PHYSICS LABORATORY MANUAL - Vidyarthiplus

PERI Institute of Technology / Department of Physics /Engineering Physics Lab

PHYSICS LABORATORY MANUAL

FOR FIRST YEAR B.E/B.Tech. DEGREE COURSES

(Common to all branches)

DEPARTMENT OF PHYSICS

PERI INSTITUTE OF TECHNOLOGY

MANNIVAKKAM, CHENNAI-48

www.Vidyarthiplus.com



LIST OF EXPERIMENTS

First semester

1. (a) Particle size determination using Diode Laser.

(b) Determination of Laser parameters-Wavelength and angle of divergence.

(c) Determination of acceptance angle in an optical fibre.

2. Determination of the thickness of a thin wire - Air wedge method.

3. Determination of velocity of sound and compressibility of liquid - Ultrasonic

Interferometer.

4. Determination of wavelength of mercury spectrum – spectrometer grating.

5. Determination of thermal conductivity of a bad conductor – Lee‘s Disc method.

6. Determination of hysteresis loss in a ferromagnetic material.

Second semester

7. Determination of Young‘s modulus of the material – Non uniform bending.

8. Determination of Band gap of a semiconductor material.

9. Determination of Specific resistance of a given coil of wire – Carey Foster Bridge.

10. Determination of Viscosity of liquid – Poiseuille‘s method.

11. Spectrometer Dispersive power of a prism.

12. Determination of Young‘s modulus of the material – Uniform bending.

13. Torsional pendulum – Determination of Rigidity modulus.




INDEX

Ex. No Date Name of the experiment Marks Staff initial Remarks

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.







PRECISION INSTRUMENTS

I. VERNIER CALIPER

The vernier Caliper consists of a main scale (fixed scale) and a moving vernier. In a

metric vernier the main scale is marked in centimeters and millimeters. The vernier is 9

millimeter long and is divided into 10 divisions each 0.9 millimeter long.

Before taking a reading with the vernier caliper the least count and zero error have to be

determined.

Least Count

10 Vernier Scale Divisions (VSD) = 9 Main Scale Division (MSD)

Value of 1 VSD = 9/10 MSD

Value of 1 MSD = 1 mm

Hence value of 1 VSD = 9/10 mm

Least Count (LC) = 1 MSD – 1 VSD

= 1mm - 9/10 mm

= 1/10 mm = 0.1 mm

Least Count (LC) = 0.01cm

When the two jaws A & B of the vernier Caliper touch each other and if the zero of the

vernier coincides with zero of the main scale division there is no zero error (Fig 1.1)

When the two jaws are in contact and if the zero of the vernier is to the right of the zero

of the main scale the error are positive. In this case let 5th vernier division coincide with definite

main scale division (Fig 1.2)

Value of zero error = + (VSC x Least Count)

= + 5 x 0.01 x 10 –2

m

= 0.05 x 10 –2

m




When the two jaws are in contact and if the zero of the vernier is to the left of the zero of

the main scale the error are negative. Let 5th vernier scale division coincides with definite main

scale divisions (fig 1.3).

Then value of zero error = - (10 – 5) x LC

= - 5 x 0.01x 10 –2

m

= - 0.05 x 10 –2

m

Determination of thickness

The object whose thickness to be measured is held between the two jaws. The main scale

reading and the vernier scale coincidence are noted.

Now observed reading (OR) = MSR + (VSC x LC)

The correct reading (CR) = OR + (zero error)

CR = OR + ZE

Thickness of the given object is measured at different places and the observations are

tabulated in Table (1) and mean value is taken.

Table 1:

Least count (L.C.) = 0.01cm Zero Error (ZE) = + ……div

Zero Correction (ZC) = + …… cm

S.No. MSR

x 10 –2

m

VSC

div

OR = MSR + (VSC x LC)

x 10 –2

m

CR = OR + ZE

x 10 –2

m

1.

2.

3.

4.

5.

Mean = x 10 –2

m







II. SCREW GAUGE

A Screw gauge can measure distances with more precision than a Vernier Calipers. The

screw gauge consists of a barrel (Pitch scale) moving within a Head Scale (Sleeve). The Head

Scale is divided into 100 divisions. The head scale moves over the pitch scale, which is

graduated in 1 mm. A ratchet knob is provided for closing the screw gauge on the object being

measured without exerting too much force.(fig.2)

To make a measurement with the screw gauge, the pitch, least count and zero error have

to be determined.

Least Count = DivisionscaleheadofNumber

divisionscalePitchofValue

....

...1..

= 100

1mm

LC = 0.01mm

When the tip of the screw and the stud are in contact if the zero of the head scale exactly

coincides with the zero of the pitch scale (Fig 2.1) there is no zero error.

If the zero of the head scale is below the reference line on the pitch scale, the error is

positive. In this case let 5h head scale division coincides with the reference line (Fig 2.2)

Then the value of zero error = HSC x LC

= 5 x 0.01 x 10 –3

m

= + 0.05 x 10 –3

m

If the zero of the head scale is above the reference line then the zero error is negative. Let

95th head scale division coincides with the reference line, in this case (Fig 2.3).

Then the value of zero error = - (100 – 95) x LC = -0.05 x 10 –3

m




Determination of thickness of an object

The object whose thickness is to be measured is held lightly between A and B . The head

scale coincidence division and the pitch scale reading are noted.

The Observed Reading (OR) = PSR + (HSC x LC)

Correct Reading (CR) = OR + ZE

Observations are repeated for different places on the object. The readings are tabulated in

Table (2) and mean value is found.

Table 2:

Least count (L.C.) = 0.01.mm Zero Error (ZE) = + ……div

Zero Correction (ZC) = + …… mm

S.No. PSR

x 10 –3

m

HSC

div

OR = PSR + (HSC x LC)

x 10 –3

m

CR = OR + ZE

x 10 –3

m

1.

2.

3.

4.

5.

Mean = x 10 –3

m







III. TRAVELLING MICROSCOPE

The Travelling microscope consists of a compound microscope that is capable of

independent vertical and horizontal movements fitted with main scales (M1 and M2) and Vernier

scales (V1 and V2). The microscope attached with Vernier V1 can be raised or lowered along M1

by a screw S1. Similarly V2 is moved to and fro horizontally with a screw S2. Two screws (L1

and L2) at the base of the microscope are used for its leveling. The main scale is marked in

millimeters and half millimeters. The vernier is divided into 50 divisions.

Least Count

Value of 1 MSD = 0.05 x 10-2

m = cm20

1

Number of divisions in vernier scale = 50

Least count verniertheindivisionsofNo

MSDValueof

......

1

= 0.05/ 50 cm

LC = 0.001 cm

In the case of travelling microscope

Correct Reading (CR) = MSR + (VSC x LC)

Measurement of diameter of a Capillary tube

The Capillary tube is held horizontally and the microscope is focused on the cross

section. The vertical cross wire is made tangential in two diametrically opposite points of the

bore of the tube and the difference in the horizontal scale readings is taken. The experiment is

repeated by making horizontal cross wire tangential to the bore and the difference in vertical

scale reading is taken and the observations are recorded in table 3.




Table 3:

L. C = 0.001cm

Position

Microscope Reading

Diameter

x 10 –2

m

MSR

x 10 –2

m

VSC

div

CR = MSR + (VSC x

LC) ×10 –2

m

Mean = x 10 –2

m







IV. SPECTROMETER

The essential parts of a spectrometer are the collimator, prism table and the telescope. A

circular graduated disc (Main scale) is attached to the telescope and hence rotates with it. The

telescope can be fixed at any desired position by a clamping screw while finer adjustments can

be made with a tangential screw.

The prism table is made up of two circular platforms one above the other connected

together by three leveling screws. A circular disc with two verniers 180 apart is mounted

coaxially with the main scale. (fig.4)

Before taking measurements using spectrometer, it is essential to make the following

adjustments.

i) Eye piece is adjusted to get a clear image of the cross wires.

ii) The telescope is turned to a distant object and is focused to get a clear image of the

object.

iii) The slit of the collimator is illuminated with monochromatic light and the image of

the slit is viewed through the telescope. The collimator screw is adjusted to get the

sharp image of the slit.

iv) The leveling of the prism table is done with a sprit level.

Least Count

The main scale is graduated in half degrees. There are 30 divisions in the vernier.

The value of 1 MSD = ½degree = 30‘

Least Count = verniertheindivisionsofNo

MSDValueof

......

1

= 30/30

Least Count = 1

In spectrometer the correct reading CR = MSR + (VSC x LC)

Illustration for measurement

The telescope is turned and the vertical cross wire is coincided with sharp image of the

illuminated slit. The main scale reading immediately before vernier zero is noted. The vernier

coincidence division with definite main scale reading is also taken. The readings are taken

making use of both the verniers.




Let one of the vernier called vernier A shows main scale reading (MSR) 50 with vernier

scale coincidence of 15th

division with one of the main scale reading (Fig4.1)

Since the correct reading (CR) = MSR + (VSC x LC)

CR = 50 + 16 x 1‘

CR = 5016‘

Now let Vernier B main scale reading is 230 and vernier scale coincidence division is 16

(Fig 4.2)

Then CR = 230 + 16 x 1‘

CR = 23016‘







1. SEMICONDUCTOR LASER

Aim:

(a) To determine the size of the micro particle using laser.

(b) To determine the wavelength of a given laser and angle of divergence.

(c) To determine the acceptance angle and numerical aperture of an optical fibre.

Apparatus Required:

Laser grating, laser source, Fine micro particles of nearly uniform size (lycopodium

powder), Glass plate, white screen, stands, and meter scale, optical fibre cables of various length,

optical fibre connections, numerical aperture jig, mandrel for optical fibre.

Formula:

(a)Size of the micro particle (diameter) nx

Dnd

metre

(b1)Wavelength of given laser source of light Nn

sin metre

(b2)Angle of divergence reedd

rrdeg

180

12

12

(c1) Acceptance angle θa =

d

r1tan degree

(c2) Numerical aperture of the optical fiber aNA sin

Explanation of symbols

Symbol Explanation Unit

θ Angle of diffraction degree

N Number of lines per metre in grating Lines/m

n Order of diffraction ---

λ Wavelength of the laser source metre

Xn Distance of the nth

order ring from the central spot of the

diffraction pattern

metre

D Distance between the glass plate and the screen metre

r1 Radius of the beam spot at a distanced d1 metre

r2 Radius of the beam spot at a distanced d2 metre

θ Acceptance angle degree

r Radius of the circle/spot metre

d Distance between the fibre end and screen metre




Determination of wavelength of a given laser:

Distance between grating and screen (D) = _________10- 2

m

Number of lines per metre in the grating (N) = ________lines/m

Order

Of

diffraction

(n)

Distance of the

Centre of the

Spot from the

Central maximum

2

21 xxx

10- 2

m

D

xtan

degree

D

x1tan

degree

Nn

sin

m

Left

(x1)

10- 2

m

Right

(x2)

10- 2

m

1

2

3

4

5

Mean λ = ----------m

= ----------10-10

m

= ----------- Å




Procedure:

(a) To determine the wavelength of a given laser:

Diode laser is kept horizontally and switched on. The grating is held normal to laser

beam. This is done by adjusting the grating in such a way that reflected laser beam coincides

with beam coming out of laser source.

After adjusting for normal incidence, the laser light is exposed to grating and it is

diffracted by it on the other side of grating on the screen the diffracted laser spots are seen.

The diameter of different orders from centre spot is measured. The distance between the

grating and screen is measured. The wavelength of laser light source is calculated using the

given formula.

(b) To find the angle of a divergence

Angle of divergence gives the angular spread of the laser beam. A simple diagrammatic

explanation of finding the angle of divergence is shown. Here the laser source and a stand is

kept at some distance say d1 and, the radius of the beam spot is measured, Now by varying

the distance to d2, the radius of the spot is again measured. By substituting these values in

the given formula, the angle of divergence can be calculated. The experiment is repeated for

various values of d1 and d2 and the mean angle of divergence is determined.

(c) To determine the size of the micro particle using laser.

Sprinkle a thin uniform layer of lycopodium powder on a glass plate. Mount the screen

and glass plate upright. The light from laser source transmitted through the layer of lycopodium

in the glass plate is adjusted to form a diffracted image in the centre of the screen. Diffracted

circular fringes of laser colour will be visible on the screen.

After adjusting the distance of the glass plate from the screen so that the first ring radius (x1)

and second ring radius (x2) are measured from the central spot. Note the distance (D) between

screen and plate. Repeat the experiment radius of the first and second rings after adjusting the

distance between screen and plate. Calculate the value of the diameter of the particle taking λ

value from the previous experiment




(d) To determine the acceptance angle and numerical aperture of an optical fibre

The numerical aperture jig consists of an iron or plastic stand with a moving screen.

In this screen, a number of concentric rings of varying diameter are present. In front of it, a

stand with a circular slit in the centre is provided which is connected to the laser light source

through the optical fibre cable. By moving the screen back and forth the laser light from the

circular slit is made to fall exactly on the circles with different diameters. The distance ‗D‘

between the circular slit in the jig and screen for various circular diameters (and hence radius

(r)) are noted on a moving scale situated at the bottom of the jig. Thus by knowing the values

of D and r, the value of the numerical aperture is calculated. The maximum divergent angle

(the acceptance angle) is also determined using the formula

Result:

(i) The wavelength of the laser λ = ------------Å

(ii) The average size of the micro particle measured using laser d = -------μm

(iii) Angle of divergence = ___________ degrees.

(iv) The acceptance angle of the given optical fiber θa = ---------degree

(v) The numerical aperture of the given optical fiber NA = -------




Angle of divergence

Determination of angle of divergence:

S.No. r1 r2 d1 d2 Angle of divergence

180

12

12

dd

rrdegrees

10- 2

m 10- 2

m 10- 2

m 10- 2

m degree




Particle size determination by LASER

Determination of size of the micro particle

S.No

Distance between

the glass plate

and the screen

(D)

cm

Order of

Diffraction

(n)

Distance between the

central spot and the nth

fringe(xn)

cm

Particle size nx

Dnd

m610

1

2

3

1

2

3

Mean d = ________10-6

m

= ________ μm




Experimental setup for Numerical aperture measurement

Numerical aperture measurement jig

Measurement of Numerical aperture and Acceptance angle of an optical fibre:

S.No

Radius of the

Circle/spot

‘r’ x 10

-3 m

Distance

between the

fibre end and

Screen (d) x 10

-2 m

Acceptance

Angle

θa =

d

r1tan

degree

Numerical

Aperture

aNA sin

1.

2.

3.

4.

5.

Mean= θa = NA =




VIVA VOCE:

1. What is semiconductor diode laser?

Semiconductor diode laser is a specially fabricated pn junction diode. It emits laser light

when it is forward biased.

2. What is laser?

The term LASER stands for light Amplication by Stimulated Emission of Radiation. It is

a device which produces a powerful, monochromatic collimated beam of light in which the

waves are coherent.

3. What is stimulated emission?

The process of forced emission of photons caused by incident photons is called

stimulated emission.

4. What are the characteristics of laser radiation?

Laser radiation has high intensity, high coherence, high monochromatism and high

directionality with less divergence.

5. Define numerical aperture.

Numerical aperture is defined as the light gathering capability of an optical fibre. It is the

sine of the acceptance angle.

aNA sin

6. What is the principle used in fiber optic communication system?

The principle behind the transmission of light waves in an optical fiber is total internal

reflection.

7. Define acceptance angle?

The maximum angle ‗θa‘ with which a ray of light can enter through one end of the fiber

and still be total internally reflected is called acceptance angle of the fiber.







1. Determination of the thickness of a thin wire - Air wedge method.

Aim:

To determine the thickness of a wire by forming interference fringes using air wedge.

arrangement.

Apparatus required:

Microscope, two optically plane glass plates, sodium vapour lamp and a thin wire.

Formula:

Thickness of the given wire,

2

lt metre



λ Wave length of sodium light metre

l Distance of the wire from the edge of contact metre

β Mean width of one fringe metre

Procedure:

Light from the sodium vapour lamp is rendered parallel by means of a condensing lens

(L). The parallel beam of light is incident on a plane glass plate (G) inclined at an angle of 45°

and gets reflected. The reflected light is normally on the glass plate in contact. Interference takes

place between the light reflected from the top and bottom surfaces of the glass plates and is

viewed through the microscope (M). Due to interference large number of equally spaced dark

and bright are formed which are parallel to the edge of the contact.

The microscope is adjusted so that the bright (or) dark fringe near the edge of contact is

made to coincide with the vertical cross wire of the microscope and this is taken as the nth

fringe. The reading from the horizontal scale of the travelling microscope is noted. The

microscope is moved across the fringes using the horizontal traverse screw and the readings are

taken when the vertical cross wire coincides with every successive 5 fringes (5, 10, 15, 20 ---).

The width of every 10 fringes is calculated. From these observations, average width of one fringe




To find the fringe width: Specimen:

L.C. = 0.001cm

Order

Of the

fringes

Microscope Reading Width of

5 fringes

(cm)

Width of one

fringe

β

(cm) MSR

(cm)

VSC

(div)

TR =MSR+(VSC×LC)

(cm)

n

n+5

n+10

n+15

n+20

n+25

n+30

n+35

n+40

n+45

n+50

β =

To find the distance between the edge of contact and the wire

L.C = 0.001cm

Position of the

Microscope

Microscope Reading

l = R1 R2

(cm) MSR

(cm)

VSC

(div)

TR

(cm)

At the edge of

Contact (R1)

At the edge of

Material of

wire (R2)

Calculation:

Wavelength of the sodium light (λ) = 5893×10-10

m

Thickness of the given wire,

2

lt metre

Thickness of the given wire, t= --------------------metre




(β) can be obtained. The cross wire is fixed at the inner edge of the rubber band and the

reading from the microscope is noted. Similarly reading from the microscope is noted by keeping

the cross wire at the edge of the material. The difference between these two values gives the

value of l. Substituting β and l. In the formula, thickness of the given material can be

determined.

Result:

Thickness of the wire, t = --------m




VIVA VOCE:

1. What do you mean by interference of light?

When the two waves superimpose over each other, resultant intensity is modified. The

modification in the distribution of intensity in the region of superposition is called interference.

2. Is there any loss of energy in interference phenomenon?

No, there is only redistribution of energy i.e. energy from dark places is shifted to bright

places.

3. What is an interference fringe?

They are alternately bright and dark patches of light obtained in the region of

superposition of two wave trains of light.

4. What type of source is required in division of amplitude?

In division of amplitude a broad source is required so that the whole firm may be viewed

together.

5. What is the shape of fringes in wedge shaped film?

The fringes in wedge-shaped film are straight line fringes.




Ultrasonic interferometer

Distance moved by the reflector Vs Crystal current




2. ULTRASONIC INTERFEROMETER

Aim:

(i) To find the velocity of ultrasonic waves in the liquid.

(ii) To find the compressibility of the given liquid.

Apparatus Required:

High frequency generator, Measuring cell and given liquid

Formula:

(i) Wave length of ultrasonic waves n

d2 metre

(ii) Velocity of ultrasonic waves

in a given liquid v= metre/second

(iii)Compressibility of the liquid 2

1K metre

2/Newton



Frequency of the generator which excites the

crystal

Hertz

d Distance moved by the reflector metre

n Number of oscillations ---

ρ Density of the liquid Kg/m3

Procedure

The high frequency generator is switched on and the alternating field from the generator

is applied to the quartz crystal. The quartz crystal produces longitudinal ultrasonic waves. These

waves pass through the liquid and get reflected at the surface of the reflector plate. If the distance

between the reflector and crystal is exactly a whole multiple of the sound wavelength, standing

waves are formed within the medium. This results in the formation of acoustic resonance and

causes a change in the potential difference at the generator which excites the crystal. Due to this,

anode current of the generator becomes maximum. The change in the anode current can be

measured from the micrometer fitted with the frequency generator.




Least count for micrometer:

L.C = scaleheadindivisionsofnototal

Pitch

......

Pitch = scaleheadtogivenrotationsofnototal

scaleheadbymovedcedis

.......

....tan

To find the velocity (V) of ultrasonic waves in the liquid using interferometer:

Type of liquid = ………….. Frequency of the Generator = ………….Hz

TR = PSR + (HSC LC)

L.C =0.01mm

S.

No.

No of

oscillation

s

Micrometer reading d

10- 3

m

= n

d2

10- 3

m

PSR

10- 3

m

HSC

div

TR

10- 3

m

1. x(initial)

2. x+10

3. x+20

4. x+30

5. x+40

6. x+50

Mean λ =

Calculation

Frequency of the generator which excites the crystal =……… Hertz

Wavelength of Ultrasonic wave generated = ………metre

Density of the given liquid = ……… kgm-3

Distance moved by the micrometer screw d =………metre

Number of oscillations n = ……….

(i) Velocity of ultrasonic wave in the given liquid metre/second

V = …………….ms-1

(ii) Compressibility of the given liquid 2

1K metre

2/Newton

K = …………… m2N

-1




The distance between the reflector and crystal is varied using the micrometer screw such

that the anode current decreases from maximum to minimum and then increases upto a

maximum.

The distance of separation between successive maxima or minima in the anode current is equal

to half the wavelength of the ultrasonic waves in the liquid. By noting the initial and final

position of the micrometer for one complete oscillation (maxima-minima-maxima), we can

determine the distance moved by the parallel reflector as shown. Thus ‗n‘ number of successive

maxima or minima are recorded for a distance‗d‘. The total distance moved by the micrometer

screw is given by

mn

d2

or wavelength mn

d2

From the value of λ, the velocity of the longitudinal ultrasonic waves is calculated using

the relation, = vλ, Where v is the frequency of the generator (2 MHz) which is used to excite

the crystal. After determining the velocity of the ultrasonic waves in liquids, the compressibility

of the liquid is calculated using the formula K=1/v2ρ where ρ is the density of the liquid. The

experiment is repeated for different liquids.

Result:

(i) Wave length of ultrasonic waves ( ) = metre

(ii) The Velocity of Ultrasonic wave in the given liquid (V) = ……………… ms-1

(iii) Compressibility of the given liquid (K) = ……………… m2N

-1




Viva voce:

1. What are ultrasonic‘s?

The sound waves having frequencies above the audible range. i.e., frequencies above

20,000 Hz or 20 kHz are known as ultrasonic.

2. What is piezo electric effect?

When mechanical pressure is applied to one pair of opposite faces of a quartz, then the

other pair of opposite faces develop equal and opposite electrical charges on the crystal.

3. What is inverse piezo-electric effect?

The piezo electric effect is reversible. If an electric field is applied to one pair of opposite

faces of the quartz crystal, alternative mechanical expansion or contraction (pressure) is

produced across the other pair of opposite faces of the crystal.

4. What is an acoustic grating?

When ultrasonic waves travel through a transparent liquid, due to alternating compression

and rarefaction, longitudinal stationary waves are formed. If monochromatic light is passed

through the liquid perpendicular to these waves, the liquid behaves as a diffraction grating. Such

a grating is known as ―acoustic grating‖.







3. SPECTROMETER – GRATING

Aim:

To find the wavelengths of the prominent spectral lines in the mercury (Hg) source

Apparatus required:

Spectrometer, Plane transmission grating, Sodium vapour lamp, Mercury vapour lamp,

and Reading lens.

Formula:

Wavelength of the prominent lines in the mercury (Hg) source λ = Nn

sinm



θ Angle of diffraction degree

n Order of diffraction (spectrum)

N Number of lines per metre in the grating. lines/metre

Procedure:

1. Adjustment of the grating for normal incidence

The initial adjustments of the spectrometer are made as usual. The plane transmission

grating is mounted on the grating table. The telescope is released and placed in front of the

collimator. The direct reading is taken after making the vertical cross-wire to coincide with the

fixed edge of the image of the slit which is illuminated by a source of light (Sodium vapour lamp

or mercury vapour lamp). The telescope is then rotated by an angle 90° (either left or right side)

and fixed. The grating table is rotated until on seeing through the telescope the reflected image of

the slit coincides with the vertical cross-wire. This is possible only when a light emerging out

from the collimator is incident at an angle 45° towards the collimator. Now light coming out

from the collimator will be incident normally on the grating.




(i) To find N

Wavelength of spectral line λ= 5893×10-10

m

TR = MSR+ (VSC×LC)

LC= 1’

Colour of

spectral line

Diffracted ray reading Difference (2θ)

deg

θ=2 θ/2

deg Mean

Θ

deg

N=

n

sin

lines/m

Left side Right side

Vernier-A Vernier-B Vernier-A Vernier-B VA A1 ˜A2

deg

VB

B1 ˜B2

deg

VA

deg

VB

deg

MSR

deg

VSC

div

TR

(A1)

deg

MSR

deg

VSC

div

TR

(B1)

deg

MSR

deg

VSC

deg

TR

(A2)

deg

MSR

deg

VSC

div

TR

(B2)

deg




2 Wavelength of the spectral lines of the mercury spectrum

The slit is now illuminated by white light from mercury vapour lamp. The central

direct image will be an undispersed image. The telescope is moved on both sides of the direct

image; the diffraction pattern of the spectrum of the first order is seen. The readings are taken

by coinciding the prominent lines namely violet, green, yellow and red with the vertical cross

wire. The readings are tabulated and from this, the angles of diffraction for different colours

are determined. The wavelengths for different line are calculated by using the given formula.

The number of lines per metre in the grating is also calculated.

Result:

(i) Number of lines drawn in the grating per metre = __________ lines/metre.

(ii) Wavelength of prominent spectral lines in the mercury spectrum are

calculated and tabulated

VIVA VOCE

1. In the present experiment, what class of diffraction does occur and how?

Fraunhofer class of diffraction occurs. Since the spectrometer is focused for parallel rays,

the source and the image are effectively at infinite distances from grating.

2. What is plane transmission diffraction grating?

A plane transmission diffraction grating is an optically plane parallel glass plate on

which equidistant, extremely close grooves are made by ruling with a diamond point.

3. How are commercial gratings made?

A commercial grating is made by pouring properly diluted cellulose acetate on the

actual grating and drying it to a thin strong film. The film is detached from the original

grating and is mounted between two glass plates. A commercial grating is called a replica

grating.




Determination of wavelength (λ) of the prominent line of the mercury spectrum

LC = 1' Order of the spectrum (n) =

N = ________lines/metre Total Reading = MSR + (VSC × LC)

Colour of

spectral line

Diffracted ray reading

Difference (2θ) θ=2 θ/2 Mean

Θ

deg

λ=Nn

sin

m

Left side

Right side

Vernier-A

deg

Vernier-B

deg

Vernier-A

deg

Vernier-B

deg

VA

A1 ˜A2

deg

VB

B1 ˜B2

deg

VA

deg

VB

deg

MSR

deg

VSC

div

TR

(A1)

deg

MSR

deg

VSC

div

TR

(B1)

deg

MSR

deg

VSC

div

TR

(A2)

deg

MSR

deg

VSC

div

TR

(B2)

deg




Calculation:

Number of lines per metre in the grating

N=

n

sin lines/m N=

Violet λ = Nn

sin =______________ m

Green λ = Nn

sin=______________ m

Yellow λ = Nn

sin=______________ m

Red λ = Nn

sin=_______________ m




Time Vs Temperature




4. Determination of thermal conductivity of a bad conductor – Lee’s

Disc.

Aim:

To determine the thermal conductivity of the bad conductor by Lee‘s Disc method

Apparatus required:

Lee‘s disc apparatus, Bad conductor, Thermometers, Stop-clock, Steam boiler, Screw

gauge and Vernier caliper.

Formula:

Thermal conductivity of the bad conductor

11

2)21

)(22(2

2

KWm

dt

d

hrr

hrMSdK



M Mass of the metallic disc kg

S Specific heat capacity of the material of the disc Jkg-1

K-1

2

dt

d

Rate of cooling at steady temperature θ2

°C/s

θ1 Steady temperature of a steam chamber °C

θ2 Steady temperature of the metallic disc °C

r Radius of the metallic disc metre

h Thickness of the metallic disc metre

d Thickness of the bad conductor metre




(i) To find the thickness of the card board (d) using screw gauge:



S.No PSR

(mm)

HSC

(div)

HSR =

(HSCxLC)

(mm)

OR=PSR+HSR

(mm)

CR=OR+ZC

(mm)

Mean thickness of the cardboard (d) =

(ii) To find the thickness of the metallic disc (h) using screw gauge:



S.No PSR

(mm)

HSC

(div)

HSR

(HSCxLC)

(mm)

OR=PSR+HSR

(mm)

CR=OR+ZC

(mm)

Mean thickness of the metallic disc(h) =




Procedure

The thickness of the bad conductor and thickness of the metallic disc are determined

using a screw gauge. The radius of the metallic disc is found using vernier calipers. The mass

of the metallic disc is also found using a common balance. The observed readings are

tabulated in the respective tabulations.

The Lee‘s disc apparatus is suspended from a stand as shown. The given bad

conductor (card board) is placed in between the metallic disc and steam chamber. Two

thermometers T1 and T2 are inserted in the respective holes.

Steam from the steam boiler is passed into the steam chamber until the temperature of

the steam chamber and the metallic disc are steady. The steady temperatures (1) of the steam

chamber and (2) of the metallic disc are noted. Now the card board is removed and the steam

chamber is placed in direct contact with the metallic disc. The temperature of the disc rapidly

rises. When the temperature of the disc rises about 10oC above 2

oC (Steady state temperature

of the disc), the steam chamber is carefully removed, after cutting off the steam supply. When

the temperature of the disc reaches 5 o

C above the steady state temperature of the disc i.e.,

(2+5) oC

, a stop clock is started. Time for every 1oC fall of temperature is noted until the

metallic disc attains the temperature (2 -5) o

C. The observed readings are tabulated.

Graph

A graph is drawn taking time along the X – axis and temperature along the Y – axis.

The cooling curve is obtained To obtain the rate of cooling

2

dt

d from this graph, slope is

taken at 2. Then the slope AB/BC gives the rate of cooling

2

dt

d the thermal

conductivity of the given bad conductor (cardboard) is calculated using the given formula.

Result:

Thermal conductivity of the given bad conductor (card board) K =…………Wm-1

K-1




(iii) ) To find the diameter of the metallic disc (D) using Vernier caliper:

L.C. = 0.01 cm Zero Error (ZE) = + ……div


S.No.

MSR

(cm)

VSC

(div)

OR=MSR + (VSCLC)

(cm)

CR=OR ZC

(cm)

1.

2.

3.

4.

5.

Mean (D) =

Mean radius of the metallic disc (r) = 2

D …..…. 10

- 2m

(iv) Determination of the rate of cooling of metallic disc

2

dt

d :

θ1 = θ2 = Room temperature:

S.

No.

Temperature

() oC

Temperature

() Kelvin Time (t)

Seconds

1

2

3

4

5

6

7

8

9

10

11




Calculation:

Mass of the metallic disc M =------------------kg

Specific heat capacity of the metallic disc (brass) S = 372 Jkg-1

K-1

Radius of the metallic disc r = ------------------10

- 2m

Thickness of the metallic disc h = ------------10- 3

m

Thickness of the bad conductor d = ------------10- 3

m

Steady state temperature of steam chamber θ1 = ------------°C

Steady state temperature of the disc θ2 = ---------°C

Rate of cooling

2

dt

d at steady state temperature θ2°C = ----------°C/s

(From graph)

Thermal conductivity of the bad conductor:

11

2)21

)(22(2

2

KWm

dt

d

hrr

hrMSdK

K = -----------------------------w/m/k




Viva-voce :

1. What is thermal conductivity? What is its unit?

It is defined as the quantity of heat conducted per second normally across unit area of

cross-section of the material per unit temperature difference. It gives the heat conducting

ability of the material. Its unit is watts/meter/Kelvin.

2. Does the value of thermal conductivity depend on the dimension of the specimen?

No, it depends only on the material of the specimen.

3. Can this method be used for good conductors?

No, in that case, due to large conduction of heat, the temperature recorded by T1 and

T2 will be very nearly the same.




Top view of the B-H curve

Hysteresis loop(B-H)




5. Determination of Hysteresis loss in a ferromagnetic material.

Aim:

To determine the hysteresis loss in the transformer core using B-H curve unit.

Apparatus Required:

B-H curve unit, Cathode Ray oscilloscope (CRO) Patch cords

Formula:

Hysteresis loss = oopareaofthelSSV

C

R

R

N

NHv

2

1

2

1 joule cycle-1

m-3

Explanations of symbols:


N1 Number of turns in the primary coil -

N2 Number of turns in the secondary coil

V Volume of the core metre3

Sv Vertical sensitivity of CRO Vm-1

SH Horizontal sensitivity of CRO Vm-1

R1&R2 Resistances in the circuit ohm

C Capacitance of the capacitor in the circuit farad

Procedure:

The experimental arrangement is shown. One of the specimens used in the unit is

made using transformer stampings. There are two windings on the specimen (primary and

secondary).

The primary is fed to low A.C voltage (50Hz). This produces a magnetic field H in

the specimen. The voltage across R1 (resistance connected in series with primary) is

proportional to the magnetic field. It is given to the output of the CRO. The A.C magnetic

field induces a voltage in the secondary coil. The voltage is proportional to dB/dt. This

voltage is applied to




Observations:

Number of turns in the primary N1 =

Number of turns in the secondary N2 =

Resistance in the primary R1 = ohm

Resistance in the secondary R2 = ohm

Capacitance of the capacitor C = μF

Vertical sensitivity of CRO Sv = Vm-1

Horizontal sensitivity of CRO Sh = Vm-1

Calculation:

Area of the loop (from the graph) = m2

Hysteresis loss = hv SSV

C

R

R

N

N

1

2

2

1 x Area of the loop joule cycle-1

m-3

Calculation of volume of the transformer core:

V = [r22-r1

2] h

r2 = outer radius of core

r1 = inner radius of core

h = thickness of core




passive integrating circuit. The output of the integrator is proportional to B and fed to

the vertical input of the CRO.As a result of the application of voltage proportional to H the

horizontal axis and a voltage proportional to B is the vertical axis, the loop is formed as

shown. A measurement of the area of the loop leads to the evaluation of energy loss in the

specimen

The top view of the unit is shown. There are 12 terminals on the panel; sin patch cords

are supplied with the kit. The value of R1 can be selected by connecting D to A, B or C (A-

D=50 ohm: B-D =150 ohm: C-D =50 ohm)

A is connected to D. The primary terminals of the specimen are connected to p, p

secondary to S, S terminals. The CRO is calibrated as per the instructions given the

instruction Manual of CRO. CRO is adjusted to work on external mode (the time base is

switched off). The horizontal and vertical position controls are adjusted such that the spot is

at the centre of the CRO screen.

The terminal marked GND is connected to the ground of the CRO. The H is

connected to the horizontal input of the CRO. The terminals V are connected to the vertical

input of the CRO. The power supply of the unit is switched on. The hysteresis loop is formed.

The horizontal and vertical gains are adjusted such that the loop occupies maximum area on

the screen of the CRO. Once this adjustment is made, the gain controls should not be

disturbed. The loop is traced on the translucent graph paper. The area of the loop is estimated.

The connections from CRO are removed without disturbing the horizontal and

vertical gain controls. The vertical sensitivity of the CRO is determined by applying a known

A.C voltage say 1 volt (peak to peak).

If the spot deflects by x cms for 1 volt, the vertical sensitivity is 1/(x×10-2

) (volt/m).

Let it be Sv. the horizontal sensitivity of CRO is determined by applying a known A.C

voltage 1 volt (peak to peak) Let the horizontal sensitivity be SH (volt/m).

The hysteresis loss is calculated by using the given formula.

Result:

Energy loss = joules cycle-1

m-3




Young’s modulus Non-Uniform bending

Determination of the depression ‘y’ using travelling microscope:

TR = MSR+ (VSC x LC)

L.C = 0.001cm

Distance between the two knife edges l = M. =50 x10-3

kg

S.No.

Load

Microscope reading Mean of

loading

&

unloading

Depression

for M kg ‗y‘

Loading Unloading

MSR VSC TR MSR VSC TR

Unit x10-3

kg cm div cm cm div cm cm cm

1 W

2 W+50

3 W+100

4 W+150

5 W+200

Mean (y) =

Determination of thickness of the beam using screw gauge (d):



S.No PSR

(mm)

HSC

(div)

HSR=(H.S.CxL.C)

(mm)

OR=PSR+HSR

(mm)

CR=OR+ZC

(mm)

Thickness of the beam d =




6. Determination of Young’s modulus of the material – Non uniform bending

Aim:

To determine the Young‘s modulus of the material of a uniform beam supported on

two knife-edges and loaded at the middle.

Apparatus Required:

1. Rectangular beam. 2. Slotted weights 3. Knife-edges 4. Travelling microscope 5.

Metre scale 6. Screw gauge and 7. Vernier calipers

Formula:

Young‘s modulus of the material:

ybd

MglE

3

3

4 Newton/metre

2



M Mass for the depression ‗y‘ kg

l Length of the beam between the knife edges metre

g Acceleration due to gravity m/s2

b Breadth of the beam metre

d Thickness of the beam metre

y Depression for mass ‗M‘ metre

Procedure:

The given specimen beam is placed horizontally on the two knife edges. Let the

distance between two knife edges be ‘l‘ (say 60×10-2

m). At the midpoint of the beam say 50

th cm, a pin is placed vertically using honey wax pointing the tip of the pin upwards and a

weight hanger with dead load is suspended at that position using twine.

The slotted weight are loaded and unloaded so that the specimen beam is brought to

elastic mood. When the twine is hanged with dead load alone, the travelling microscope

screws




Determination of breadth of the beam using vernier calipers (b):

Least count (L.C.) = 0.01.cm Zero Error (ZE) = + ……div


S.No. MSR

(cm)

VSC.

(div)

VSR = VSC x

LC

(cm)

TR = (MSR + VSR) + ZC

(cm)

1.

2.

3.

4.

5.

Breadth of the beam (b)

=

Calculation:

Mass for the depression ‗y‘ M = ………..x 10-3

kg

Length of the beam between the knife-edges l = ……….x 10-2

m

Acceleration due to gravity g = 9.8 m/second2

Breadth of the beam b = ………. x 10-2

m

Thickness of the beam d = ……… x 10-2

m

Depression for ‗M‘kg y = ……… x 10-2

m

Young‘s modulus of the material

ybd

MglE

3

3

4 Newton/metre

2

E = …………..N/m2




are well adjusted to focus the tip of the pin just touch the horizontal crosswire. At this time,

the reading is noted for W gram as shown by the vertical scale. Add 50gm slotted weight to

the dead

load. Now when viewed through the travelling microscope, due to non-uniform bending of

the beam, the tip of the pin does not touch the horizontal crosswire. To make the tip of the pin

to touch the horizontal crosswire, the fine adjustment screw on the top of the vertical scale is

adjusted carefully in one direction (by screwing).

Then the readings are noted for w+50 gram. Another 50 gram is added to the weight

hanger and similar adjustments are made. The reading is noted for w+100 gram. Also the

readings are noted for w+150, w+200 and w+250 grams each time during loading. The

experiment is repeated for unloading also and the readings are noted with similar procedure.

But the fine adjustments screw which is on the top of the vertical scale is adjusted carefully in

the direction opposite to the direction made during loading. The total reading is determined

for both loading and unloading. The difference between the values is found for depression (y)

made by certain mass (M).

The distance ‗l‘ between the knife-edges A and B is measured. The thickness of the

beam is measured with the help of screw gauge at different places on the beam; from the

observed values the mean thickness of the beam is found. The breadth of the beam is

measured using vernier caliper. Hence Young‘s modulus ‗E‘ can be calculated using the

given formula.

Result:

The Young‘s modulus of the given material of the beam E = ……….N/m2




Viva voce:

1. What is Young‘s modulus?

Young‘s modulus is defined as the ratio of longitudinal stress to longitudinal strain

2. How are longitudinal strain and stress produced in your experiment?

Due to depression, the upper or the concave side of the beam becomes smaller than the

lower or the convex side of the beam. As a result, longitudinal strain is produced. The change

in length will be due to the forces acting along the length of the beam. These forces will give

rise to longitudinal stress.

3. How do you ensure that in your experiment the elastic limit is not exceeded?

The consistency in the readings of depressions both for increasing load and decreasing

load indicates that in the experiment the elastic limit is not exceeded.

4. Which dimension—breadth, thickness, or length of the bar-should be measured very

carefully and Why?

The thickness of the bar should be measured very carefully since its magnitude is small

and it occurs in the expression ‗E‘ in the power of three. An inaccuracy in the measurement

of the thickness will produce the greatest proportional error in E.

5. Why do you place the beam symmetrically on the knife edges?

To keep the reaction at the knife edges equal in conformity with the theory.




Circuit for band gap determination

Model graph

1000/T

Y X

A

B c Slope=BC

AB

Log Is

Variation of current with inverse temperature in a reverse biased pn- diode.




7. DETERMINATION OF BAND GAP SEMI CONDUCTOR

Aim:

To find the width of the forbidden energy gap in a semiconductor material taken in the

form of a pn – diode.

Apparatus required:

0-15V DC power supply, heating arrangement to heat the diode, Thermometer (0°c to

100°c) Micrometer (0-50μA) and Germanium diode.

Formula:

The width of the forbidden energy gap is given by

eV (electron volt)

Circuit

The circuit diagram for conducting the experiments is shown. The diode is reverse

biased with the help of dc voltage obtained from a dc power supply and the current that flows

through the reverse biased diode is measured with a micrometer. A heating system (heating

coil and oil bath) helps to raise the temperature of the diode. The circuit is available in a

ready-to-use training form

Procedure

Sufficiently long wires are soldered to the diode terminals and diode is connected in

to the circuit as shown. The diode is immersed in an oil bath which in turn is kept in a heating

mantle. A thermometer is also kept in the oil bath such that its mercury bulb is just at the

height of the diode. The power supply is switched on and the voltage is adjusted to say 0.5

volt. The heating mantle is switched on and the oil bath is heated upto 60°c (say 62°c).

The heating mantle is switched off when the temperature of oil bath reached 62°c.

The oil is stirred and is allowed to cool. The temperature of oil bath decreases to say at 60°c

and corresponding current through the diode are noted. The oil bath is allowed to cool slowly.

As its temperature falls, the current through the diode decreases, The current through diode

Eg =0.397× slope




Observations

Room temperature =

Volt =

Reading of temperature vs reverse saturation current

S.No

Temperature

(t)

Temperature

T

1000

T

Reverse

Saturation

Current Is

Log Is

°C K K-1

×10-6

A A

1.

2.

3.

4.

5.

6.

7.

8.

9.

10.

11.

12.

13.

14.

15.

Calculation:

Eg =0.397× slope

=0.397×BC

AB

= __________eV




corresponding to each 2°c (i.e. 60°c, 58°c, 56°c, 54°c etc) fall of temperature is noted down

in the table. A graph is plotted taking 1000/T on X – axis and log I s on Y- axis. A straight

line is

Obtained. The slope of the straight line is determined and using it in the formula, bandgap Eg

is calculated.

Result

The width of the forbidden gap in germanium semiconductor is__________eV

Viva-voce:

1. Define Fermi level.

―Fermi level‖ is that level at which the probability of electron occupation is ½ at any

temperature above 0K and also it is the level of maximum energy of the filled levels at 0K.

2. What are intrinsic semiconductors? Give examples.

Intrinsic semiconductors are semiconductors in pure form. These materials are having

an energy gap of the order of 1 eV. Charge carries are generated due to breaking of covalent

bond. ‗Ge‘ and ‗Si‘ are some examples of intrinsic semiconductors.

3. What are extrinsic semiconductors? Give examples.

A semiconducting material in which the charge carriers originate from impurity atoms

added to the material is called‖ extrinsic semiconductors‖. The addition of impurity increases

the carrier concentration and hence the conductivity of the conductor.

Phosphorous, arsenic or antimony added to either germanium or silicon gives n-type

semiconductors, while aluminum, gallium or indium added results in p-type

semiconductors




Carey Foster Bridge

To find the thickness of the wire(d) using screw gauge:



S.No PSR

(mm)

HSC

(div)

HSR =

(HSCxLC)

(mm)

OR=PSR+HSR

(mm)

CR=OR+ZC

(mm)

Mean (d) =




8. Carey Foster Bridge

Aim:

To determine the specific resistance of the material of the given wire.

Apparatus Required:

Carey Foster Bridge, Two equal resistance, Fractional resistance, Key, Galvanometer,

Thick copper strip, Leclanche cell.

Formula:

Specific resistance of the wire =l

rX 2ohm metre



X Resistance of the wire ohm

r Radius of the wire metre

l Length of the wire metre

Procedure:

The circuit connections are made as shown. The two equal resistance P and Q in the

inner gaps 1 and 2. The known low resistance S is connected in the end gap on the right hand

side. The unknown resistance X is connected in the end gap on the left hand side. A

Leclanche cell and a galvanometer are connected as shown. The circuit is closed by putting

the key. The jockey is pressed near one end A of the bridge wire and then the other end B.

The connections are correct only when deflections in the galvanometer are in opposite

direction.




To find the resistance of the bridge wire per metre (ρ):

Trial

No.

Resistance

In box S

Balancing length '

2

'

1 ll

Resistance

Of the wire

per metre

)('

1

'

2 ll

S

Before

interchanging

After

interchanging

unit ohm metre metre metre Ohm/metre

1 0.1

2 0.2

3 0.3

4 0.4

5 0.5

6 0.6

7 0.7

8 0.8

9 0.9

10 1.0

Mean




The balancing length (l1) is measured by pressing the jockey along the bridge wire for null

deflection in the galvanometer. The second balancing length (l2) is found by interchanging X

and S. For different values of S, l1 and l2 are found. The readings are tabulated, and using the

formula X = S+ ρ (l1-l2) the unknown resistance value is very easily to find ρ, a thick copper

strip of zero resistance is connected instead of X. The balancing lengths l1‘and l2'‘ are

determined as above.

Then, the formula X= )('

1

'

2 llS

In this case, X = 0 or ρ = )(

'

1

'

2 ll

S

The experiment is repeated with different values of S and then the mean value of ρ is taken.

The specific resistance of the wire is found using the given formula by measuring r the radius

of the wire and l the length of the wire.

Result:

(i) Resistance of the wire = ohm.

(ii) Specific resistance of the given wire = ohmmeter

Questions:

1. What is Carey-Foster Bridge?

2. What is the maximum difference in the resistance that you can measure with a Carey-

Foster Bridge?

3. Is it essential to have the value of resistance P and Q equal?

4. What will happen if the bridge wire is not-uniform?







9. VISCOSITY OF A LIQUID BY POISEUILLE’S METHOD

Aim:

To determine the co-efficient of viscosity of the given liquid (water) by poiseuille‘s

method.

Apparatus:

Graduated burette, capillary tube, beaker, water, stop clock, metre scale, rubber tube,

pinch cock and travelling microscope.

Formula:

Coefficient of viscosity of water 24

8

)( Nsmlv

htgr

Where 1 2

02

h hh h metre

Explanation of the symbols:


Density of water

---

g

Acceleration due to gravity

metre/s2

r

Radius of the capillary tube

metre

l

Length of the capillary tube

metre

h

height of the liquid level.(Pressure head) metre

h1

Height from table to initial level of water in the burette

metre

h2

Height from table to final level of water in the burette

metre

h0

Height from table to mid-portion of capillary tube metre

t

Time taken for the flow of 5 cc liquid second

v Volume of the liquid m3




Observation:

Height from table to midpoint of capillary tube h0 = ……………… x 10-2

m

Length of the capillary tube l = ……………… x 10-2

m

Volume of the liquid v = 5cc =5 x 10-6

m3

S.

N

o

Burette

reading

Time

of

flow

Range Time for

flow of

5 cc

liquid

Height

of

initial

reading

h1

Height

of final

reading

h2

Pressure

head

1 2

02

h hh h

ht

v

ht

cc s cc s x10-2

m x10-2

m x10-2

m m-second x106 s/m

2

1. 0 0 - 5

2. 5 5 – 10

3. 10 10 – 15

4. 15 15 – 20

5. 20 20 – 25

6. 25 25 – 30

7. 30 30 – 35

8. 35 35 – 40

9. 40 40 – 45

10

.

45 45 – 50

11 50

Mean v

ht= ……………. s/m

2




Procedure:

The burette is fixed vertically on the stand. The capillary tube is held in the horizontal

direction. So, that the flow of liquid through it is streamlined. Fill the burette with a given

liquid above 0 cc.Open the knob of the burette and allow the liquid to flow. When the

liquid level reaches 0 cc, start the stop clock and for each 5 cc flow of liquid the time is

noted

Continuously upto 50 cc.The time taken for the flow of 5 cc of liquid is determined.

The initial height of the liquid (0 cc) is measured from the surface of the table and is

taken as ‗h1‘.Similarly, the final height of the liquid (after 5 cc flow from the initial level)

is again measured from the surface of the table and is taken as ‗h2‘.Repeat the

measurement of initial and final height of the liquid for each 5 cc flow upto 50 cc and

enter it as ‗h1‘ and ‗h2‘ respectively. Measure the height from table to mid-portion of

capillary tube ‗h0‘.Pressure head is calculated using the given formula. The value of ht is

calculated.

Measurement of radius of capillary tube:

Mount the capillary tube over a stand in the horizontal direction. By adjusting the

distance between capillary tube and traveling microscope, the bore of the capillary tube is

focused. The horizontal cross-wire of the traveling microscope is made to coincide with

top of the bore of the capillary tube using vertical scale fine adjustment screw.

Reading is noted from the vertical scale of the traveling microscope similarly, the

bottom of the bore of the capillary tube is made to coincide with horizontal cross-wire

and the corresponding readings are noted from the vertical scale of the traveling

microscope. Again, the left and right side of the bore of the capillary tube is made to

coincide with the vertical cross-wire using horizontal scale fine adjustment screw and the

corresponding readings are measured from the horizontal scale of the traveling

microscope. Using above values diameter and radius of the capillary tube is calculated.

The coefficient of viscosity of given liquid (water) can be calculated by given formula.

Result:

The coefficient of viscosity of water = …………… Ns/m2




Find the radius of the capillary tube:


L.C= 0.001 cm

Horizontal scale reading Vertical scale reading

Position MSR VSC TR Position MSR VSC TR

Unit x 10-2

m div x 10-2

m x 10-2

m div x 10-2

m

Left

R1

Top

R3

Right

R2

Bottom

R4

Difference d1=R1R2 = …… x 10-2

m Difference d2=R3R4 = ….. x 10-2

m

Average diameter of the capillary tube 1 2

2

d dd

d = ………… x 10-2

m

Radius of the capillary tube r = 2

d………… x 10

-2 m




Viva voce:

1. What is Viscosity and define the coefficient of viscosity?

In the presence of a relative motion between two layers of a liquid, an opposing

tangential force sets in between the layers to destroy the relative motion. This property of the

liquid is termed viscosity and is analogous to friction. The tangential force acting per unit

area over two adjacent layers of the liquid for a unit velocity gradient is referred to as the

coefficient of viscosity.

2. How does the coefficient of viscosity changes with temperature?

The coefficient of viscosity decreases with rise in temperature in the case of liquids,

but for gases it increases with rise in temperature.

3. Which quantity requires greatest care in its measurements? Why?

The radius of the capillary tube requires greatest care in its measurements since it

occurs in the fourth power in the expression of η. Thus a small error in the measurement of r,

which itself is small, will contribute to a large proportional error in η. The tube selected must

therefore be uniform and its radius be measured very carefully.

4. Can you use this method for all the types of liquids?

No, this method can be suitably applied for liquids of low viscosity. For highly

viscous liquids, Stoke‘s method can be used.

5. Is there any difference between friction and viscosity?

Friction and viscosity have some similarities and same differences between them. For

liquids at rest, friction works but viscosity does not because viscosity arises only when there

is a relative motion between the layers of a liquid.




Position of the Spectrometer-Reflection of light from prism

to find angle of the prism

Position of the Spectrometer-Refraction of light from prism

to find angle of minimum deviation




10. DISPERSIVE POWER OF A PRISM-SPECTROMETER

Aim:

To find the dispersive power of the material of the prism using spectrometer.

Apparatus Required:

Spectrometer, Mercury vapour lamp, Glass prism, and Reading lens

FORMULAE

Refractive index of the prism

2sin

2sin

A

DA

(no unit)

Dispersive power of the material of the prism 1

rv (no unit)

Explanation of the symbols


A Angle of the prism degree

D Angle of minimum deviation degree

v Refractive index of the prism for violet line -

r Refractive index of the prism for red line -

2/)( rv -

PROCEDURE

The initial adjustment of the spectrometer namely adjustment of eye piece for distinct

vision of cross wires, adjustment of telescope for the distant object and collimator for parallel

rays are made as usual. The slit of the collimator is illuminated by the mercury vapour lamp.

i) Determination of angle of prism (A)

The given prism is mounted vertically at the centre of the prism table with its

refracting edge facing the collimator. Now the parallel rays of light emerging out from the

collimator falls almost equally on the two faces of the prism ABC as shown in fig. The

telescope is turned to catch the reflected image from one face of the prism and fixed in that

position. The tangential screw is adjusted until the vertical cross-wire coincides with the fixed

edge of the image of the slit.




To find the angle of prism (A) – Spectrometer readings

TR = MSR + (VSC X LC) LC = 1'

Reflected ray Vernier A Vernier B

Lines of the spectrum MSR

(deg)

VSC

(div)

TR

(deg)

MSR

(deg)

VSC

(div)

TR

(deg)

Readings of reflected

image from face 1

(left)

(R1)

(R1)

Readings of reflected

image from face

2(right)

(R2)

(R2)

2A= R1~R2 =

2A= R1~R2 =

Mean 2A =

Mean A =

To find the angle of minimum deviation (D) – Spectrometer readings

TR = MSR + (VSC X LC) LC = 1'

S.

No

.

Refracted

ray

readings

Vernier A

R1

Vernier B

R2

Angle of

minimum

deviation (D)

Mean D =

2

VerBVerA

(deg)

µ

(No

unit) Lines of

the

spectrum

MSR

(

deg)

VS

C

(div

)

TR

( deg)

MSR

( deg)

VSC

(div)

TR

( deg)

Ver A

R1~D1

( deg)

Ver B

R2~D2

( deg)

1. Violet

2. Green

3. Yellow

4. Red

Direct ray

(D1)

(D2)




The readings on both the vernier are noted. Similarly the readings corresponding to

the reflected image of the slit on the other face are also taken. The difference between the two

readings of the same vernier gives twice the angle of the prism. Hence, the angle of the prism

‗A‘ is determined.

(ii) Determination of angle of minimum deviation (D)

The prism table is rotated so that the beam of light from the collimator is incident on

one face of the prism and emerges out from the other face. The telescope is rotated to catch

the refracted image of the yellow slit. The prism table is rotated in such a direction so that the

refracted images move towards the direct beam. The telescope is rotated carefully to have the

image in the field of view. At one stage, the image stops momentarily and turns back. This is

the position of the minimum deviation. The telescope is rotated and made to coincide with the

violet slit. The telescope is fixed in this position and refracted ray reading of the telescope is

noted. The experiment is repeated for red slit. The prism is removed and the direct reading of

the slit is taken. The difference between the direct reading and the refracted ray reading

corresponding to the minimum deviation gives the angle of minimum deviation ‗D‘. The

dispersive power is calculated using the given formula.

RESULT

1. Angle of prism = ………….. degree

2. Dispersive power of the prism = …………… (no unit)




Calculation:

Refractive index

2sin

2sin

A

DA

2sin

2sin

A

DA

v =

2sin

2sin

A

DA

g

2sin

2sin

A

DA

y

2sin

2sin

A

DA

r

Dispersive power of the material of the prism 1

rv




VIVA-VOCE

1. What is the condition for obtaining minimum deviation?

The deviation is minimum when the angles of incidence and the emergence are equal.

2. Define refractive index

The ratio of the sine of the angle of incidence of the sine of angle of refraction is constant

for any two media, i.e., r

i

sin

sin a constant known as refractive index.

3. How does refractive index change with wavelength of light?

Higher the wavelength, smaller is the refractive index.

4. Does the deviation depend on the angle of the prism?

Yes, greater the angle of the prism more is the deviation.

5. Define dispersive power of a prism.

Dispersive power indicates the ability of the material of the prism disperse the light

rays. It is defined as the ratio of the angular dispersion to the deviation of the mean ray.




Young’s Modulus –Uniform Bending

Determination of the Elevation ‘y’ using travelling microscope:


L.C = 0.001cm

Distance between the two knife edges l = a = M. =50 x10-3

kg

S.No.

Load

Microscope reading Mean of

loading &

unloading

Elevation for

M kg

‗y‘ Loading Unloading

MSR VSC TR MSR VSC TR

Unit x10-3

kg cm div cm cm div cm cm cm

1 W

2 W+50

3 W+100

4 W+150

5 W+200

Mean(y)=

Determination of thickness of the beam using screw gauge (d):

Least count (L.C.) = 0.01.mm Zero Error (ZE) = +

……div Zero Correction

(ZC) = + …… mm

S.No PSR

(mm)

HSC

(div)

HSR=(H.S.CxL.C)

(mm)

OR=PSR+HSR

(mm) CR=OR + ZC

(mm)

Thickness of the beam d =




11. Young’s modulus– Uniform bending.

Aim:

To determine Young‘s modulus of the material of the beam by uniform bending

method.

Apparatus Required:

Rectangular beam. 2. Slotted weights 3. Knife-edges 4. Travelling microscope 5.

Metre scale 6. Screw gauge and 7. Vernier calipers

Formula:

Young‘s Modulus of the material by Uniform bending method:

ybd

aMglE

3

2

2

3 Newton/metre

2



M Mass for the depression ‗y‘ kg

l Length of the beam between the knife edges metre

g Acceleration due to gravity m/s2

b Breadth of the beam metre

d Thickness of the beam metre

y Elevation for mass ‗M‘ metre

Procedure

The given beam is supported symmetrically on two knife-edges such that the distance

between them is about 0.7m. Two weight hangers are suspended, one each on either side of

the knife edges so that their distances from the nearer knife edge are equal. A pin is fixed

vertically at the centre of the beam.

A traveling microscope is placed in front of this arrangement and adjusted so that the

tip of the pin is seen clearly when viewed through the eyepiece of the microscope. With the

two weight hangers as dead load, the microscope is adjusted till the tip of the pin just touches




Determination of breadth of the beam using vernier calipers (b):

Least count (L.C.) = 0.01.cm Zero Error (ZE) = + ……div


S.No. MSR

(cm)

VSC.

div

VSR = VSC x LC

(cm)

TR = (MSR + VSR) + ZC

(cm)

1.

2.

3.

4.

5.

Breadth of the beam (b) =

Calculation:

Mass for the Elevation ‗y‘ M = ………..x 10-3

kg

Length of the beam between the knife-edges l = ……….x 10-2

m

Acceleration due to gravity g = 9.8 m/second2

Breadth of the beam b = ………. x 10-2

m

Thickness of the beam d = ……… x 10-2

m

Elevation for ‗M‘kg y = ……… x 10-2

m

Young‘s modulus of the material: ybd

aMglE

3

2

2

3 Newton/metre

2

E = -------------- N/.m2




the horizontal cross-wire. The reading on the vertical scale is noted. Equal weights

are loaded to the hangers in steps of 0.05kg and each time reading on the vertical scale of the

Vernier microscope is taken, ensuring that the tip of the pin just coincides with the horizontal

cross-wire. The procedure is followed until the maximum load is reached. Now weights are

removed from both

the hangers in steps of 0.05kg and in each case of the reading are taken. For each load, the

mean reading taken while loading and unloading is obtained. The difference between the

mean reading for a particular load and that of the dead load gives elevation ‗s‘ for that load.

Result:

The Young‘s modulus of the given material by Uniform bending (E) = ……….N/m2

Viva voce:

1. What is Young‘s modulus?

Young‘s modulus is defined as the ratio of longitudinal stress to longitudinal strain

Young‘s modulus of elasticity (E) = alstrainLongitidin

alstressLongitudin

2. What is elasticity?

The property of the body to regain its original shape or size, after the removal of

deforming force is called elasticity.

3. What are elastic bodies?

Bodies which regain its original shape or size, after the removal of deforming force

are called elastic bodies.

4 .What is uniform bending?

The beam is loaded uniformly on its both ends, the bent beam form an arc of a circle.

The elevation in the beam is produced. This bending is called uniform bending.




To find the time period of oscillation:

Length of the suspension wire (l) = …………….x10-2

metre

Position of the equal masses Time for 10 oscillations Time period (time

for one oscillation)

Trial – 1 Trial – 2 Mean

Seconds Seconds Seconds Seconds

Without masses T0 =

With masses at

d1 = ………….x10-2

m

T1 =

With masses at

d2 = …………x10-2

m

T2 =




12. Torsional pendulum

Aim:

To determine the moment of inertia of the given disc and the rigidity modulus of a

given wire by the method of torsional oscillations.

Apparatus:

Uniform circular disc, thin wire, two equal masses, stop-clock, screw gauge and metre

scale.

Formula:

1. Moment of Inertia of the disc

2 2 2

2 1 2

2 2

2 1

2.

om d d TI kg m

T T

2. Rigidity modulus of the material of the wire

2

2 4

8Newton/metre

o

Iln

T r



m Mass of one of the equal masses placed on the disc. Kg

d1 Minimum distance between centre of suspension wire and the

centre of any one of the masses

metre

d2 Maximum distance between centre of suspension wire and the

centre of any one of the masses.

metre

T0 Time period when there is no mass placed on the disc seconds

T1 Time period when equal masses are placed at a distance d1 seconds

T2 Time period when equal masses are placed at a distance d2 seconds

l Length of the suspension wire metre

r Radius of the given wire metre




Measurement of radius of the wire (r):

Least count (L.C.) = 0.01mm Zero Error (ZE) = + ……div


S.No PSR

(mm)

HSC

(div)

HSR

(HSCxLC)

(mm)

OR=PSR+HSR

(mm)

CR=OR+ ZC

(mm)

Mean diameter of the wire (d) =

Radius of the wire (r) = d/2 = …………. mm

Calculation:

Time period of oscillation (without masses) T0 = seconds

Time period when masses are at distance ‗d1‘ T1 = seconds

Time period when masses are at distance ‗d2‘ T2 = seconds

Closest distance between suspension wire

and the center of cylindrical mass d1 = x 10-2

m

Farthest distance between suspension wire

and the center of cylindrical mass d2 = x 10-2

m

Mass of one of the cylinders m = x 10-3

kg

Length of the suspension wire l = x 10-2

m

Mean radius of the wire r = x 10-3

m

Moment of Inertia of the disc

2 2 2

2 1 2

2 2

2 1

2.

om d d TI kg m

T T

I = ……………….kg m2

Rigidity modulus of material of the wire 2

2 4

8Newton/metre

o

Iln

T r

n = ……………….N/m2




Procedure:

One end of a long, uniform wire whose rigidity modulus is to be determined is

clamped by a vertical chuck. To the lower end, a heavy uniform circular disc is attached by

another chuck. The length of the suspension wire ‗l‘ (from top portion of chuck to the clamp)

is fixed to a particular value (say 60 or 70 cm). The suspended disc is slightly twisted so that

it executes torsional oscillations. Care is taken to see that the disc oscillates without

wobbling.

The first few oscillations are omitted. By using the pointer (a mark made on the disc)

the time taken for 10 oscillations is noted. Two trials are taken and the mean time period T

(time for one oscillation) is found.

Two equal masses are placed on the disc symmetrically on either side, close to the

suspension wire (at the minimum distance). The closest distance ‗d1‘ from the center of the

cylindrical mass and the center of the suspension wire is found. The disc with masses at

distance d1 is made to execute torsional oscillations by twisting the disc. The time taken for

10 oscillations is noted. Two trials are taken and the mean time period ‗T1‘ is determined.

Two equal masses are now moved to the extreme ends, so that the edges of the masses

coincide with the edge of the disc and the centers are equi-distant. The distance ‗d2‘ from the

center of the cylindrical mass and the center of the suspension wire is noted. The disc with

masses at distance ‗d2‘ is allowed to execute torsional oscillations by twisting the disc. The

time taken for 10 oscillations is noted and the mean time period ‗T2‘ is calculated. The mass

of one of the cylinders placed on the disc is found. The diameter of the wire is accurately

measured at various places along its length with screw gauge. From this, the radius of the

wire is calculated. The moment of inertia of the disc and the rigidity modulus of the wire are

calculated using the given formulae.

Result:

Moment of inertia of the disc (I) = …………………..kg m2

Rigidity modulus of the wire (n) = …………………..N/m2




VIVA VOCE:

1. What is torsional pendulum?

A body suspended from a rigid support by means of a long and thin elastic wire is called

torsional pendulum.

2. What is the type of oscillations?

This is of simple harmonic oscillation

3. on what factors does the time period depend?

It depends upon (i) Moment of inertia of the body

(ii) Rigidity of wire i.e. length, radius and material of the wire.

4. How will you determine the rigidity of fluids?

As fluids do not have a shape of their own, hence they do not possess rigidity. Hence

there is no question of determining it.

5. State Hooke‘s Law.

There is a simple relationship between stress and strain discovered by Hooke and is

called Hooke‘s Law. According to this law strain is proportional to the stress producing it

within elastic limits




PHYSICAL CONSTANTS

Young’s modulus of the material

Material 1010

Nm-2

Brass 9.8

Copper 11.7

Glass 6

Wood 0.8 – 1.2

Rigidity modulus of the material of the wire

Material of the wire 1010

Nm-2

Brass 3.45

Copper 3.6 – 3.9

Steel (cast) 7.4 – 7.6

Steel (mild) 8.9

Coefficient of Viscosity

Liquid Nsm-2

Water 0.8x10-3

Kerosene 2x10-3

Castor Oil 986x10-3

Glycerine 309x10-3

Thermal conductivity

Substance Wm-1

K-1

Cardboard 0.04

Glass 1.0

Rubber 0.15

Wood 0.15

Compressibility of liquids

Liquid m2N

-1

Benzene 9.1 x 10-10

Kerosene 7.5 x 10-10

Water 4.5 x 10-10

Castor Oil 4.7 x 10-10



http://www.ligo.caltech.edu/~vsanni/ph3/ToolsTables/NISTPhysicalConstants.pdf


Band gap of a semiconductor

Material eV

Germanium 0.7

Silicon 1.1

GaAs 1.44

Refractive Index of the material

Material No unit

Air 1.0

Glass(crown) 1.50

Glass(flint) 1.56

Water 1.33

Specific resistance of the material

Material µΩ-m

Aluminium 0.026

Brass 0.068

Copper 0.017

Constantan 0.49

Wavelength of the Mercury Spectral lines

Colour Å

Violet I 4044

Violet II 4078

Blue 4358

Bluish green 4916

Green 5461

Yellow I 5769

Yellow II 5791

Red 6234



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