

= ______F _____s ; C =Rτ

= ______F

Formula: The capacitance and dielectric constant of the given capacitor are given below:

1. Capacitance τC= (F)R

; 2. Dielectric Constant r0

Cdε =ε A

Where τ : time constant, εr: relative permittivity or the dielectric constant of the dielectric, oε : Absolute permittivity of free space = 8.854x10-12F/m., C: capacitance of the capacitor

(F), A: area of each plate (m2), d: thickness of the dielectric (m). The value of C = obtained from equations 1 & 2 should be calculated separately.

The average value of the capacitance is determined. Thickness of dielectric d (m) Area of each plate A (m2)

=τ =τ

Rτ

0 20 40 60 80 100 1200.0

0.5

1.0

1.5

2.0

2.5

3.0

63.2%Vm Growth

Vm

Volta

ge (V

)

Time (s)τ 0 20 40 60 80 100 120

0.0

0.5

1.0

1.5

2.0

2.5

36.8%VmDecay of voltage

VmVo

ltage

(V)

Time (s)

τ


DIELECTRIC CONSTANT Experiment No: 09 Date: Aim: 1. To determine the capacitance of the parallel plate capacitor.

2.To determine the dielectric constant of the dielectric material in an electrolytic capacitor by the method of charging and discharging of the capacitor.

Apparatus and materials required: Power supply, electrolytic capacitor, Multimeter, two

way key and stop clock. Principle: In an electrical circuit containing a capacitor and resistance the growth as well as decay of voltage are opposed by the induced emf. Therefore, it takes some finite time for voltage to reach its maximum value when the circuit is switched on. Similarly when the circuit is switched off the electric voltage takes again some finite time to become zero. Formula: The capacitance and dielectric constant of the given capacitor are calculated by using the formulae given below:

1. C = 𝜏𝑅 (F) 2.

where : Time constant. : Relative permittivity or the dielectric constant of the dielectric.

oε : Absolute permittivity of free space = 8.854x10-12F/m. C: Capacitance of the capacitor (F). A: Area of the dielectric (m2). d: Thickness of the dielectric (m). Procedure:

( I ) Charging of the capacitor: • Make the circuit connection as shown in the diagram. • Open the circuit at ‘a’ and ‘b’. • Keep the current knob for the maximum position, keep the course volts and fine volts

knobs at the minimum position before switching ON the RPS. • Keep the battery voltage to suitable value (say 2 to 4V). • Keep the multimeter position in 20 V DC. • Set the stopwatch for zero. • Close the circuit at ‘a’ and ‘b’ and simultaneously start the stop watch. • Note down the voltage across the capacitor using multi meter for every thirty seconds of

charging time till the voltage reaches saturation. • Plot a graph of voltage versus time. Find out the time constant by taking 63.2% of the

saturation voltage as shown in the graph.

ACd

or ε

ε =

τrε


Table R = ___Ω; Battery voltage=___V

Time in seconds

(s)

Voltage during

Charging (V)

Voltage during

Discharging (V)

30 60 90 120 150 180 210 240 270 300 330 360 390 420 450 480 510 540 570 600

.

. Result: 1. Capacity of parallel plate capacitor C = _________________F.

2. Dielectric constant of the given dielectric material rε = _____ .

Note: Data

C=3300µF C=4700 µF R=47 kΩ R=47 kΩ L=47 cm L=55 cm B=1.5 cm B=2.5 cm d=80µm d=80 µm


(II) Discharging of the capacitor:

• Set the stop watch to zero. • Disconnect the circuit at ‘a’ and ‘b’ and connect the circuit at ‘a’ and ‘c’ and

simultaneously start the stop watch. • Switch off the power supply (RPS). • Note down the voltage across the capacitor using multi meter for every thirty seconds of

time of discharge till the voltage reaches minimum value. • Plot a graph of Voltage versus time of discharge. Find the time constant by taking 36.8%

of the saturation voltage as shown in the graph.

(III) Calculation of Capacitance and Dielectric constant

• Calculate the average time constant ‘τ = RC’ from charging and discharging graph. • Note down length, breadth and thickness of the dielectric for the given capacitor and

hence calculate its dielectric constant. Note: Multiply the dielectric constant value with the correction factor 10-6. This is because the dielectric in the electrolytic capacitor is alumina on a paper, the total thickness of the dielectric is not the alumina. Result: 1. Capacity of parallel plate capacitor C = _________________F.

2. Dielectric constant of the given dielectric material rε = _____ .


LEE’S AND CHARLTON’S METHOD OBSERVATIONS:

Diagram: Experimental Setup Model Graph:

H : Steam chamber, S: Sample disc (bad conductor), B: metallic disc, T1 and T2 are the Thermometers Formula:

1 2

dθmct dtK=A(θ -θ )

Where, Mass of the metallic disc B, m = …. kg Specific heat of the material of B, c = ….. J/kg/K Thickness of the poor conductor S, t = ……. m Radius of the poor conductor S, r= ………. m Area of cross section poor conductor A= ……….. m2 Steady temperature of steam chamber ‘H’, θ1= ⁰C Steady temperature of disc B, θ2= ⁰C

Rate of cooling from the graph dtdθ =

A

B

T

Time (t) in seconds

θ2

C


THERMAL CONDUCTIVITY OF A POOR CONDUCTOR BY LEE’S AND CHARLTON’S METHOD


Aim: To determine the thermal conductivity of a given poor conductor by Lee’s and Charlton’s method.

Apparatus: Lee's disc apparatus, a brass disc (B) of large diameter compared to its thickness, sample of poor conductor in the form of the disc (S) with same diameter, thermometers, stop watch, screw gauge and a balance.

Principle: At steady state there will be constant temperature gradient across the sample and the rate of flow of heat across the sample is constant. In the determination of thermal conductivity of the poor conductor the rate of flow of heat through the poor conductor is equated to the rate of loss of heat through the brass disc. i.e.,

1 2(θ -θ ) dθQ=KA = m×ct dt

Where k - Thermal conductivity of the sample, A - Surface area of the sample disc θ1 – θ2- Temperature difference across the sample, t - Thickness of the sample disc where ‘m’ and ‘c’ are the mass and specific heat of the disc ‘B’ respectively.


Thickness of the given sample using screw gauge: Zero error =

Pitch = distance moved on the pitch scale /No. of rotations givens given to head scale

LC = Pitch/ Total number of head scale divisions = ____mm

Table 1:

Trial No. PSR mm HSD TR=PSR+(HSR-ZE)×LC mm 1 2 3 4

Thickness of the poor conductor =……………….m Table 2: Rate of cooling of the metallic disc

Sl. No. Time (min) Time in s Temperature of metallic disc T ⁰C

1 1 2 2 3 3 4 4 5 5 6 6 7 7 8 8 9 9 10 10

Result: The thermal conductivity of a given poor conductor by Lee’s and Charlton’s method is found to be…………………...Wm-1K-1


Procedure: • Determine the mean thickness of the given sample disc ‘S’ using a screw gauge and

tabulate the values in Table 1. • Keep the given sample disc of poor conductor ‘S’ between metal disc ‘B’ and the steam

chamber ‘H’. • Introduce the thermometerT1through hole in the steam chamber (H) and thermometer

T2in the metal disc (B).Pass the steam through the chamber ‘H’ until the temperature indicated by thermometers T1 and T2 become steady and note the steady temperature θ1 and θ2 respectively.

• Remove the poor conductor ‘S’ and keep the steam chamber in direct contact with the brass disc till its temperature rises about 8ºC to 10ºC above θ2.

• Remove the steam chamber and cover the top surface of the brass disc with filt. • Note down the decrease in temperature of the brass disc for every one minute using a

stop clock, till the temperature of the disc is 10 0C below it's steady state θ2and enter the readings in Table 2.

• Plot a graph of time versus temperature. Draw a tangent to the curve at the point corresponding to its steady temperature θ2and find the slope =

dtdθ of the tangent.

• Determine the thermal conductivity of poor conductor using the given formula. Result: The thermal conductivity of a given poor conductor by Lee’s and Charlton’s method is found to be…………………...Wm-1K-1


FOUR PROBE’S METHOD OBSERVATIONS Diagram: Model Graph:

Figure 1. Four probe unit

Formula:

Resistivity of a semiconductor Vρ= 2S mI

Ω

Where V is the voltage across two probes I is the current across two probes S is the inter probe spacing

Energy band gap of a semiconductor g -194.606 k mE = eV1.6×10

Where k is the Boltzmann constant = 1.38X10-23 J/K m is the slope of the graph

Figure 2. Arrangements of four probes Table 1:

Probe spacing S = 0.2 cm, Current ( I ) = ……………. mA. S.No Temperature

(°C) Temperature

T (K) Voltage (mV)

Resistivity ρ (Ω cm)

Log10 ρ 1/T (K-1)

1. 150 2. 145 3. 140 4. 5. 6. 7. 8. 9. 10 11

Result:

1. The resistivity of a given material……….. with decreasing temperature and hence the material is found to be…………………

2. The energy band gap of the given semiconductor is ………………….eV

1/T

Log

ρ

A

B C

G

B

R

Y


Electrical resistivity by Four Probe’s method Experiment No: 11 Date: Aim: To determine, a) The resistivity of Semiconductor, b) The energy band gap. Apparatus: Four Probe arrangement, Oven, Semiconductor and thermometer

Formula: Resistivity of a semiconductor Vρ= 2S mI

Ω

Energy band gap of a semiconductor g -194.606 k mE = eV1.6×10

Where, V is the voltage across two probes, I is the current across two probes, S is the inter probe spacing, k is the Boltzmann constant = 1.38X10-23 J/K, m is the slope.

Principle: The two-probe method gives an average value of the resistivity. If the specimen ingots are of irregular shape, resistivity determination becomes difficult due to the difficulty in measurement of cross sectional area. The four probe method avoids the necessity of measurement of cross sectional area and is suitable for arbitrary shaped samples. Experimental Setup: The four probe unit consists of four pointed pressure contacts which are equally spaced about 2mm apart. The four probes are loaded over the sample so that all the four probes which are of equal height make contact with the sample. Procedure: 1. Ensure that the specimen is placed on the base plate of the four probe arrangement. 2. Connect the pair of green and yellow leads to the constant current power supply and the

black and red leads to the voltage terminals. 3. Place the four probe arrangement in the oven and place the thermometer in the oven

through the hole provided. 4. Adjust the current to 5 mA and keep it constant throughout the experiment. 5. Switch on the oven and heat the sample up to 150 0C. 6. Unplug the oven connection from the four probe setup after reaches 150°C and note

down the value of voltage for every 50 fall in temperature starting from 150 °C. 7. Plot a graph of Log10ρ versus 1/T and calculate its slope. 8. Energy gap of a given semiconductor is calculated by substituting the value of slope in

the above formula. RESULT:

1. The resistivity of a given material……….. with decreasing temperature and hence the material is found to be…………………

2. The energy band gap of the given semiconductor is ………………….eV Precautions: 1. Do not apply pressure on the electrical contacts to the specimen as it is very brittle.


SEARLE’S METHOD OBSERVATIONS: Experimental setup:

Table 1: Determination of the cross-sectional area ‘A’ of the bar and the distance x between the holes C1 and C2

Sl No

Diameter (D) of the bar Cross-sectional area 𝐀 = 𝛑𝐃𝟐

𝟒

(cm2)

MSR (cm)

CVD TR= MSR + (CVDx LC) Mean diameter D

(cm) 1. 2. 3. 4.

Table 2: Determination of temperature gradient

Distance of the thermometers (cm)

Thermometer readings (°C)

X1 θ1 X2 θ2 X3 θ3 X4 θ4

Model Graph:

mKdxd

BCABSlope /........===

θ

x

θ A

B C

dxd

BCAB θ

=

T1 T2 T3 T4 T6 T5

C1 C2 C3 C4

X1 X2 X3

Water

X4


THERMAL CONDUCTIVITY OF A GOOD CONDUCTOR BY SEARLE’S METHOD


Aim: To determine the thermal conductivity of a given conductor using Searle’s method. Apparatus: Searle’s apparatus, sample of conductor in the form of a bar, thermometers, digital stop watch, water tank, beaker, vernier calipers, meter scale, measuring cylinder etc. Principle: At steady state there will be constant temperature gradient across the specimen bar and the rate of flow of heat across the specimen bar is constant. In the determination of thermal conductivity of the good conductor, the rate of flow of heat through the good conductor is equated to the rate of loss of heat by circulating water through the specimen bar.

Therefore )( 56 θθθ−= mct

dxdKA

Theory: In this Searle’s method there are two conditions to be satisfied (i) The rate of heat flow through the conductor must be measurable, i.e., a sufficient amount of heat must flow through it per second and (ii) The temperature gradient along it must be steep enough to be accurately measured. In a metal, the heat flow is fast enough hence the first condition is satisfied. If the length is very much greater than cross sectional area of the bar, the second condition is satisfied. In the laboratory, we are conducting the experiment satisfying both conditions cited above. The experimental set up is shown in the figure. The set up consists of a cylindrical bar ‘Y’ of a metal with its one end inside a steam chamber and the other end, a copper tube is wound helically around it through which cold water enters at inlet I1 from a constant water storage or from direct tap and exits at water outlet I2. Thermometers T1, T2, T3 and T4 placed in four mercury filled cups C1, C2, C3 and C4 measures the temperatures θ1, θ2, θ3 and θ4 respectively. Care is taken to see that placing the thermometers do not alter the rectilinear flow of heat. The bar is lagged with felt and enclosed in a wooden box, this helps to reduce heat loss from the bar and maintains a constant temperature gradient. This arrangement permits us to measure temperature gradient in one part of the bar and the rate of heat flow in the other part without much error. The temperatures θ5 and θ6 of the incoming and outgoing water in the copper coil are measured by the thermometers T5 and T6 respectively. The water flowing out of the copper tube is collected for a certain period of time ‘t’.


Table 2: Recording of Temperatures

Observation Time (minutes)

Thermometer readings in 0C Difference (θ5~θ6) 0C θ1 θ2 θ3 θ4 θ5 θ6

After passing the steam

0 2 4 6 8 10 12 14

Table 3: Time of flow (t) and the mass (m) of the water collected

Table 4: Determination of K No. of

sets

Rate of flow m/t

(g/s)

(θ6 ~ θ5)°K Cross-sectional area A

Specific heat of water at θ6

‘c’

Value of K cal/°C

1 Thermal conductivity of a given conducting bar is given by

avtm

dxdA

cK

−

=θθθ

.

).( 56

Where, K is the thermal conductivity of a given good conductor =K= --------- J K-1m-1s-1

c is the mean specific heat of water at θ6 = -----------------J kg-1 K-1. Steady state temperature between (θ6 – θ5) = ----------- º C. (m/t) is the mass of water flowing/ second = -------------g/c =----------x10-3 kg/s

No. of

obs.

Time of collection

t (sec)

Volume of water

collected V (cm3)

Mean Temp. θ6°C

Density of water ρ for the Mean

Temp. θ6°C (Kg/m3)

Mass of water

m=ρV (g)

m/t (g/s)

Mean m/t (g/s)

1 60 2 120 3 180 4 240 5 300


When the steady state is reached i.e., when the temperature readings are stationary, then the heat flowing through any section of the rod is the same. In this state, heat flowing through the rod is taken up entirely by the circulating water at the farther end of the bar. Heat flowing through the rod in a time ‘t’ is Q.

Q= KA tdxdθ …………(i)

Where Q is the heat flowing through the bar, ‘A’ is the area of cross section of the bar, dxdθ

is the temperature gradient.. If ‘m’ is the mass of water collected in ‘t’ second and ‘c’ is the mean specific heat of water at the mean temperature of θ5- θ6. ratio of heat gained by water is given by Q = mc(θ6- θ5) ………….(ii) Under steady state conditions, the two quantities of heat are equal

Therefore KA tdxdθ = mc(θ6- θ5) Or

avtm

dxdA

cK

−

=θθθ

.

).( 56

Where, K is the thermal conductivity of a given good conductor =K= --------- J K-1m-1s-1

c is the mean specific heat of water at θ6 = -----------------J kg-1 K-1.

Steady state temperature between (θ6 – θ5) = ----------- º C.

(m/t) is the mass of water flowing/ second = -------------g/c =----------x10-3 kg/s


(Not to be written in the practical record) REFERENCE DATA

Table 5: Specific heat of water at various temperature

Table 6: Variation of density of water with temperature Temp. (0C)

Density (kg m-3)

Temp. (0C)

Density (kg m-3)

Temp. (0C)

Density (kg m-3)

Temp. (0C)

Density (kg m-3)

20 998.2071 31 995.344 42 991.4394 53 986.6537

21 997.9955 32 995.0292 43 991.0388 54 986.1791 22 997.7735 33 994.706 44 990.6310 55 985.6981

23 997.5414 34 994.3745 45 990.2162 56 985.2111 24 997.2995 35 994.0349 46 989.7944 57 984.7178

25 997.0479 36 993.6872 47 989.3657 58 984.2185 26 996.7867 37 993.3316 48 988.9303 59 983.7131

27 996.5161 38 992.9683 49 988.4881 60 983.2018 28 996.2365 39 992.5973 50 988.0392 61 982.6846

29 995.9478 40 992.2187 51 987.5839 62 982.1615 30 995.6502 41 991.8327 52 987.1220 63 981.6327


Procedure: 1. Measure the diameter of the rod at different places using a slide calipers. Also the

measure the distance ‘x’ of the centers of the holes C1, C2, C3 and C4 from the hot end of the rod with a meter scale.

2. Six thermometers are inserted in their proper positions (T1,T2,T3, T4, T5 and T6 as shown in the figure).

3. Pass the steam continuously through the steam chamber. As the temperature of the bar rises, the readings of the thermometers T1, T2, T3 and T4 will increase.

4. Arrangement for the circulation of water through the spiral copper tube at a constant rate. The constancy of the rate of flow of water is achieved by keeping the water level in the water tank.

5. Take the readings of T1, T2, T3, T4, T5 and T6 as θ1, θ2, θ3, θ4, θ5 and θ6 at intervals of 2 minutes, until they exhibit steady reading at least for ten minutes. These steady values indicate that the bar has attained steady state and the steam end of the bar has attained the steam temperature.

6. Plot a graph of the temperature θ of the thermometers as a function of the distance ‘x’ of the thermometers from hot end of the copper rod.

7. Find the slope ‘M’ of the graph and it is the temperature gradient dxdθ in the copper rod.

8. In the steady state, collect a volume V of the water in measuring cylinder for a time t, measured with a stop watch. Multiply the volume V by the density of water at the mean temperature of ‘θ5 & θ6’ to obtain the mass ‘m’ of the water collected. Find the value of water collected per unit time (m/t).

9. Repeat step 6, for four more value of ‘t’ and find the mean (m/t). Calculate the values of K using the relevant formula.

Result: The thermal conductivity of the given conducting bar is found to be ‘K’ =…………….J K-1m-1s-1.

Precautions to be taken. 1. Water should be collected after the steady state has been attained.

2. Avoid air bubbles while collecting water

3. See that there is no leakage of water at the corks through which the thermometers pass.


NUMERICAL APERTURE AND LOSS IN OPTICAL FIBER OBESERVATIONS: Diagram: Experimental Setup: Part A: Numerical aperture measurement

Formula:

𝑁.𝐴. = sin𝜃0 = 𝑊(4𝐿2+𝑊2)

,

W→ diameter of the beam spot, L→ distance from the Optical Fiber to the screen Part B: Measurement of Transmission loss

LED

PHOTODIODE

V/10 in dBm

DPM Pout Knob

Power meter

O F

LED

PHOTODIODE

V/10 in dBm

DPM Pout Knob

Power meter

L

W

O F


NUMERICAL APERTURE AND ESTIMATION OF LOSS IN OPTICAL FIBERS


Aim: Part A: To determine the Numerical aperture of the given Optical Fiber

Part B: To measure the transmission loss in the given Optical Fiber

Apparatus: Optical Fiber Kit, Optical fiber cables, In-line adapter , Numerical Aperture Jig.

Part A: To determine the Numerical aperture of Optical Fiber

Principle:

Optical fibers are wave guides that transmit light from one point to another. The principle behind the propagation of light in the optical fiber is Total Internal Reflection (TIR) at the core-cladding interface.

Acceptance angle (θ0) is the maximum angle from the axis of the optical fiber at which the light ray may enter the fiber so that it may propagate via TIR in the core.

Numerical Aperture (NA): It is the light gathering capacity of the optical fiber. Sine of acceptance angle gives the numerical aperture.

2 221 21 2

0 20 1 0

n nn nSinθ 1n n n

−= − =

Where, n1 and n2 are the refractive indices of the core and cladding of the optical fiber respectively, n0 is the refractive indices of the surrounding medium (n0=1)

Formula: 𝑁.𝐴. = sin𝜃0 = 𝑊(4𝐿2+𝑊2)

,

W→ diameter of the beam spot, L→ distance from the Optical Fiber to the screen

Procedure:

• Connect one end of the OF cable (1-meter/2 meter/5 meter) to LED and the other end to the numerical aperture jig as shown in the figure

• Plug the kit to the AC mains and switch on the circuit board. Light should appear at the end of the fiber on the NA jig

• Turn the Pout knob clockwise to set to maximum Pout. The light intensity should increase

• Hold the white screen with the concentric circles to make the red spot from the emitting fiber coincide with the center of the circle. Avoid any bends in the optical fiber

• Note down the diameter of the beam spot for different distances as per table A • Compute the numerical aperture from the given formula for the different distances.


Table A:

S. No W (mm) L (mm) NA θ = sin-1(NA) in degree

1.

2.

3.

4.

𝑁.𝐴. = sin𝜃0 = 𝑊(4𝐿2+𝑊2)

,

Table B:

Sl. No Length (cm) (A) Loss in dBm

Length (cm) (B) Loss in dBm

αL (A-B)

Loss in dBm for 3m

1 1 4 2 2 5 3 3 6

Avg Loss (α L) =__________ dBm

Attenuation coefficient

Average Attenuation lossα = dB/m

α =------------------ dB/km

Length=


Part B: Measurement of Transmission loss in the given Optical Fiber

Principle:

Attenuation can be defined as the loss of power due to the total length of the fiber while transmitting light. The major factors contributing to the attenuation in optical fiber are

i) Absorption loss, ii) Scattering loss, iii) Bending loss, iv) Intermodal dispersion loss and

v) coupling loss.

These losses are a consequence of material, composition, structural design of the fiber and can be minimized by taking proper care in selection of materials, design and the operating wavelengths.

Attenuation in fiber is measured in terms of attenuation coefficient. It is denoted by symbol α. mathematically attenuation of the fiber is given by,

10 log( / )α out inP PL

×= − dB/km

Where Pout and Pin are the power output and power input respectively, and L is the length of the fiber in km.

Procedure:

• Connect one end of OF cable (1 meter) to the OF LED and the other end to the photo diode.

• Connect the Digital Panel Meter (DPM) to the power meter as shown in the figure B • Plug in AC mains, after relieving all twists and strains on the fiber,fix the output (PO)

knob to -15dBm. Note this as transmission loss in 1m OF. • Repeat the above procedure for different cable lengths (as given in table) by joining

the 1m OF cables using the inline adapter (ILA)

Result: The numerical aperture of the given optical fiber is =

The acceptance angle θ is = _______ °

The transmission loss in optical fiber (α/L) = _______ dB/m


VIVA VOCE CHARACTERISTICS OF LASER BEAM 1. What is meant by diffraction?

Bending of waves round the edges of an obstacle is called diffraction. 2. What is the condition for diffraction?

Size of the obstacle should be comparable with that of the wave length of the light source. Since grating constant and wave length are of the same order (10-6 metre), diffraction takes place within the grating.

3. Distinguish between diffraction and dispersion? Diffraction: Bending of light round the edges of an obstacle is called diffraction. In this case lower the wave length lesser will be the deviation. Dispersion: When white light passes through a prism it splits into its constituent colors. This phenomenon is called dispersion. In this case lower the wave length higher will be the deviation.

4. Distinguish between polychromatic & monochromatic source.

Polychromatic source a source having different wave lengths. Ex. Mercury vapour lamp. Monochromatic source is a source having single wave length. Ex Sodium vapour lamp.

5. What does LASER stands for?

The term LASER stands for Light Amplification by Stimulated Emission of Radiation.

6. What are the characteristics of laser radiation? Laser radiations have high intensity, high coherence, high monochromaticity and high directionality with less divergence.

7. What is population inversion?

When the number of atoms are more in higher energy state than in the lower energy state, this condition is known as population inversion, it is essential for stimulated emission.

8. What is pumping in a laser?

It is the process in which atoms are excited to higher energy states by continuously supplying energy.

9. What is meant by the term coherence?

The state of vibration, same phase or constant phase difference is known as coherence.

10. What is an active medium? A solid, liquid, or gaseous medium in which population inversion can be achieved is called an active medium.

11. What is the action of an optical resonator? It gives the directionality to the laser beam and amplifies the laser beam.


STEFAN’S LAW

1. What is a ‘black body’? Black body is a body which absorbs all wavelengths which are incident on it and emits all those wavelengths on heating. This total energy emitted or absorbed is independent of the nature of the body and depends on its temperature.

2. How does this law differ from Newton’s law of cooling?

Newton’s law of cooling is applicable only when the difference of temperature between the body and the surroundings is very small. There is no such conditions for Stefan’s law.

3. Can the value of Stefan’s constant be determined from this method?

Yes, taking the value of ‘n’ as 4, the value of σ can be calculated from the value of the intercept C of the graph.

4. What for Stefan’s law is used? It has the advantage like finding the temperature of Sun.

5. What are the limitations of Stefan’s law? This method is even though not very precise and accurate, it has some advantages. The bulb is never truly a black body and at steady state, the power radiated is never equal to V.I exactly. The working theory in this method is to some extent approximate, nevertheless, the method is very simple and the accessories are easy to procure. It gives an approximate idea about Stefan’s Law, Stefan’s constant and the verification of the law.

6. State Stefan’s law.

It states that total radiant energy emitted per second per unit surface area of a perfectly black body is proportional to the fourth power of its absolute temperature.

7. State Kirchoff’s law of black body radiation.

It states that at any temperature, the ratio of emissive power to the absorptive power for a given wavelength is same for all black bodies. A good emitter is a good absorber of heat.

8. What is emissive power and absorptive power?

Emissive power: At a particular temperature and for a given wavelength, it is defined as the radiant energy emitted per unit time per unit surface area of the body within a unit wavelength range. Absorptive power: At a particular temperature and for a given wavelength, it is defined as the ratio of the radiant energy absorbed per second per unit surface area of the body to the total energy falling per unit time on the same area.

9. State Wien’s law?

The wavelength corresponding to the maximum energy emitted from a black body is inversely proportional to its absolute temperature (λmaxT = Constant).

10. Explain the energy distribution in a black body radiation spectrum? The energy radiated by a black body is not uniformly distributed over the entire wavelength range emitted by the body but it is maximum for a particular wavelength and


decreases on either side. The value of wavelength at which the energy radiated peaks depends on the temperature of the black body.

11. Mention the properties of black body radiation?

Black body radiation has all the property of electromagnetic radiation.

12. What are the modes of transmission of heat? Conduction, convection and radiation. Conduction is the mode of transmission of heat in which heat travels from the hot part of the body to its cold part without actual motion of heated atoms. Convection is the mode of transmission of heat from one part of the medium to the other by the bodily motion of heated atoms. Radiation is mode of transmission of heat from one body to another without the intervening medium.

13. Distinguish between Heat and Temperature?

Sensation of hotness is called heat. Degree of hotness is called temperature. WAVELENGTH OF LED’s 1. What is LED?

Light Emitting Diode is a heavily doped p n junction, where voltage yields a flow of current.

2. What is quatntum?

According to Quantum theory, it is a bundle of energy given by E = hυ = h cλ

Where E is the energy of the quantum, h is the Planck’s constant and υ is the frequency of the light emitted.

3. What is ‘ħ’ (h cross)?

It represents angular momentum.2πhp ==

4. Discuss the mechanism How light is produced from LED?

A photo P-N junction can convert absorbed light energy into an electrical energy. The same process is reversed here (i.e. the P-N junction emits light when electrical energy is applied to it). This phenomenon is generally called electroluminescence, which can be defined as the emission of light from a semi-conductor under the influence of an electric field. The charge carriers recombine in a forward-biased P-N junction, as the electrons cross from the N-region and recombine with the holes existing in the P-region light is emitted. Free electrons are in the conduction band of energy levels, while holes are in the valence energy band. Thus the energy level of the holes will be lesser than the energy levels of the electrons. Some portion of the energy must be dissipated in order to recombine the electrons and the holes. This energy is emitted in the form of heat and light

5. What is semiconductor diode laser?

Semiconductor diode laser is a specially fabricated p-n junction. It emits laser light when it is forward biased and the applied current must be above the threshold value.

https://en.wikipedia.org/wiki/Electroluminescence

https://en.wikipedia.org/wiki/Semi-conductor

https://en.wikipedia.org/wiki/Electric_field

https://en.wikipedia.org/wiki/Electric_field

https://en.wikipedia.org/wiki/Conduction_band

https://en.wikipedia.org/wiki/Energy_band


TORSION PENDULUM

1. What is meant by Torsion Pendulum? A pendulum in which the oscillations are due to the torsion in the suspension wire.

2. Describe Torsion pendulum?

It consists of a rigid body (disc, rod, etc.,) attached to the lower end of the wire, whose top end is fixed to the rigid support and it is subjected to rotational oscillations.

3. Define moment of inertia?

It is the opposition for the rotational motion. When rigid body rotates about an axis, it has tendency to oppose the change in its state of rest or of uniform rotation about its axis. This tendency is called moment of inertia of a body about the axis of rotation. Quantitatively it is the product of the mass and radius of gyration.

4. Define inertia of a body. What is the measure of the inertia of a body?

The property of a body by virtue of which every body tries to stay in its state of rest or uniform motion along a straight line unless compelled by an external force. Mass is the measure of the inertia of a body.

5. What are the factors on which moment of inertia of a body depend? Moment of inertia depends on a) mass of the body b) The distribution of mass about the axis of rotation.

6. Why (I/T2) is a constant for a given wire?

For a torsion pendulum, the period of torsion oscillation is Where I = the

moment of inertia about its axis, C = couple per unit twist of the suspended wire. Rearranging (I/T2) = (C)/ (4Π2) Since “C” (couple per unit twist) is a constant for a given wire; I/T2 is a constant.

SERIES RESONANCE

1. What is an Inductor? A non resistive coil of wire in which there will be an opposing emf when there is varying current is passing through it. It is a passive component used to store energy in the form of magnetic field.

2. What is Resistance?

The resistance of a conductor is the opposition offered by the conductor to the flow of electric current through it. The opposition is due to the collision of electrons with ion cores of the conductor. It is independent of frequency.

3. What is Impedance?

Impedance measure the effective opposition to the flow of current due to the reactance and resistance. It is frequency dependent.

4. What is Inductive reactance?

The opposition offered by the inductor to the flow of AC is called inductive reactance (XL). XL = ωL = 2πfL.

CIT π2=


5. What is capacitive reactance? The opposition offered by the capacitor to the flow of AC is called capacitive reactance (XC). XC == 1/ωC = 1/C2πf

6. What do you mean by resonance in LCR series circuit?

The condition at which the current is maximum due to the matching of inductive and capacitive reactances.

7. What is Quality factor? Explain the variation of quality factor with change in resistance of

the circuit. It is defined as the ratio of resonant frequency to the bandwidth of the circuit. Quality factor

measures the sharpness of resonance CL

RQ 1

= or 0

2 1

fQ=f -f

The smaller the value of resistance, the greater is the current at resonance and the resonance curve is sharper. As the resistance is increased, the sharpness of resonance decreases and the circuit becomes less selective.

8. What is Bandwidth?

It is the difference between upper and lower cut off frequencies. Bandwidth is the applicable range of frequencies.

9. What are the applications of LCR resonant circuits?

They are used as tuning circuits in radio and television receivers.

FERMI ENERGY

1. What is Fermi energy of a metal? It is the energy of the highest occupied level at absolute zero temperature..

2. What is meant by Fermi factor? It is the probability of occupation of given energy state by a charge carrier.

3. What is meant by Fermi temperature TF? What is the relation between EF& TF? It is the temperature at which the average thermal energy of the free electrons in a solid becomes equal to the Fermi energy at 0°K. EF = kBTF

4. What is meant by Fermi velocity?

It is the velocity of those electrons which occupy the Fermi level. It is given by EF = 1

2 mvF2

5. What is Fermi Dirac distribution?

It gives distribution of electrons / fermions among the various available energy levels of a material under thermal equilibrium conditions.

6. What are the factors on which EF depend? EF depends on the material and the temperature.

7. How many electrons will be there in each energy level? According to Paul’s exclusion principle, there will be two electrons in each energy level.


8. State Paul’s exclusion principle? It states that no two electrons can have all the four quantum numbers same.

9. What is meant by free electron? Free electron is the electron which moves freely in the absence of external field. These electrons collide with each other and also with the lattice elastically and hence there is no loss in energy.

HALL EFFECT 1. What is Hall Effect, Hall voltage and Hall field?

If a metal or a semiconductor carrying a current I is placed in a transverse magnetic field B, a potential difference is produced in the direction normal to both the current and magnetic field directions. This phenomenon is called Hall Effect. The corresponding potential difference generated is called Hall voltage and the electric field generated is called the Hall field.

2. Define Hall co-efficient. It is numerically equal to Hall electric field induced in the specimen crystal by unit current when it is placed perpendicular in a magnetic field of 1 W/m2.

3. Define mobility.

It is the ratio of average drift velocity of charge carriers to applied electric field. 4. What is Fleming’s Left Hand Rule?

Stretch thumb, first finger, middle finger at right angles to each other such that fore finger points in the direction of magnetic field, middle finger in the direction of current then thumb will point in the direction of the force acting on it.

5. How does mobility depend on electrical conductivity?

It is directly proportional to conductivity. 6. Which type of charge has greater mobility in semiconductors?

In semiconductors, electron has greater mobility than holes. 7. What happens to the hall coefficient when number of charge carriers are decreased?

Hall coefficient increases with decrease in number of charge carriers per unit volume. 8. What are the applications of Hall effect?

It is used to verify whether a substance is a semiconductor, conductor or insulator. Nature of charge carriers and mobility of charge carriers can be studied. This experiment can be used to measure magnetic field strength.


BAND GAP OF THERMISTOR 1. What is a band gap?

This is the energy gap between the conduction and valence bands of a semiconductor (or insulator).

2. How do you differentiate between a conductor, an insulator and a semiconductor in relation to energy gap? In conductors, the valence and conduction bands overlap each other. In insulators, there is large energy gap between valence and conduction bands, while in semiconductors this energy gap is not too large so that at room temperature the thermal energy gained by some of the electrons in the valence band is sufficient to make them jump to conduction band, crossing this energy gap.

3. What is intrinsic and extrinsic semiconductors? A pure or natural semiconductor free of impurities is called an intrinsic semiconductor e.g. silicon and germanium. But it has low electrical conductivity. When some pentavalent (like arsenic) or trivalent (like boron) impurity is added to it, then it is called an extrinsic semiconductor.

4. Why a semiconductor behaves as an insulator at zero degree Kelvin? At 0K, electrons in valence band do not have sufficient energy to cross the energy band gap so as to reach to conduction band and to make them available for conduction. Thus semiconductor behaves as an insulator.

5. What do you mean by doping and what is the effect of doping on depletion region? Doping is a process of introducing a small amount of impurity into the intrinsic semiconductor. As the doping concentration increases width of the depletion region decreases.

6. Explain the effect of doping in a germanium crystal. The conductivity of germanium crystal is enhanced by adding the impurities from V or III group of periodic table e.g., arsenic or antimony from V group or indium from III group. When antimony or arsenic having five electrons in their outermost shell are introduced into the germanium crystal, the fifth electron of the impurity atom does not find a place in the symmetrical covalent bond structure and is free to move through the crystal. These free electrons are then available as current carriers. When a III group elements like indium is added each atom leaves behind a hole.

7. On what factors does the conductivity of an extrinsic semiconductor depend? The conductivity of an extrinsic semiconductor depends upon the type and amount of impurity added.

8. What do you mean by a hole? Hole indicates the deficiency or absence of an electron. It effectively behaves like a positively charged particle when an electric field is applied across the crystal.

9. What are values of energy gap of silicon & germanium? Silicon & Germanium energy gap values are 1.1eV & 0.7eV respectively at 20°C.


10. What is the effect of temperature on the energy gap of a semiconductor? As temperature increases energy gap decreases.

11. What are the factors on which energy gap of semiconductor depend? It depends on the material & temperature.

DIELECTRIC CONSTANT 1. What is a capacitor?

Capacitor is a device used to store charge.

2. What is meant by capacitance? Ability to store charge in a capacitor is called capacitance and it is measured in farad.

3. What is the relation between Q, C & V?

Q=CV, Q, charge stored (coulomb), C capacitance (farad), V (voltage).

4. Define one farad. It is the amount of charge required to raise the potential by 1 volt.

5. What is a dielectric?

Dielectric is an insulator which is used to increase the capacitance of the capacitor.

6. Classify dielectrics. Dielectrics are classified into polar and non-polar dielectrics.

7. On what factors the dielectric constant depend?

It depends on frequency, material and temperature.

8. What is ‘time constant’ of a circuit? The time constant of a circuit consisting a resistor and a capacitor, is the time taken by the capacitor to acquire 63% of its maximum charge.

9. What is the expression for time constant?

If the capacitance of the capacitor is C farad and resistance in the circuit is R ohm, then the time constant is equal to RC.

10. How charging and discharging depend on time constant?

The smaller the time constant, more rapid is the charging and discharging of capacitor.

11. What is dielectric constant of a dielectric? The dielectric constant of a dielectric is the factor by which the capacitance of the capacitor is increased when vacuum between the two plates of the capacitor is replaced by the dielectric material.

12. What is the effect of an alternating field on the dielectric constant of a material?

Dielectric constant depends on the frequency of the applied field. Dielectric constant decreases as the frequency of the applied field increases.


LEE CHARLTON’S METHOD 1. Define thermal conductivity. Does the value of K depend on the dimension of the

substance? Thermal conductivity is defined as the quantity of heat flowing per second, normally a unit cube when the temperature difference between the two opposite faces of the cube is unity. No. The value of K depends only on the material of the body.

2. What is the unit of K? Give its dimensions. J. K-1 m-1s-1→ [K]=[MLT-3θ-1]

3. Can you measure the thermal conductivity of a good conducting disc by this method? This method cannot be used to determine the thermal conductivity of a good conducting disc since the temperatures θ1 and θ2 will be nearly equal and the measurement of their difference will be very difficult.

4. Can you apply this method to determine the thermal conductivity of a liquid? The present method can be used to determine the thermal conductivity of a liquid if the disc S is replaced by a thin-walled flat-end copper container with the edges made of bad conductors. The experimental liquid is taken in this container.

5. What is the largest source of error in this experiment? The largest source of error lies in the measurement of rate of cooling.

6. Why is the specimen taken in the form of a disc? For a disc the thickness is small whereas the cross-sectional area is large. Hence the amount of heat that is conducted is large.

7. Is it absolutely necessary that the steam chamber be placed at the top? No. it can also be placed below the bad conductor.

8. Why do you measure the thickness of the disc in situ? When the disc is placed between the two slabs its effective thickness decreases due to pressure. Therefore the thickness is measured under the experimental condition of heat flow.

ELECTRICAL RESISTIVITY BY FOUR PROBE’S METHOD 1. What is four probe method?

An experiment in which four contacts probes are used.

2. What is four probe method used for? It is used to accurately determine the resistivity of materials of irregular films.

3. What is the principle used in four probe method?

Current is forced through outer probes 1 & 4 & a drop voltage across probes 2 & 3 is measured.


4. What is an ohmic material? A material whose resistance is a constant is called ohmic material or a material which obeys ohms law is an ohmic material.

5. What is meant by electrical resistivity (ρ)? It is the property of the material by the virtue of which it opposes the flow of current. It is also defined as the reciprocal of electrical conductivity (Ωm)-1. Its unit is Ωm i.e., ρ= l/σ = RA/L.

SEARLE’S METHOD 1. Why is the experimental bar kept in an enclosure?

To minimize the loss of heat by conduction and radiation

2. Is the method suitable for bad conductors? No, in this case, the temperature difference (θ1-θ2) will be very large so that there will be practically no flow of heat to the other end of the tube.

3. What would happen if the bar is thin?

The error in the measurement of its cross-sectional area would be large.

4. What is the temperature gradient in this experiment? It is the ratio of change in temperature to the distance between corresponding temperature measurement point.

5. How can the temperature be more accurately measured?

By replacing the thermometers by thermocouples

6. Does the conductivity of metals or alloys depend on temperature? How? Yes. The conductivity of pure metals decreases with increase of temperature whereas that of alloys increases.

OPTICAL FIBER

1. What is the basic principle of an optical fiber? Optical fiber works based on the principle of total internal reflection [TIR]

2. What is an optical fiber? An optical fiber is a wave guide system through which light signals are carried over longer distances without loss of energy.

3. What is Numerical aperture (NA)? The light gathering capacity of an optical fiber is called Numerical Aperture. NA= Sinθ Where θ=acceptance angle.

4. What is an acceptance angle of an optical fiber? It is the maximum angle of incidence at the core of an optical fiber so that the light can be guided through the fiber. Acceptance angle θ=Sin-1(NA).


5. Why optical fibers do not pick up electricity? Optical fibers are made by pure non-metallic materials hence they won’t allow electricity.

6. What is Total Internal Reflection (TIR)?

Total internal reflection is the phenomenon in which complete reflection of a light ray occurs into the same medium, when a propagated wave strikes a medium boundary at an angle larger than a critical angle with respect to the normal to the surface.

7. What is an acceptance angle? The angle θ0 is called the wave guide acceptance angle or the acceptance cone half-angle which is the maximum angle from the axis of optical fiber at which light ray may enter the fiber so that it will propagate in core by total internal reflection.

8. What is an attenuation in optical fiber? The total power loss offered by the total length of the fiber in the transmission of light is called attenuation.

9. What are the different types of optical fiber? The optical fibers are classified under three categories. They are 1) Single Mode Step Index Fiber, 2) Step Index Multi Mode Fiber and 3) Graded index Multi Mode Fiber

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DEPARTMENT OF PHYSICS R.V.COLLEGE OF ENGINEERING · PDF fileto enable engineering graduates to understand, leverage and appreciate the role of physics for the development of sustainable