1.Refractive index of the prism. Aim To determine the angle of prism and the angle of minimum deviation and hence to determine the refractive index of the prism using spectrometer.

Formula

2

2

A DSin

ASin

μ = refractive index of the prism

A = angle of the prism D=angle of minimum deviation

Procedure

To determine the angle of prism.

The preliminary adjustment of the spectrometer are made. The prism is placed at the prism table with its refracting edge facing the collimator.

The telescope is moved to view the reflected image from one side of the prism.

The reading is noted from both the vernier scale. Let it be R1.

The telescope is moved to the other side to view the image. The reading is calculated from both vernier. Let it be R2.

The difference between two the reading R1 and R2 gives 2A. Hence A is calculated.

To determine the angle of minimum deviation.

The edge of the prism is turned away from the collimator.

Telescope is adjusted to view the refracted image.

The prism table is rotated in such a direction the angle of deviation decreases. At a particular position the image remains stationary for a moment and moves in the opposite

direction.

This position is called is known as minimum deviation position. The reading is noted for both vernier (R3).

Prism is removed and direct image is viewed. The readings is noted (R4 )

The difference between R3 and R4 is the angle of minimum deviation D.

The refractive index of the prism is calculated using the formula 2

2

A DSin

ASin

Observation: Least count of the spectrometer. 1 MSD= ______

Number of divisions in vernier scale(N) =_______

Least count = 1

1x MSDN

Angle of prism

Ray Vermier I Vernier II

main scale

reading

VC Total reading= MSR+ (V.CxL.C)

Main scale reading

VC Total reading= MSR+ (V.CxL.C)

Ray reflected

from one

side of the prism

Ray

reflected

from other side of the

prism

.

Angle of minimum deviation

Ray Vermier I Vernier II

main scale

reading

VC Total reading= MSR+ (V.CxL.C)

Main scale reading

VC Total reading= MSR+ (V.CxL.C)

Reading of

minimum

deviation

position

Direct ray

.

Result:

1. The angle of the prism:

2. The angle of minimum deviation:

3. The refractive index of the prism:

2.Spectrometer – grating. Aim To determine the wavelength of the spectral lines of mercury spectrum using spectrometer.

Formula:

Sin

Nm

λ = wavelength of the spectral line

θ = angle of diffraction

m= order of the spectrum N= number of lines per unit length in the grating element..

Procedure:. The preliminary adjustment of the spectrometer are made

The grating is placed on the prism table and direct image is viewed.

The telescope is rotated through an angle 900 and fixed.

The prism table is rotated until the image of the slit reflected by the grating is seen. Now the vernier table is rotated in appropriate direction through 45

0 so that the rays of light from

the collimator falls normally on the grating.

The telescope is released and adjusted to view the image of spectral line (blue). The reading from both the vernier is noted. Let it be R2

The telescope is moved in same direction to view the yellow and green light and the corresponding

readings are noted.

The telescope is moved to view the direct image. Let it be R1. The difference between R2 and R1 gives the angle of diffraction θ

The wavelength of the spectral line of the mercury spectrum is calculated using the formula

Sin

Nm

Observation:

Number of lines per unit length in the grating element =N= ________

Order of the spectrum = m= _________

Vernier I Vernier II

Direct reading (R1)

Result: The wavelength of the line spectral lines of mercury spectrum are determined.

1. Wavelength of blue colour = _________ 2. Wavelength of green colour = _________

3. Wavelength of yellow colour = _________

Colour

Vermier I(R1) Vernier II (R2) Vernier I

Vernier II

MSR VC Total reading =

MSR+

(V.CxL.C)

MSR VC Total reading =

MSR+

(V.CxL.C)

θ1 = R1 – R2

θ2 = R1 –R2

θ = θ1 + θ2 2

Blue

Green

Yellow

3. Metre bridge.

Aim To determine the resistance and specific resistance of the wire using metre bridge.

Formula

X

R

lX R

l

X= Resistance of the wire

R = known resistance

Xl = balancing length of the unknown resistance

Rl = balancing length of the known resistance.

2r X

l

ρ = specific resistance of the wire

r = radius of the wire

X= resistance of the wire

l = length of the wire.

Before interchanging

After interchanging

Procedure: A resistance box R is connected in the left gap. The unknown resistance X is connected in the right gap.

Now a suit able resistance is included in the resistance box R.

Let J be the position of the jockey where there is null deflection in the galvanometer.

The balancing length AJ( l1 ) is measured. The length BJ is( l2.) The resistance in the resistance box is increased in the order of 1Ω and each time the balancing

length is determined.

The resistance box R and unknown resistance X is inter changed. The experiment is repeated as before by including the same values of resistance. In this case the

balancing length AJ gives l4 and l3 is calculated. The values of lx and lR is calculated. The unknown

resistance is calculated by using the formula X

R

lX R

l .

The radius of the wire is calculated using the screw gauge.

The specific resistance of the wire is calculated using the formula

2r X

l

observation:

Length of the wire = _______

Resistance of the wire :

S.No

R

(ohm)

Balancing length before

interchanging

Balancing length

after

interchanging

Mean X

R

lX R

l

ohm

l1 cm l2 = 100-l1

cm

l4 cm

l3 =100-l4

cm

1 3

2R

l ll

cm

2 4

2X

l ll

cm

1

2

3

4

5

Radius of the wire:

Least count = ______ Zero error = _______

Zero correction = _______

Result :

1. The resistance of the given wire = ___________ 2. The specific resistance of the wire = ___________

S.No P.S.R (mm) H.S.C C.H.S.R Total reading = PSR +

(CHSR X LC) (mm)

1

2

3

4

4.potentiometer. Aim To compare the emf of the two given cells using potentiometer.

Formula :

1 1

2 2

E l

E l

E1 = Electromotive force of leclanche cell E2 = Electromotive force of Daniel cell

l1 = balancing length of leclanche cell

l2 = balancing length of Daniel cell

Circuit diagram:

Procedure: Connection are given as in the circuit diagram.

The leclanche cell is included in the secondary circuit by putting the DPDT switch in the

appropriate position. Contact is made with the jockey at various points on the potentiometer wire.

The point of contact at which the galvanometer shows null deflection is found out. The balancing

length (l1) is measured. With out changing the position of rheostat the DPDT switch is put on the other side so that the

Daniel cell is connected and balancing length (l2) is noted. The readings are tabulated.

The experiment is repeated by changing the position of the rheostat.

For each position of rheostat , the balancing lengths for both the cells is determined. The mean

ratio of e.m.f is calculated using the formula 1 1

2 2

E l

E l

Observation:

Result: The mean ratio of emf of two cells =

S.No Balancing length of leclanche

cell (cm)

Balancing length of Daniel

cell (cm) 1 1

2 2

E l

E l

1

2

3

4

5

6

5. Tangent galvanometer. Aim To determine the horizontal component of earth magnetic field using tangent galvanometer.

Formula:

0

2 tanH

n IB

a

BH = Horizontal component of magnetic field

μ0 = permeability of free space

n = number of turns in T.G I = current

a = radius of the T G coil

θ = mean deflection in TG

Circuit diagram :

Preliminary adjustments

1. The plane of the coil is kept in the magnetic meridian 2. The compass box is rotated till the aluminium pointer reads 0

0-0

0

The connections are made as shown in the circuit diagram. A suitable current is allowed to pass through a circuit by adjusting a rheostat such that the pointer

readings lie about 300-30

0. The readings are noted down as θ1 , θ2

The commutator is reversed such that the current flows in the opposite direction to the circuit.

The deflection of the T.G in the opposite two quadrants are noted down θ3 , θ4 The average of θ1 , θ2 ,θ3 ,θ4 is calculated as θ.

The experiment is repeated by changing the current flowing through the circuit . The readings are

tabulated . In each case I/tanθ is calculated.

The radius of the coil (a) is calculated from the circumference of the coil.

The horizontal component of magnetic induction of earth magnetic field 0

2 tanH

n IB

a

is determined.

Observation:

Circumference of the coil (2πa) = _______

Radius of the coil = a = 2πa/2π = _______

Number of turns n = _____

Result :

The horizontal component of earth’s magnetic field = _________

6.Sonometer.

Aim To determine the frequency of the alternating current using sonometer.

Formula:

1 1

2

Tn

l m

n = frequency of the ac mains

T = tension of the sonometer wire

l = resonating length

m = linear density of the wire

Procedure: By means of suitable step down transformer the ac mains voltage is brought down to 6 volts.

A secondary of the transformer is connected to the ends of the sonometer wire.

A bar magnet is held at the centre of the sonometer wire. A load (m) is added to the sonometer wire.

On passing an AC current, the wire is acted upon by both magnetic field and electric field which

are perpendicular to each other.

Therefore the wire is set into forced vibrations. The resonating length between two bridges is measured as (l1).

The experiment is repeated for same load and the resonating length is measured as (l2).

The experiment is repeated for different loads and the resonating length are tabulated. The radius of the wire (r) is calculated using screw gauge.

The linear density of the wire can be determined using the formula m = πr2ρ

The frequency of the ac mains can be calculated by using the formula.

1 1

2

Tn

l m

S.No Current

I (A)

Deflections in T.G (degrees) I / tanθ

(A) θ1 θ2 θ3 θ4 Average θ

1

2

3

4

Observation:

Radius of the wire: Least count = ______ Zero error = _______

Zero correction = _______.

S.No P.S.R (mm) H.S.C C.H.S.R Total reading = PSR + (CHSR X

LC) (mm)

1

2

3

4

Diameter of the wire d = _________

Radius of the wire r = __________

Density of the wire ρ = _________

Linear density m = πr2ρ

Result :

The frequency of a.c. mains __________

7.Junction Diode and Zener diode.

Aim To determine (i) the resistance of the diode and (ii) zener breakdown voltage of the zener diode.

Formula

FF

F

VR

I

RF = Resistance of the diode in forward bias

VF = Forward bias voltage IF = Forward bias current.

S.No Load

(kg)

Length of the vibrating

segment

Mean l

(cm)

T

newton T

l N/m

l1 (cm) l2 ( cm)

1

2

3

4

5

Diode Zener diode

Junction diode.

Connect the circuit as shown in figure.

Turn on the supply V1 and adjust the forward voltage Vf as 0.1V. Record Vf and forward current If through the diode.

Increase Vf in steps of 0.1 V upto 0.8V. Record Vf and If from the voltmeter and the milliammeter.

Draw the graph by plotting the Vf in x-axis and If in Y-axis.

Obtain the slope of the curve. The inverse of the slope gives the forward resistance of the diode.

Zener diode.

Connect the circuit as shown in the figure. Turn the power supply V1 and adjust such that V0 is 0.5V.

Note IZ from the milliammeter.

Increase V0 in steps of 0.5V upto 6 volts and note the corresponding IZ values.

Tabulate the readings . Draw the graph by plotting V0 in negative x-axis and Iz valves in negative y – axis.

Determine the zener breakdown voltage VZ from the graph.

Diode Zener diode

S.No V0 (volt) IF (mA)

1 0.5

2 1.0

3 1.5

4 2.0

5 2.5

6 3.0

7 3.5

8 4.0

9 4.5

10 5.0

11 5.5

12 6.0

Result:

1. Forward bias resistance of the diode = __________

2.Zener breakdown voltage = ___________

S.No VF (volt) IF (mA)

1 0.1

2 0.2

3 0.3

4 0.4

5 0.5

6 0.6

7 0.7

8 0.8

8. NPN Transistor. Aim:

To determine the (i) input impedance (ii) output impedance and (iii) current gain of the transistor.

Formula:

Input impedance

CE

BEi

B V

Vr

I

Output impedance 0

B

CE

C I

Vr

I

Current gain

CE

C

B V

I

I

VBE = change in emitter base voltage

VCE = change in collector emitter voltage IB = change in base current

IC = change in collector current

Circuit diagram

Input characteristic.

Keep VCE as constant at 5 volts. Change VBE to set 0.1 v. Note down IB in microammeter.

Iincrease VBE in steps of 0.1 v upto 1 V. Note down the corresponding values of IB in

microammeter. Plot the graph with VBE along x-axis and IB along the y-axis .

Obtain the slope of the curve above the knee point.

The reciprocal of the slope gives the input impedance.

Output characteristic. Keep IB as constant at 20µA.

Set VCE as 0.1 V and note down Ic in milliammeter.

Increase VCE in steps of 0.1 V upto 1 V. Note down the corresponding values of Ic in mA. Repeat the experiment keeping IB as constant 40µA,60μA.

Plot the graph with VCE along the x-axis and Ic along the y-axis.

Obtain the slope of the curve .

The reciprocal of the slope gives the input impedance

Transfer characteristic. It indicates how Ic varies with changes in IB when VCE is constant.

Plot the graph with IB in x-axis and the corresponding Ic in the y-axis .

Obtain the slope which the current gain.

Observation.

Input characteristic

VCE = 5V current gain

VCE = 5 V

S.No` IB (μA) IC (mA)

1

2

3

4

.

Output characteristic

.

Result :

1. Input impedance = _________ 2.output impedance = ____________

3. current gain = ._____________

S.No VBE ( volt) IB (μA)

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

S.No IB 20μA 40μA 60μA 80μA

VCE (volt) IC (mA) IC (mA) IC (mA) IC (mA)

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

11 2.0

12 3.0

13 4.0

14 5.0

9.Operational amplifier.

Aim 1. To construct the inverting amplifier and non inverting amplifier using operational amplifier

and to determine the voltage gain.

2. To construct the summing amplifier and to determine the output voltage.

Formula:

Voltage gain in inverting amplifier

O FV

in S

V RA

V R

Voltage gain in non inverting amplifier

1O FV

in S

V RA

V R

Output voltage in inverting amplifier

1 2OV V V

VO = output voltage

V1,V2,Vin = input voltage

RF , RS external resistance

Circuit diagram

Inverting amplifier Non inverting amplifier

Summing amplifier

Inverting amplifier.

Connect the circuit as shown in the figure using op-amp IC 741.

Keep Rf=10 KΩ and RS =22KΩ Set input voltage Vin=1 V and measure output voltage V0.

For various values of Vin=1.5V,2V and 2.5V measure output voltage V0.

For second set keep Vin=1 V and RS =22KΩ. For Rf=10 KΩ,33KΩ and 47KΩ measure output voltage V0.

Compare and verify experimental value with the theoretical value.

Non inverting amplifier.

Connect the circuit as shown in the figure using op-amp IC 741. Keep Rf=10 KΩ and RS =10KΩ

Set input voltage Vin=1 V and measure output voltage V0.1

For various values of Vin=1.5V,2V and 2.5V measure output voltage V0.

For second set keep Vin=1 V and RS =22KΩ. For Rf=10 KΩ,33KΩ and 47KΩ measure output voltage V0.

Compare and verify experimental value with the theoretical value

Summing amplifier.

Connect the circuit as shown in the figure using op-amp IC 741.

Keep R1 =R2=10 KΩ and Rf =10KΩ Set input voltage V1=1 V and V2=1V and measure output voltage V0.

Measure output voltage V0 for different values of input voltage.

Tabulate the readings and verify the summing action of the amplifier

Inverting amplifier.

S.No RS

(Ω) RF

( Ω) Vin

(volt)

VO

(volt)

Experimental

voltage gain

OV

in

VA

V

Theoretical

voltage gain

AV= F

S

R

R

1 10K 22K 1

2 10K 22K 1.5

3 10K 22K 2

4 10K 22K 2.5

5 10K 10K 1

6 10K 22K 1

7 10K 33K 1

8 10K 47K 1

Non inverting amplifier.

S.No RS

(Ω) RF

(Ω) Vin

(volt)

VO

(volt)

Experimental voltage gain

OV

in

VA

V

Theoretical voltage gain

AV=1 F

S

R

R

1 10K 10K 1

2 10K 10K 1.5

3 10K 10K 2

4 10K 10K 2.5

5 10K 10K 1

6 10K 22K 1

7 10K 33K 1

8 10K 47K 1

Summing amplifier.

S.No V1 (volt) V2 (volt) Experimental output voltage VO (volt)

Theoritical output voltage VO (volt)

1

2

3

4

Result:

1.The inverting and non inverting amplifier are construted using operational amplifier and the voltage gain is determined.

2. Summing amplifier is constructed and the output voltage is equal to the sum of the input

voltage.

10.Integrated logic gate circuit.

Formula:

Logic gate IC No Expression

AND 7408 Y=A.B

OR 7432 Y = A+B

NOT 7404 Y A

NAND 7400 .Y A B

NOR 7402 Y A B

EX-OR 7486 Y A B

Y= Output voltage

A & B are input voltage

Pin configuration

IC 7400(NAND), 7408(AND),7432(OR),7486(EXOR)

IC 7402(NOR)

IC 7404(NOT)

Procedure

Logic 1 represents High voltage

LED is ON Logic 0 represents low voltage

LED is OFF

To verify the Truth table of IC 7432,7408,7400,7486.

Connect 5V to the pin 14 and the ground to pin 7 of the IC.

Connect the two inputs A and B to the pin 1 and 2.

Connect the output to the pin 3. Keep both the inputs as 0 and check whether the LED is ON or OFF.

Keep the inputs A and B as 0,1 &1,0&and 0,0 and check whether the LED is ON of OFF.

Verify the output with the logic equation.

To verify the IC 7402.

Connect 5V to the pin 14 and the ground to pin 7 of the IC. Connect the two inputs A and B to the pin 2 and 3.

Connect the output to the pin 1.

Keep both the inputs as 0 and check whether the LED is ON or OFF.

Keep the inputs A and B as 0,1 &1,0 & and 0,0 and check whether the LED is ON of OFF. Verify the output with the logic equation.

To verify IC 7404 Connect 5V to the pin 14 and the ground to pin 7 of the IC.

Connect the two inputs A and B to the pin 2 and 3.

Connect the input A to the pin 1. Connect the output pin to the pin 2.

Keep the input logic as 0 and check the output.

Keep the output logic as 1 and check the output.

Observation:

OR

A A+B

B

AND

A B Y= A.B

0 0

0 1

1 0

1 1

NOT

A Y A 0

1

NOR

A B Y A B 0 0

0 1

1 0

1 1

NAND

EXOR

A B Y A B

0 0

0 1

1 0

1 1

Result:

Logic gate OR, AND, NOT, NAND,NOR,EXOR gate is constructed using integrated chips and

their truth table is verified.

A B Y= A+B

0 0

0 1

1 0

1 1

A B .Y A B

0 0

0 1

1 0

1 1

NAME: ___________________

STD:_____________________

SCHOOL:_____________________