Physics practicals

here you can find out the stuff related to the physics practicals during the first year of engineering students.

1. Experiment No. 01

Experiment Title: Newton’s Rings Experiment



Aim:-

To study interference phenomenon using Newton‟s ring and to determine the
wavelength of Sodium light (source) using Newton‟s ring.

Apparatus:-
Sodium light source, Plane glass plate, Plane convex lens, Traveling microscope,
Convex lens


Diagram:-

Theory : -
The formation of Newton‟s ring is due to the interference of light waves
reflected from the upper & lower surface of the air film enclosed between the
lens & plane glass plate . The thickness of the air film at point of contact is
zero & increases as well as move from the point of contact towards the
periphery of the lens. The thickness of the air film will be uniform for all points
lying on a circle with the centre at the point of contact. Thus using reflected

2. light, the central point will be dark surrounded by bright circles separated by dark
rings .
If R is radius of curvature of the lens & „t‟ the thickness of the air film at A & B.
The necessary condition for bright fringes using reflected light is as -

2µtcosθ = 2
)12(n

Where n = 0,1,2,3, --------------

Since θ is very small cos θ = 1 ( θ is the angle of refraction )

2 µt = 2
)12(n ( For air , µ = 1 )

For dark Fringes,

2µt = 2
)2(n

= nλ Where n = 0,1,2,3,------------


D





A B

From geometry of figure ,
EB x AE = EO x ED
EB = AE = r = The radius of interference fringe
EO = BQ = t = Thickness of the film
 r
2
= t ( 2Rt )
R


C


E
P O R

3. = 2Rt - t
2

= 2Rt
Neglecting t
2
in comparison to 2Rt

 t = R
r
2
2
i) For bright fringe,
2 µ R
r
2
2 = 2
)12(n
r = Rn)12( /2
ii) For dark fringe,
r = 2/Rn
When n =0 the radius of dark fringes is zero & the bright fringe is R
Therefore alternately dark & bright fringes are produced with dark center .
Suppose the diameter of the n
th
dark fringe be on then

Rn
D
n
4
2

& diameter of (n+p)
th
ring be Dn+p , then
2
pnD
 = 
Rpn)(4
2
pnD
 = 2
nD = 
Rp4



For bright fringes also ,
RPDD
npn 4
22


Procedure:-
1) Clean the glass plate G1 ,G2 & the Plano convex lens.
2) Illuminate the center of lens L2 by adjusting the inclination of glass plate
λ = pR
DD
npn
4
22



4. G1 at 45
0

3) Focus the eyepiece on the cross wire & move the microscope in the vertical
Plane by means of rack & Pinion arrangement till the rings are quite distinct
clamp the microscope in the vertical side.
4) Move the microscope in a horizontal direction to the one side of the ring. Fix
up the cross wire tangential to the ring & note this reading .Again the microscope
is moved in the horizontal plane &the cross wire is fixed Tangentially to the
Successive bright rings noting the vernier readings till the other side is reached.
This is shown in following figure 3.







Dn

Dn+p
Fig. 3
Formulae:-
pR
DD
or
pR
SLOPE
npn
4
4
22






Where λ = Wavelength of monochromatic light used

R = Radius of curvature of the spherical surface of the Plano
convex lens
D n + p = Diameter of (n+p)
th
dark ring
Dn = Diameter of the n
th
dark ring

5. Observations:-

1. Least count of traveling microscope:

i) Smallest division on main scale (X) = -------------cm

ii) Total no. of division on vernier scale (Y) = -------------

Y
X
CLLeastCount.).(
---------------- = --------------cm
2. Observation table to determine diameter of ring :

Calculations :- λ = pR
DD
npn
4
22










Calculations by graph :



No. of
ring
(n)

Microscope reading

Diameter
D
(a-b) cm

D
2
=
(a-b)
2

cm
2


D
2
n + p
- D
2
n
cm
2


Mean
(D
2
n + p
- D
2
n)
cm
2

Left end
‘a’ cm
Right end
‘b’ cm
M.S.R V.S.R. Total M.S.R. V.S.R. Total
1 (D
2
6-D
2
1)
= 2
3 (D
2
7-D
2
2)
= 4
5 (D
2
8-D
2
3)
= 6
7 (D
2
9-D
2
4)
= 8
9 (D
2
10-D
2
5)
= 10

6. Draw a graph between diameter D
2
n along Y- axis & No. of rings (n) along X axis.
Nature of the graph is as follows –


Y



(Diameter) D
2
n

D
2
n + p

D
2
n


n n+p X

No. of rings

Result:
i) The mean wavelength of sodium light = --------------- A
0


ii) Standard mean wavelength = --------------- A
0

iii) Percentage error = --------------- %

Conclusion:-

i)______________________________________________________________________

ii)______________________________________________________________________

iii)_____________________________________________________________________

References:

i) Engineering Physics By M. N. Avadhanulu & P. G. Kshirsagar

ii) Engineering Physics By S. L. Gupta

Experiment No. 02

7. Experiment Title: Newton’s Rings Experiment



Aim:-

To study interference phenomenon using Newton‟s ring and to determine the
Refractive index of liquid using Newton‟s rings experiment.

Apparatus:-
Sodium light source, Plane glass plate, Plane convex lens, Traveling microscope,
Convex lens, liquid


Diagram:-

Theory : -
The experiment is performed when there is an air film between the plano-convex
lens and the optically plane glass plate.The diametre of the mth and the
(m+p)th dark rings are determined with the help of a travelling microscope.

For air
As shown in figure arrange the lens with glass plate. Pour one or two drops of
liquid whose refractive index is to be determined without disturbing the
arrangement. Now the air film between the lens and glass plate is replaced by the
liquid. The diameters of m+pth and mth rings are determined.

8. For liquids,

,for dark rings

For normal incidence cosr =1,so

But
Rearranging the above equation ,we get


We have
for liquids,

From these two euations the refractive index of the given liquids is given by


Procedure:-
1) Clean the glass plate G1 ,G2 & the Plano convex lens.
2) Illuminate the center of lens L2 by adjusting the inclination of glass plate
G1 at 45
0

3) Focus the eyepiece on the cross wire & move the microscope in the vertical
Plane by means of rack & Pinion arrangement till the rings are quite distinct
clamp the microscope in the vertical side.
4) Move the microscope in a horizontal direction to the one side of the ring. Fix

9. up the cross wire tangential to the ring & note this reading .Again the microscope
is moved in the horizontal plane &the cross wire is fixed Tangentially to the
Successive bright rings noting the vernier readings till the other side is reached.
This is shown in following figure 3.







Dn

Dn+p
Fig. 3
5) You have to take the reading of rings on either side of the centre dark ring.
6) Enter the readings in the tabular column.
7) Calculate the refractive index of the medium by using the given formula .


Observations:-

1. Least count of traveling microscope:

i) Smallest division on main scale (X) = -------------cm

ii) Total no. of division on vernier scale (Y) = -------------

Y
X
CLLeastCount.).(
---------------- = --------------cm




2. Observation table to determine diameter of ring :
Medium : Air

10. Medium: Water




Result:


No. of
ring
(n)

Microscope reading

Diameter
D
(a-b) cm

D
2
=
(a-b)
2

cm
2


D
2
n + p
- D
2
n
cm
2


Mean
(D
2
n + p
- D
2
n)
cm
2

Left end
‘a’ cm
Right end
‘b’ cm
M.S.R V.S.R. Total M.S.R. V.S.R. Total
1 (D
2
6-D
2
1)
= 2
3 (D
2
7-D
2
2)
= 4
5 (D
2
8-D
2
3)
= 6
7 (D
2
9-D
2
4)
= 8
9 (D
2
10-D
2
5)
= 10


No. of
ring
(n)

Microscope reading

Diameter
D
'

(a-b) cm

D
'
2
=
(a-b)
2

cm
2


D'
2
n + p
- D'
2
n
cm
2


Mean
(D'
2
n + p
- D'
2
n)
cm
2

Left end
‘a’ cm
Right end
‘b’ cm
M.S.R V.S.R. Total M.S.R. V.S.R. Total
1 D'
2
6 - D'
2
1
= 2
3 D'
2
7 - D'
2
2
= 4
5 D'
2
8 - D'
2
3
= 6
7 D'
2
9 - D'
2
4
= 8
9 D'
2
10 - D'
2
5
= 10

11. i) The mean wavelength of sodium light = --------------- A
0


ii) Standard mean wavelength = --------------- A
0

iii) Percentage error = --------------- %

Conclusion:-

i)______________________________________________________________________

ii)______________________________________________________________________

iii)_____________________________________________________________________

References:

i) Engineering Physics By M. N. Avadhanulu & P. G. Kshirsagar

ii) Engineering Physics By S. L. Gupta

12. Experiment No – 3

Experiment Title - LASER Wavelength & Angular Divergence

Aim : - To find wavelength and angular divergence of Diode Laser
Apparatus:- Diode laser, diffraction grating, screen, lens, optical bench
arrangement

Fig. :

















Theory :-
(i)Wavelength
When monochromatic radiation of wavelength λ is diffracted by a
dffraction grating the n
th
order principal maxima is formed at an angle θn
given by
(a+b) sin θn = n λ
Where (a+b) is grating constant which is equal to
(a+b)= 1/N (N= No. of lines per unit length)


Diffraction
grating
Diode laser
2
nd
ordere
2
nd
ordere
1
st
ordere

13. Laser light is incident on the diffraction grating and diffraction is obtained on the
screen usually a graph paper
sin θn = yn / √ x
2
+ yn
2


λ = yn / √ x
2
+ yn
2
(1 / n.N)
A graph is plotted between sin θn Vs. n for n = 1, 2, 3, 4, .. and from the
slope of the graph Δsin θn / Δn wavelength of the laser light is calculated
(ii) Angular divergence
θ = d /D
where d = diameter of spread. and
D = distance between source and screen.
It is assumed that θ is extremely small as actually the case is.

Procedure:-

1. Place the diffraction grating in front of the laser beam so that a diffraction pattern
is obtained on a graph paper which serves as screen placed at a distance of around
2m from the grating.
2. The central maxima (bright one ) is due to the undeviated rays. Measure the
distance of first order diffraction spot from the central spot. Also, measure the
distance from the grating to the screen. Calculate sin θ1 .
3. Repeat the above procedure for second and third order diffraction spot and find sin
θ2. plot a graph between sin θn Vs n and calculate λ by using formula.
4. Remove the diffraction grating and measure the spread d of the laser on the screen
and the distance D between the laser gun and the screen as shown in the fig.
Calculate the angular spread by using formula.
5. Repeat the above procedure for three different D and find the mean.
Observations and Calculations:

14. (i) Measurement of λ
Grating constant (a + b ) = 1/N…………..
N = 1/ (a + b)
N= 500 lines/mm
Distance of screen from grating =

Sr. No. Order of diffraction Position of
screen x
Cm (distance
between Grating
and screen)
Distance from
central spot y
cm
sin θn = sin θn
= yn / √ x
2
+ yn
2



1 0 60 cm -
2 1 60 cm 22 cm
3 2 60 cm 57 cm

Slope of sin θ Vs n , Δsin θ / Δn = ………………


Actual λ of a diode laser light = …6320 A
o

% Error = …………

(ii)
Measurement of angular spread


Sr. No.

Distance from
laser to screen (D)
Linear spread d Angular spread
θ = d / D

140 8
150 7.5


Mean θ = ………..radian

=…………degree

15. Result:-
(i) Wavelength of Diode laser light = ………….

(ii) Angular spread of laser light = ……….


Conclusion:
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------


Sources of errors and precautions :-
1. The graph used as screen should not have folding. If present it may introduce error.
2. The laser should neither be too close nor to far from the screen. Keep a distance of
a few meters.
3. The spread should be obtained on a wide screen.
4. Do not stare laser source directly, it may damage your eye.

References:

i) Engineering Physics By M. N. Avadhanulu & P. G. Kshirsagar

ii) Engineering Physics By S. L. Gupta

16. Experiment No. 04

Experiment Title: Energy Band Gap Of Semiconductor


Aim:
To determine forbidden energy gap of semiconductor.

Apparatus:
P-N junction diode, oven, thermometer, micrometer, etc.


Experimental Arrangement:



Formula:
Eg= 2.k.β

Where,
Eg= Forbidden energy gap
K= Boltzmann‟s constant
Β= slope of graph of ln RT Vs 1/T

17. Theory:
In the atom of solid there are mainly three bands, namely valence band,
Formed by valence electrons, conduction band, formed by conduction
electrons and forbidden gap which is gap between valence and conduction
band. The minimum energy required for valence electrons to cross over to
the conduction band is called energy gap. On the basis of forbidden energy
gap solids are classified into three groups,
1.Conductors : Eg = 0eV
2.Semiconductors : Eg = 1eV
3.Insulators : Eg = 15eV
In semiconductors, as temperature increases, the valence electrons gain energy,
and cross over forbidden gap to move into conduction band. Therefore, in
semiconductors, as temperature increases, resistivity decreases. We have,
ρT = exp (Eg/2kT)
Or
RT = exp(Eg/2kT)
Taking logarithm on both sides, we get
Ln RT = ln R0 + Eg/2kT
This equation has the form y=mx+c. Thus the graph of in RT Vs 1/T is straight
line with slope equal to Eg/2k. This consideration is used to estimate the
forbidden energy gap of given semiconductor material.


Procedure :

1. Note down voltage of a given cell o battery.
2. Connect the circuit as shown in the diagram and get it checked by
teacher.
3. Put power on and put the oven switch to „on‟ position and allow the
oven temperature to increase unto .
4. When temperature reaches , switch of the oven.
5. Measure current at and then start to take the current readings as
the diode cools down to , in the steps of each.
6. Plot a graph of Ln RT Vs 1/T and find the slope of graph, as β.
7. Substitute it in the given formula to find „Es‟ in joules and then in
electron volt.
8. Compare it with the standard value for given diode,

18. Observations:
Battery voltage (V) = volts.


Observation Table :

Sr.
No.
Temperature

I
(μA)
T = (t+
273)K
RT = V/I 1/T Ln RT
1. 80 50*10
-6

2. 75 46.3*10
-6

3. 70 34.4*10
-6

4. 65 24.2*10
-6

5. 60 19.2*10
-6

6. 55 11.7*10
-6

7. 50 7.1*10
-6

8. 45 5.2*10
-6

9. 40 3.9*10
-6

10.

19. Graph:
Y



lnRT





1/T
0
K X

Calculations:
Eg = 2.k.β
=2× ×(slope of graph of ln RT Vs 1/T)
= ……………….Joules
=………………..eV
Result:
The energy gap of given semiconductor diode is…………………..eV.

Conclusions:
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
Precautions:
1.The maximum temperature should not exceed
2.Bulb of thermometer and the diode should inserted well in the oven.


References:

i) Engineering Physics By M. N. Avadhanulu & P. G. Kshirsagar

ii) Engineering Physics By S. L. Gupta

20. Experiment No.05

Experiment Title: Width of Single Slit By Using Laser

Aim: To measure width of single slit from the study of Fraunhofer diffraction
using Laser source.

Apparatus: Diode, Laser, single slit, lens, screen and meter scale.

Experimental Arrangement:

Formula: a Sinθ = nλ

Where,
a =width of single slit
λ =wavelength of laser source
n =order of diffraction pattern
θ =angle of diffraction for order „n‟
Sinθn =dn/D
dn =distance of minima on either side of central maxima
D =distance of screen from the slit
Theory:
LASER stands for Light amplification by Stimulated Emission of Radiation.
It is very intense, monochromatic, highly unidirectional, and highly coherent
beam of light. The phenomena of bending of light round the edges of obstacle
in the path of light and spreading of light in the region of geometrical shadow is
called as diffraction.
When a narrow single slit is illuminated by laser light, diffraction pattern is
Observed on the screen. It consist of central maximum surrounded by alternate
minima and maxima of varing intensity. The intensity distribution in single slit

21. diffraction pattern is as shown in fig.



Width of single slit can be calculated by using condition of minimum intensity,
a Sinθ = nλ

a = nλ/ Sinθ

22. Procedure:
1. Mount the slit on the stand and place it in front of laser, so that laser beam
falls on single slit.
2. View diffraction pattern on the screen. Adjust the height and position of
the slit in front of laser device so that a good diffraction pattern with
distinct maxima and minima is obtained on the screen.
3. Measure distance between slit and screen with the help of meter scale
(Dcm)
4. Measure distance between mid point of the central maximum and the
first, second third minima accurately on either side of it (dn cm)
5. Calculate width of single slit by using formula.


Observations:
Wavelength of laser =…………6320. =……6320*10
-8
cm


Observation Table:


Sr.
N
o.
Order of
diffraction
pattern
n
Distance
between centre
of central
maximum and
minimum
dn cm
Distance
between
slit and
screen
D cm
Sinθ=
dn/D
a = nλ/
Sinθ
Mean
a
1 1 0.35 74.9
2 2 0.7 74.9
3 3 1 74.9
4 4 1.4 74.9
5 6 1.8 74.9

23. Calculations: a = nλ/ Sinθ








Result: The width of single slit using the Laser source=……………………..cm

Conclusion:
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------


Precautions:
1. Screen should be placed normal to the incident beam.
2. „d‟ should be measured accurately.
3. The Laser beam should not be seen directly. a = nλ/ Sinθ



References:

i) Engineering Physics By M. N. Avadhanulu & P. G. Kshirsagar

ii) Engineering Physics By S. L. Gupta

24. Experiment No – 6

Experiment Title - Hall Effect

Aim:- Study of Hall effect in semiconductor and determination of allied parameters.

Apparatus:- 1) Hall Effect Board (Digital) 2) Hall Probe 3) Electromagnet 4) Constant
current power supply 5) Digital Gauss meter with Hall probe 6) Hall
probe stand.

Diagram:-





Formulae:-
I) Hall coefficient RH = .
II) No of charge carriers n = (e=electric charge in coulomb i.e. 1.6 × 10
-19
)
III) Mobility of charge carriers µ = RH [ ] .
IV) Hall angle tan Ø = µ ( c = Velocity of light)
V) Conductivity σ =

25. Procedure:-
(1) Connect the widthwise (red color) contacts of the Hall probe to the terminal
marked „Voltage‟ and lengthwise (green color) contacts to the terminal
marked “Current” of the Hall effect Board.
(2) Put the switch marked ON/OFF of Hall effect Board to ON position.
(3) Put the meter selector switch towards 20mA and adjust current (not exceeding
8 mA) by current adjust pot.
(4) Put the meter selector switch towards 200mV. There may be some Voltage
reading even outside the magnetic field. This is due to imperfect alignment of
the four contacts of the Hall probe and is generally known as „Zero field
potential‟. In case its value is comparable to the Hall voltage, it should be
adjusted to a minimum possible {for Hall probe (Ge) only}. In all cases, this
error should be subtracted from the Hall voltage reading.
(5) Now place the probe in the magnetic field as shown in fig.
(6) Connect the constant current power supply with electromagnet. Switch on the
supply and adjust the current to any desired value. Rotate the Hall probe till it
become perpendicular to magnetic field. Hall voltage will be maximum in this
adjustment.
(7) Measure Hall voltage for both the direction of the current by interchanging the
current leads and magnetic field by changing power supply leads.
(8) Measure the Hall voltage as a function of current by changing the current
through the coil. Plot a graph of VH / Hz and determine its slope.
(9) Calculate the value of RH

Observations:-
1) Distance between poles of electromagnets = 1X = 1cm
2) Breadth = b 0.5cm
3) Thickness = d = 0.51mm

26. Observation Table I :-
A) Measurement of magnetic Field:
Sr.
No.
Current (A) Magnetic field (Gauss) Hz
1
2
3
4
5



Observation Table II :-

Measurement of Hall Voltage:


Trial
No:
Magnetic Field
(Tesla T)
Thickness (t)
m
Hall current,
mA
Hall Voltage
mV
R
H
1
2
3
4



Result:-


Sr.
No
.
Hall Coff.
RH cm
3
/
coulomb
No. of charge
carriers per unit
volume(n) cm
-3

Mobility of
charge Carriers
(µ) cm
2
Volt
-1
sec
-
1

Hall angle Conductivity
σ
Tan Φ Φ

27. Conclusion:
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------
----------------------------------------------------------------------------------------------------

Precautions:
(1) Properly connect the widthwise (red color) contacts of the Hall probe to the
terminal
marked „Voltage‟ and lengthwise (green color) contacts to the terminal
marked “Current” of the Hall effect Board.

28. Experiment No.7

Aim:- To draw the characteristic curves of a forward & reverse Biased P-N junction diode and
to determine the static resistance of the given diode.

Apparatus Required: Semiconductor P-N junction diode kit and connecting wires.

Theory: The diode is a device formed from a junction of n-type and p-type semiconductor
material. The lead connected to the p-type material is called the anode and the lead connected to
the n-type material is the cathode. In general, the cathode of a diode is marked by a solid line on
the diode.



.

Function of a P-N junction diode in Forward Bias


The positive terminal of battery is connected to the P side(anode) and the negative terminal of
battery is connected to the N side(cathode) of a diode, the holes in the p-type region and the
electrons in the n-type region are pushed toward the junction and start to neutralize the depletion
zone, reducing its width. The positive potential applied to the p-type material repels the holes,
while the negative potential applied to the n-type material repels the electrons. The change in
potential between the p side and the n side decreases or switches sign. With increasing forward-
bias voltage, the depletion zone eventually becomes thin enough that the zone's electric field
cannot counteract charge carrier motion across the p–n junction, which as a consequence
reduces electrical resistance. The electrons that cross the p–n junction into the p-type material
(or holes that cross into the n-type material) will diffuse into the nearby neutral region. The
amount of minority diffusion in the near-neutral zones determines the amount of current that
may flow through the diode.

29. Function of a P-N junction diode in Reverse Bias


The positive terminal of battery is connected to the N side(cathode) and the negative terminal of
battery is connected to the P side(anode) of a diode. Therefore, very little current will flow until
the diode breaks down.




The positive terminal of battery is connected to the N side(cathode) and the negative terminal of battery
is connected to the P side(anode) of a diode, the 'holes' in the p-type material are pulled away from the
junction, leaving behind charged ions and causing the width of the depletion region to increase.
Likewise, because the n-type region is connected to the positive terminal, the electrons will also be
pulled away from the junction, with similar effect. This increases the voltage barrier causing a high
resistance to the flow of charge carriers, thus allowing minimal electric current to cross the p–n junction.

30. The increase in resistance of the p–n junction results in the junction behaving as an insulator.
The strength of the depletion zone electric field increases as the reverse-bias voltage increases. Once the
electric field intensity increases beyond a critical level, the p–n junction depletion zone breaks down and
current begins to flow, usually by either the Zener or the avalanche breakdown processes. Both of these
breakdown processes are non-destructive and are reversible, as long as the amount of current flowing
does not reach levels that cause the semiconductor material to overheat and cause thermal damage.

Procedure:

1. Forward Bias-Si Diode
1. Set DC voltage to 0.2 V .
2. Select the diode.
3. Set the resistor.
4. Voltmeter is placed parallel to Silicon diode and ammeter series with resistor.
5. The positive side of battery to the P side(anode) and the negative of battery to the N
side(cathode) of the diode.
6. Now vary the voltage upto 5V and note the Voltmeter and Ammeter reading for
particular DC voltage .
7. Take the readings and note Voltmeter reading across Silicon diode and Ammeter reading.
8. Plot the V-I graph and observe the change.
9. Calculate the dynamic resistance of the diode. rd=ΔV/ΔI
10.Therefore from the graph we see that the diode starts conducting when the forward bias
voltage exceeds around 0.6 volts (for Si diode). This voltage is called cut-in voltage.



2. Reverse Bias-Si Diode
1. Set DC voltage to 0.2 V .
2. Select the diode.
3. Set the resistor.
4. Voltmeter is placed parallel to Silicon diode and ammeter series with resistor.
5. The positive terminal of battery is connected to the N side(cathode) and the negative
terminal of battery is connected to the P side(anode) of a diode.
6. Now vary the voltage upto 30V and note the Voltmeter and Ammeter reading for DC
voltage .
7. Take the readings and note Voltmeter reading across Silicon diode and Ammeter reading.
8. Plot the V-I graph and observe the change.



Observation Table: Forward bias

Sr.
No.
Voltage (Volts) Current in (mA)

31. Observation Table: Reverse bias

Sr.
No.
Voltage (Volts) Current in (mA)














Result:- The I/V Characteristic of P-N Junction diode is shown in the graph.

Conclusion:-






Precaution and source of error:
1. Voltmeter and ammeter of appropriate ranges should be selected.
2. The variation in V should be done in steps of 0.1 V.
3. The battery connections of p-n junction diode should be checked and
it should be ensured that p is connected to positive and n to the
negative of the battery.
4. Never cross the limits specified by the manufacturer otherwise the
diode will get damaged.