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

A D
2
A
Sin
2
Sin
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 R 1.
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 (R 4 )
The difference between R3 and R4 is the angle of minimum deviation D.
The refractive index of the prism is calculated using the formula  
Observation:
Least count of the spectrometer.
1 MSD= ______
Number of divisions in vernier scale(N) =_______
Least count =
1
x1MSD 
N
A D
2
A
Sin
2
Sin
Angle of prism
Ray
main
scale
reading
VC
Vermier I
Total reading=
MSR+ (V.CxL.C)
Main scale
reading
VC
Vernier II
Total reading= MSR+
(V.CxL.C)
Main scale
reading
VC
Vernier II
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
main
scale
reading
VC
Vermier I
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 R 2
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 R 1.
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)
Vermier I(R1)
Vernier II
(R2)
Vernier I
Vernier II
θ1 = R1 – R2
θ2 = R1 –R2
Colour
MSR
VC
Total
reading =
MSR+
(V.CxL.C)
MSR
VC
Total
reading =
MSR+
(V.CxL.C)
Blue
Green
Yellow
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 = _________
θ = θ 1 + θ2
2
3. Metre bridge.
Aim
To determine the resistance and specific resistance of the wire using metre bridge.
Formula
X R
lX
lR
X= Resistance of the wire
R = known resistance
l X = balancing length of the unknown resistance
lR = balancing length of the known resistance.

 r2 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
.
lR
The radius of the wire is calculated using the screw gauge.
The specific resistance of the wire is calculated using the formula  
 r2 X
l
observation:
Length of the wire = _______
Resistance of the wire :
S.No
R
(ohm)
Balancing
length before
interchanging
l1 cm
l2 =
100-l1
cm
Balancing
length
after
interchanging
l4
l3
cm
=100-l4
cm
Mean
X R
ohm
lR 
cm
l1  l3
2
lX 
l2  l4
2
cm
1
2
3
4
5
Radius of the wire:
Least count = ______
Zero error = _______
Zero correction = _______
S.No
P.S.R (mm)
H.S.C
C.H.S.R
1
2
3
4
Result :
1. The resistance of the given wire = ___________
2. The specific resistance of the wire = ___________
Total reading = PSR +
(CHSR X LC) (mm)
lX
lR
4.potentiometer.
Aim
To compare the emf of the two given cells using potentiometer.
Formula :
E1 l1

E2 l2
E1 =
E2 =
l1 =
l2 =
Electromotive force of leclanche cell
Electromotive force of Daniel cell
balancing length of leclanche cell
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
E1 l1

E2 l2
Observation:
S.No Balancing length of leclanche
cell (cm)
1
2
3
4
5
6
Result:
The mean ratio of emf of two cells =
Balancing length of Daniel
cell (cm)
E1 l1

E2 l2
5. Tangent galvanometer.
Aim
To determine the horizontal component of earth magnetic field using tangent galvanometer.
Formula:
BH 
0 n
I
2a tan 
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 00-00
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 300-300. 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 θ 3 , 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 BH 
is determined.
0 n
I
2a tan 
Observation:
Circumference of the coil (2πa) = _______
Radius of the coil = a = 2πa/2π = _______
Number of turns n = _____
S.No
Current
I (A)
Deflections in T.G (degrees)
θ1
θ2
θ3
θ4
Average θ
I / tanθ
(A)
1
2
3
4
Result :
The horizontal component of earth’s magnetic field =
_________
6.Sonometer.
Aim
To determine the frequency of the alternating current using sonometer.
Formula:
n
1 T 1
2 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. n 
1 T 1
2 l
m
Observation:
S.No
Load
(kg)
Length of the vibrating
segment
l1 (cm)
l2 ( cm)
Mean l
(cm)
T
newton
T
N/m
l
1
2
3
4
5
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
RF 
VF
I F
RF = Resistance of the diode in forward bias
VF = Forward bias voltage
IF = Forward bias current.
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
VF (volt)
IF (mA)
S.No
V0 (volt)
1
0.5
2
1.0
3
1.5
4
2.0
5
2.5
6
3.0
0.6
7
3.5
7
0.7
8
4.0
8
0.8
9
4.5
10
11
5.0
5.5
12
6.0
1
0.1
2
0.2
3
0.3
4
0.4
5
0.5
6
Result:
1. Forward bias resistance of the diode = __________
2.Zener breakdown voltage = ___________
IF (mA)
8. NPN Transistor.
Aim:
To determine the (i) input impedance (ii) output impedance and (iii) current gain of the transistor.
Formula:
 VBE 

 I B VCE
Input impedance ri  
 VCE 

 I C  I B
Output impedance r0  
 I C 

 I B VCE
Current gain   
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 I c 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
IB (μA)
current gain
VCE = 5 V
S.No
VBE ( volt)
1
0.1
2
0.2
3
0.3
4
0.4
2
5
0.5
3
6
0.6
4
7
0.7
8
0.8
9
0.9
S.No`
IB (μA)
IC (mA)
1
.
Output characteristic
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
Result :
1. Input impedance = _________
2.output impedance = ____________
3. current gain
= ._____________
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
AV 
VO  RF

Vin
RS
Voltage gain in non inverting amplifier
V 
R 
AV   O   1   F 
 Vin 
 RS 
Output voltage in inverting amplifier
VO  V1  V2
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
VO
Experimental
voltage gain
(volt)
(volt)
AV 
Vin
VO
Experimental
voltage gain
(volt)
(volt)
AV 
1
2
3
10K
10K
10K
22K
22K
22K
1
1.5
2
4
10K
22K
2.5
5
6
10K
10K
10K
22K
1
1
7
10K
33K
1
8
10K
47K
1
VO
Vin
Theoretical
voltage gain
AV=
 RF
RS
Non inverting amplifier.
S.No
RS
(Ω)
RF
(Ω)
1
2
3
10K
10K
10K
10K
10K
10K
1
1.5
2
4
10K
10K
2.5
5
6
10K
10K
10K
22K
1
1
7
10K
33K
1
8
10K
47K
1
VO
Vin
Theoretical
voltage gain
AV= 1 
RF
RS
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
Logic 0 represents
High voltage
LED is ON
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
0
0
1
1
B
0
1
0
1
Y= A+B
B
0
1
0
1
Y= A.B
A
A+B
B
AND
A
0
0
1
1
NOT
A
0
1
YA
NOR
B
0
1
0
1
Y  A B
A
B
Y  A.B
0
0
1
1
0
1
0
1
A
0
0
1
1
NAND
EXOR
A
0
0
1
1
B
0
1
0
1
Y  A B
Result:
Logic gate OR, AND, NOT, NAND,NOR,EXOR gate is constructed using integrated chips and
their truth table is verified.
NAME: ___________________
STD:_____________________
SCHOOL:_____________________