Practical Paper-2
Practical Paper-2
Practical Paper-2
Exp. No.2.1
Spectrometer- i-d curve
Aim: To study the relationship between the angle of incidence ‘i’ and the angle of deviation ‘d’
of a glass prism and hence to determine the refractive index of the material of the prism by
drawing the graph of ‘i’ and ‘d’.
Apparatus: Spectrometer, sodium vapor lamp, prism, reading lens etc.
Theory: For a given prism, corresponding to a
given angle of deviation there are two possible A
angles of incidence i1 and i2. These two angles
are such that if one of the angles is the angle of
incidence, the other angle will be the angle of d
emergence. i1 r1 i2
r2
Let i1 and i2 be the two angles of incidence
and r1 and r2 be the corresponding angles of
refraction for the given angle of deviation d.
Fig:a
Then,
i1 i 2 = A + d (1)
r1 r2 = A (2) d
Fig.b gives the variation of angle
of deviation d with angle of
incidence i. When the angle of
deviation is minimum, i1 = i2 = i,
r1 = r2 = r and d = D. Then, from
eqn.1 we get,
D
2i = A + D (3)
i
AD i1 i1=i2 i2
i = (4)
2
From eqn.2, Fig.b : i-d curve for an equilateral prism of = 1.62
A
r = (5)
2
AD
sin
Refractive index of the material of the prism, =
sin i
= 2 (6)
sin r A
sin
2
Procedure: All the preliminary adjustments of the spectrometer are made (refer Exp. No. 1.11
practical-I). The prism is mounted on the prism table with its base is parallel and close to the
clamp. The prism table is leveled either by observing the reflected images from both the sides of
the prism or by using a spirit level. Then the prism is removed.
Adjustment to set the prism for a particular angle of incidence, say ‘i’ : Now the telescope is
brought parallel to the collimator and the direct image of the slit is obtained at the vertical wire
4 Optics & Electricity Practical II
of the cross wire. The reading on one of the verniers is noted. Now the telescope is released and
turned through an angle = 1802i as shown in the fig.c (dashed arrow represents the motion of
the telescope) and it is clamped there. Place the prism on the prism table with one of the faces,
say AB, facing the collimator. Then the vernier table (or prism table) is rotated to and fro so that
the image of the slit reflected from the face AB is obtained on the vertical cross wire. The vernier
table is clamped there. Now the prism is set for the angle of incidence i. [It may be convenient to
set the two verniers at 0-180 for the direct image. But in this case the prism table alone is rotated
to find the reflected image.]
The telescope is then released, turned towards the refracted ray and the refracted image of
the slit is obtained at the vertical cross wire. The readings on both the verniers are noted. Let it
be ‘a’. Again the telescope is brought in a line with the collimator and direct image of the slit is
obtained at the cross wire. The readings (b) on both the verniers are taken. The difference
between ‘a’ and ‘b’ gives the angle of deviation. The angles of deviation for different angles of
incident, say, i = 40, 45, 50 etc. [For prisms with 1.5 to 1.6, the range of i is 35 to 65
and for prisms with, 1.6 to 1.7, the range of i is 40 to 70].
A graph is plotted with ‘i’ along the X-
axis and ‘d’ along the Y-axis as shown in
Collimator
fig.b. The angle of incidence corresponding to
Incident ray
the angle of minimum deviation can be
determined from the graph. Then by using
eqns.1 and 3 the angle of the prism can be B
calculated and the refractive index is Reflected ray
i i
Precautions:
The vernier table and the prism table A
are initially adjusted at the proper
positions (both the verniers are in a 1802i
Direct ray
MSR
MSR
MSR
MSR
VSR
VSR
VSR
VSR
Ver II
Mean
Total
Total
Total
Total
Ver I
35
40
45
50
55
60
65
70
Result
Angle of the prism A = ………….
Refractive index of the material of the prism, = ………….
Standard data*
Refractive index against air for mean sodium line (589.3 nm)
Crown glass 1.48 1.61
Flint glass 1.53 1.96
6 Optics & Electricity Practical II
Exp.No.2.2
Spectrometer-i1-i2 curve
Aim: To study the relationship between the two angles of incidence (one is the angle of
incidence i1 and the other is the angle of emergence i2) for a given angle of deviation. We also
aim to study the variation of angle of emergence with angle of incidence and to draw the i1-i2
curve.
Apparatus: Spectrometer, sodium vapor lamp, prism, reading lens etc.
Theory:
Let i1 and i2, respectively, be the angle A
of incidence and the angle of emergence of a
prism of angle A corresponding to the angle d
of deviation d. Then, i1 r1 r2 i2
i1 i 2 = A + d (1)
r1 r2 = A (2)
Fig:a
Fig.b gives the variation of angle of
emergence i2 with angle of incidence i1.
Angle of emergence i2
By eqn.4, 35
AD i1=i2
i = (6) O 35 40 45 50B 55 60 65
2 Angle of incidence i1
From eqn.2, Fig.b: i1-i2 curve for an equilateral prism of = 1.56
A See that the scales are same for X and Y axes
r = (7)
2
AD
sin
Refractive index of the material of the prism, =
sin i
= 2 (8)
sin r A
sin
2
Procedure: Preliminary adjustments of the spectrometer are made (refer Exp. No. 1.11 of
practical-I) and the prism is mounted on the prism table with its base is parallel and close to the
clamp. The prism table is leveled either by observing the reflected images from both the sides of
the prism or by using a spirit level.
Practical-II MCT 7
The angle of the prism is found out either by the supplementary angle method (refer Exp.
No. 2.6 of practical-II) or by observing the reflected rays from both sides of the prism.
Now the prism is set for a particular
angle of incidence, say i1 = 40 as in the i-
Collimator
d curve experiment. The telescope is then
Incident ray
released and is brought in a line with the
refracted ray. (The dashed curves indicate
the motion of the telescope). The telescope
is clamped there. By using the tangential i2 B
Reflected ray 1
screw of the telescope, the refracted image
i1 i1
Direct ray
direction of direct ray). Continue the C
rotation of the vernier table in the same d
direction till the refracted image is
returned at the vertical wire of the 2=1802i2
telescope. The vernier table is then
clamped and the tangential screw of it is Refracted ray
Fig:c
adjusted to get the refracted image on the
vertical cross wire.
The telescope is released and is rotated towards the reflected ray. (Reflected ray 2 in the
fig.c). By adjusting the tangential screw of the telescope, the reflected image of the slit is
obtained exactly on the cross wire. The readings on both the verniers are noted. (Use the reading
lens). Let it be ‘a’.
The telescope is again released. It is brought in the line of the collimator and the direct
image of the slit is obtained on the cross wire. Again readings (b) on both the verniers are taken.
The difference between the reading of the reflected image and the direct image gives 2=1802i2.
From that i2 can be calculated.
The experiment is repeated for different values of i1 = 45, 50, 55 etc. In each case i2 is
calculated. A graph is drawn between i1 and i2. From the graph, the angle of incidence ‘i’
corresponding to angle of minimum deviation D can be found by using eqn.3. Hence D can be
calculated using the equation D = 2i A.
Finally, the refractive index of the material of the prism is calculated using the eqn.8.
Precautions:
See all the precautions given in the i-d curve experiment.
Draw the i1-i2 curve with same scale in both the axes. Then it is easy to find out the i = i 1
= i2 for minimum deviation from graph by simply drawing the bisector of the angle
between X and Y axes (diagonal of squares on the graph).
Observation and tabulation
Value of one main scale division (1 m s d) = ……………
8 Optics & Electricity Practical II
Determination of i2
Reading corresponding to Reading corresponding to
reflected ray for second direct ray after the second
Angle of incidence ‘i1’
180 θ 2
2
vernier
MSR
MSR
MSR
MSR
VSR
VSR
VSR
VSR
Ver II
Mean
Total
Total
Total
Total
Ver I
35
40
45
50
55
60
65
70
Result
Angle of the prism A = ……….
Angle of incidence corresponding to minimum deviation, i = ……….
Angle of minimum deviation D = ……….
Refractive index of the material of the prism = ……….
Standard data*: Same as in the experiment for i-d curve.
Practical-II MCT 9
Exp.No.2.3
Spectrometer-Cauchy’s constants
Aim: To determine the constants in the Cauchy’s dispersion formula for the material of the
prism.
Apparatus: Spectrometer, mercury vapor lamp, prism, reading lens etc.
Theory: Considering the microscopic properties of the bound charged particles of a transparent
medium Cauchy developed a relation connecting the refractive index of the material and the
wavelength of light passing through it. Cauchy’s relation between the refractive index of the
material and the wavelength of the light is given by,
B
Refractive index, = A (1)
λ2
where, A and B are constants for a transparent material and are called the Cauchy’s constants.
(Do not confuse with constant A and the angle of the prism A). These constants can be
determined by a method as follows. Let 1 and 2 be the refractive indices corresponding to the
wavelengths 1 and 2 respectively. Then, (if 1 > 2)
B
1 = A (2)
λ12
B
2 = A 2 (3) P
λ2
1 1 λ2 λ2
μ1 μ 2 = B 2 2 = B 2 2 2 1 R Q
λ1 λ 2 λ1 λ 2
S PQ
B tan θ
B =
μ1 μ 2 λ λ 2
1
2
2
(4) QR
λ 22 λ12 A
From eqn.2 and 3,
B
A = μ1 (5a) 1
λ12 O
Fig.a
λ2
B
Or, A = μ2 (5b)
λ 22
If D is the angle of minimum deviation for a wavelength when it passes through a prism of
angle A , the refractive index is given by,
A D
sin
= 2 (6)
A
sin
2
10 Optics & Electricity Practical II
1
Cauchy’s constants can also be determined graphically. A graph is drawn with along
λ2
the X axis and along the Y axis (fig.a). The graph will be a straight line. Its slope gives the
constant B and the Y intercept gives the constant A.
Procedure: As usual, the preliminary adjustments, including the leveling of the prism table, of
the spectrometer are made. The angle of the prism A is determined as described in Exp.No.1.
11 of practical-I or Exp. No. 2.6 of practical-II.
The prism is then adjusted to obtain the refracted spectrum. It consists of different spectral
lines with violet being deviated most and red the least. The prism is then adjusted to be in the
minimum deviation position for the violet line as described in experiment number 11 and 12 of
Part 1. Readings on both the verniers are taken. The prism is removed carefully and the readings
on both the verniers for direct ray are noted. The difference between these readings gives the
angle of minimum deviation for violet light. Similarly the angles of minimum deviation for the
other colours are found out. Refractive indices of the material of the prism for various colours
are calculated using eqn.6. The Cauchy’s constants are determined by the calculation and
graphical methods.
Precaution:
See all the precautions given in the i-d curve experiment.
The prism is set to the minimum deviation position for each spectral line and in each case
the direct reading is to be taken.
‘a’ deviation
Ver 1 Ver II Ver 1 Ver II D = ab 1012
λ2
MSR
MSR
MSR
MSR
VSR
VSR
VSR
VSR
Ver II
Mean
Total
Total
Total
Total
Ver I
Calculation of Cauchy’s constants
1 2 1 2 B A
Mean
Exp.No.2.4
Spectrometer – Diffraction grating-normal incidence
Aim: To determine the grating element of the given diffraction grating by using the light of
known wavelength (green) and hence to determine the wavelength of the prominent lines of
mercury spectrum by normal incidence method.
Apparatus: Spectrometer, mercury vapor lamp, diffraction grating, reading lens etc.
Theory: An arrangement consisting
of a large number of parallel slits of X
S1
equal width and separated from one
another by equal opaque spaces is M1
called a diffraction grating. A grating S2 P
is made by ruling a very large number M2
(about 15000 lines per inch) of fine, S3 C
equidistant and parallel lines with a M3
diamond point on an optically plane
glass plate. The ruled lines act as Mn1
Sn
opaque region and the space between Fig.a
any two lines is transparent and it acts Y
as slits. Such a grating is called a
plane transmission grating.
The directions of the principal maxima
of a grating in normal incidence are given by,
b c sin θ = n ; n = 0, 1, 2…. (1)
Next the grating is mounted on the prism table with its ruled surface facing the collimator such
that the rulings parallel to the slit and the prism table (or vernier table) is rotated till the reflected
image from one face of the grating coincides with the vertical
wire. If necessary, the leveling screws of the prism table are
adjusted such that the reflected image is divided by the
horizontal wire and again the
vernier readings are noted. Then
the vernier table is rotated
exactly through 45(or 135) so
that the ruled surface of the
grating faces the incident light.
In this position the grating is
normal to the incident light. Ruled face
Determination of grating element: The grating can be
standardized as follows. Use the light of known wavelength
(green of mercury spectrum) to illuminate the slit and find out
the angle ‘’ for the corresponding spectral line. Then,
nλ 1
b+c = = (3)
sinθ N
sinθ
Or, N = (4)
nλ
2
Determination of angle of diffraction for the unknown
wavelength: The slit is illuminated with the light whose
wavelength is to be determined. Now the telescope is rotated to the
left of the direct image till the vertical wire coincides with the first
order spectral line of the unknown wavelength and the readings of
both the verniers are noted. Then rotate the telescope to the right Fig.c
side till the vertical wire coincides with the spectral line of the same
order on the right side of the direct image and the vernier readings
are again noted. The difference between the two readings of the same vernier gives 2 for first order.
sin θ
Wavelength can be calculated using = b c sin θ = . While determining the wavelengths
Nn
of spectral lines of mercury, take readings successively from the red line on one side to the red line
on the other side. Experiment can be repeated for other orders of the spectrum.
Precautions
See all the precautions given in the i-d curve experiment.
If the spectral lines are not bright and sharp rotate slightly the slit in its plane so as to make
the rulings parallel to the slit.
Observation and tabulation
Value of one main scale division (1 m s d) = ……………
Number of divisions on the vernier n = ……………
Value of 1 m s d
Least count (L C) = = ……………
n
[One degree = 60 minute, (1 = 60 )]
14 Optics & Electricity Practical II
Wavelength
Colour of Ver 1 Ver II Ver 1 Ver II
the spectral
lines
MSR
MSR
MSR
MSR
VSR
VSR
VSR
VSR
Ver II
Mean
Total
Total
Total
Total
Ver I
Yellow I
Yellow II
Blue-green
Blue
Violet I
Violet II
Wavelength of green line of mercury = 546.07 nm N
Green
Yellow I
Yellow II
Blue-green
Blue
Violet I
Violet II
Wavelength of green line of mercury = 546.07 nm N
Green
Result
Number of lines per metre of the grating N = ………..
Grating element 1/N = ………..
The prominent lines of mercury spectrum are determined and are recorded in the tabular
column.
Standard data*
Mercury spectral lines
Colour Wavelength in nm
Yellow I 579.06
Yellow II 576.96
Green 546.07
Greenish blue 491.60
Blue 435.83
Violet I 407.78
Violet II 404.65
Practical-II MCT 15
Wavelength
Colour of Ver 1 Ver II Ver 1 Ver II
the spectral
lines
MSR
MSR
MSR
MSR
VSR
VSR
VSR
VSR
Ver II
Mean
Total
Total
Total
Total
Ver I
Yellow I
Yellow II
Left side of direct image
Blue-green
Blue
Violet I
Violet II
Green
Yellow I
Yellow II
Right side of direct image
Blue-green
Blue
Violet I
Violet II
Exp.No.2.6
Small angled prism- normal incidence & normal emergence
Aim: To find the refractive index of the material of the given small angled prism by setting the
prism for (1) normal incidence and (2) normal emergence.
Apparatus: Spectrometer, sodium vapor lamp, small angled prism, reading lens etc.
Theory: For a given prism, corresponding to a given angle of deviation there are two possible
angles of incidence i1 and i2. These two angles are such that if one of the angles is the angle of
incidence, the other angle will be the angle of emergence. So these two angles are
interchangeable. In the following experiment we make use of this property.
Let i1 and i2 be the two angles of incidence and r1 and r2 be A
the corresponding angles of refraction for the given angle of
deviation d. Then,
i1 i 2 = A + d (1)
r1 r2 = A (2) d
i1
Normal incidence: In this case the incident ray is normal to one i2
of the refracting faces (AB) of the prism. Then, i1 = 0 and r1 = 0.
Thus, by eqn.1,
i2 = A+ d (3) B C
And by eqn.2, r2 = A (4) Fig.a
sin i 2 sin A d
Refractive index, = = (5)
sin r2 sin A
Normal emergence: In this case the emergent ray is normal to the face (AC) of the prism. Then,
i2 = r2 = 0. Hence, i1 = A+ d and r1 = A. Thus,
sin i1 sin A d sin i1
Refractive index, = = = (6)
sin r1 sin A sin i1 d
Procedure: All the preliminary adjustments of the spectrometer are made. The small angled
prism is then mounted on the prism table with its base parallel and close to the clamp.
A
C
B
B
C A
Fig.b Fig.c
Practical-II MCT 19
To find the angle of the prism by supplementary angle method: The telescope is clamped
nearly normal to the collimator. The vernier table is slowly rotated until the reflected image of
the slit from one of faces, say AB, is obtained on the cross wire of the telescope (fig.b). The
vernier table is then clamped and using its tangential screw the image is made to coincide exactly
with the vertical wire. The readings on both the verniers are noted. Then the vernier table is
released and is rotated through an angle as shown in fig.b such that the reflected image from
the other face (AC) is obtained on the cross wire of the telescope (fig.c). After making the fine
adjustments of the vernier table, the readings on both the verniers are taken. The difference
between the two readings gives and (180) gives the angle of the prism.
To set the prism for normal incidence: (We follow the same method as in the case of grating
normal incidence). The prism is removed from the prism table. The telescope is brought in a line
with the collimator and the direct image is made to coincide with the vertical cross wire. This
position of the telescope is noted by taking reading on one of the verniers. The telescope is now
turned 90 and is clamped.
Next the prism is
mounted on the prism table
with one of its refracting
surface facing the collimator
and the prism table (vernier
table) is rotated till the
reflected image from that face 45 A
of the prism coincides with the 45
45 B 90
vertical wire (fig.d). If A
necessary the leveling screws B
C
of the prism table are adjusted d
C
such that the reflected image is
divided by the horizontal wire Fig.d
and again the vernier readings
are noted. Then the vernier
table is rotated exactly through
45 in the proper direction so Fig.e
that the surface facing the
collimator now becomes
normal to the incident light
(fig.e). The vernier table is clamped in this position.
Determination of the angle of deviation in normal incidence: The telescope is released and is
brought in the line of refracted ray (fig.e). By making the fine adjustments with the tangential
screw of the telescope the refracted image is obtained exactly on the vertical cross wire. The
readings on both the verniers are noted. The prism is then removed carefully without changing
the vernier table. The telescope is released and is made to coincide with the direct image. The
readings on both the verniers are again noted. The difference between these two readings gives
the angle of deviation. The refractive index is calculated by the eqn.5.
Normal emergence: To find the angle of deviation and the angle of incidence at the first face
when the ray undergoes normal emergence, we make use of the property of the prism that the
incident and the emergent rays are interchangeable.
20 Optics & Electricity Practical II
The prism is again set for normal incidence and the refracted image is made to coincide
with the vertical cross wire. After clamping the telescope the vernier table is released and is
rotated such a direction that the refracted image moves towards the minimum deviation position.
The rotation of the vernier is continued in the same direction until the refracted image returns to
the cross wire. The vernier table is then clamped at this
position and the fine adjustment of the vernier is done if
necessary. Now the prism is set for normal emergence at
Incident ray
the second face (fig.f). The readings on both the verniers
are taken. Let it be ‘a’.
The telescope is then released and is brought in the
line of the reflected image from the first face. By Reflected ray
B
making fine adjustments the reflected image is made to i1
i1
coincide with the vertical cross wire and readings on
C
both the verniers are noted. Let it be ‘b’. 90
The prism is then removed carefully. The A
telescope is brought in the line of the direct ray and the = 180 2i1
readings on both verniers are noted (c). The difference
between the refracted image readings and the direct Refracted ray d
image readings (ac) gives the angle of deviation
corresponding to the normal emergence. The difference
in the readings between reflected image and the direct
image (bc) gives = 1802i1, from which i1 can be Fig.f
calculated. Finally the refractive index of the material of
the prism is calculated using eqn.6.
If the image is not in field of view of the telescope make sure that the prism table is
leveled. Looking through the prism with naked eye (without telescope) you can see the
image and judge its direction which helps to know whether it passes through the field of
view of the telescope and the leveling of vernier is needed.
Observation and tabulation
Value of one main scale division (1 m s d) = ……………
Number of divisions on the vernier n = ……………
Value of 1 m s d
Least count (L C) = = ……………
n
[One degree = 60 minute, (1 = 60 )]
Angle of the prism A (Supplementary angle method)
Ver I Ver II Mean
MSR VSR Total MSR VSR Total A=180
Reflected image from first
face ‘a’
Reflected image from second
face ‘b’
Difference between the above readings = ab = ab
Practical-II MCT 21
sin A d
Refractive index of the material of the prism, = = ………….
sin A
To set prism again for normal incidence (for normal emergence method)
Ver I Ver II
Direct reading
Reading at which telescope is to be set = Direct reading + 90 =
Reading corresponding to the reflected image
Reading at which vernier is to be set = Reflected reading 45 =
To find angle of deviation ‘d’ and angle of incidence i1 for normal incidence
Reading corresponding to Ver I Ver II Mean Mean
MSR VSR Total MSR VSR Total d
Refracted image ‘a’
Reflected image ‘b’
Direct image ‘c’
Difference between the readings ac = d ac = d
Difference between the readings bc = bc =
180 θ
Angle of incidence at the first face for normal emergence, i1 = = ……..
2
sin i1
Refractive index of the material of the prism = = ……..
sin i1 d
Result
Mean refractive of the material of the prism, = ……….
Standard data*: Same as given for normal incidence method.
22 Optics & Electricity Practical II
Exp.No.2.7
Air Wedge-Diameter of a thin wire
Aim: To find the diameter of a thin wire by measuring the width of the interference band
formed by an air wedge arrangement with this thin wire and two plane glass plates.
Apparatus: An air wedge, sodium vapour lamp, travelling microscope, reading lens etc.
Theory: An air wedge is produced by two optically plane
rectangular glass plates in contact with one pair of the edges and a
thin wire, whose diameter is to be determined, is kept in between
the plates near the other end parallel to the line of contact of the two
plates. d
If the angle between the two glass plates is small and the ray yn
incident normally, the approximate path difference between the two l
reflected rays from the upper and lower surfaces of the air film is Fig.a
given by,
λ λ
= 2tin medium = 2tin air
2 2
Since when the reflection takes place at the boundary of an optically denser medium (lower
surface of the air film) the reflected ray undergoes a phase change or an equivalent path
difference /2. ( is the refractive index of the thin wedge shaped film in between the glass
plates. For air wedge = 1). For constructive interference path difference ‘’ is an even multiple
of /2. Thus,
λ λ
2t = 2n where, n = 0, 1, 2, 3…
2 2
λ
i.e., 2t = 2n+1 , where, n = 0, 1, 2, 3… (1)
2
For destructive interference path difference ‘’ is an odd multiple of /2 . Thus,
λ λ
2t = 2n 1
2 2
λ
Or, 2t = 2n = n, where, n = 0, 1, 2, 3… (2)
2
If yn is the distance of the nth dark fringe from the line of contact of the two glass plates, t = yn.
Then, eqn.2 becomes,
2 yn = n (3)
For (n+1)th dark fringe, 2 yn+1 = n 1 λ (4)
Subtracting eqn.3 from eqn.4, we get,
2μ yn+1 yn θ =
λ
Therefore, fringe width, = yn+1 yn = (5)
2μθ
Practical-II MCT 23
From eqn.1 we can also show that the distance between two consecutive bright fringes
λ
x n+1 x n . If an air wedge is formed by placing a thin wire of diameter ‘d’ in between the
2μθ
glass plates at a distance ‘l’ from the line of contact of the two glass plates, the fringe width is
given by,
λ λl M
= = (6)
2θ 2d
d
Since = and for air = 1. G1
l Light from
Procedure: Light from the sodium lamp is allowed to Sodium lamp
fall on the glass plate G1 kept at 45 with the horizontal.
The air wedge is placed such that the reflected light from
the glass plate G1 falls normally on it. The interference
pattern is viewed from above by the travelling
d Fig.b
microscope as shown in fig.b. The pattern consists of
large number of equally spaced alternate dark and bright
bands as shown in fig.c. The microscope is l
moved towards one of the sides, say left, n0 n n+6 n+12 n+18 n+24
and one of the cross wires is made to
coincide with any of the dark line, say n0.
The microscope is then moved in the
opposite direction. (Remember now
onwards the tangential screw is rotated only
in one direction to avoid the backlash
error). It is then made to coincide with the
nth dark line and the microscope reading on n+3 n+9 n+15 n+21 n+27
the horizontal scale corresponding to it is Fig.c
noted. Then the microscope is moved and
the cross wire is made to coincide with the dark lines (n+3), (n+6), (n+9), ……….upto (n+27)
and the corresponding readings are noted. From these observations find out mean band width .
To find the distance l between wire and the line of contact of the glass plates take
microscope readings corresponding to the line of contact and the wire. The difference between
these readings gives ‘l’. (Microscope measurement is not essential).
Knowing the values of and l and assuming the wavelength of sodium light (589.3 nm) the
λl
diameter of the wire can be calculated using the equation d .
2β
Precautions
The glass plate G1 should be 45 with the light from the sodium lamp.
The glass plate G1 should be oriented such that the light from the sodium lamp incident at
the inner side of it. This helps the incident light to reflect towards the air wedge.
To see the interference bands clearly focus the microscope. The objective lens of the
microscope must be at a certain distance from the air wedge. This can be achieved by
adjusting the main clamping screw and the rack and pinion arrangement of the
microscope.
24 Optics & Electricity Practical II
In order to avoid the backlash error of the travelling microscope, initially the tangential
screw of the microscope is rotated in a direction and the cross wire is moved from back
of the nth line and is then made to coincide with the nth line. The tangential screw should
not be rotated in the opposite direction (or to and fro) while coinciding with the first line
and throughout the experiment. By mistake, if you have moved the microscope in the
opposite direction while taking the reading give up all the readings taken and do the
experiment from the beginning itself.
Before starting to take readings ensure that we can move the microscope from the n0th
line to more than n+30 lines. If it is not so loose the main screw of the vernier and loosen
or tighten sufficiently the tangential screw and then tighten the main screw.
Observation and tabulation
Value of one main scale division (1 m s d) = ………. cm
Number of divisions on the vernier n = ……….
1msd
Least count = = ……… cm
n
Determination of band width
Number Microscope readings Width of Mean width Band width
of bands MSR VSR Total 15 bands of 15 bands w = w/15
cm cm cm cm m
n
n+3
n+6
n+9
n+12
n+15
n+18
n+21
n+24
n+27
Distance between the wire and the line of contact of the plates l = ………. m
λl
Diameter of the wire, d = = = …….. m
2β
d λ
Angle of the wedge, = = = = ……… radian
l 2β
Result
Diameter of the wire, d = ………. M
Angle of the wedge, = ………. radian
Standard data*
Wavelength of sodium light, = 589.3 nm.
Practical-II MCT 25
2Rt
the wedge shaped film (for air = 1). The
additional path difference /2 is due to the
phase change occurs when the reflection L L
takes place at the lower surface of the film. For A
P Q t r t P
normal incidence r = 0, and if is small, the O G2 G2 O
effective path difference,
Fig.a Fig.b
λ
= 2t (2)
2
Diameter of dark rings: The condition for dark ring is that the path difference is equal to odd
λ
multiple of
2
λ λ
i.e. = 2t = 2n 1 , where, n = 1, 2, 3…….
2 2
Or, 2t = n, where, n = 1, 2, 3…….
For air, 2t = n, where, n = 1, 2, 3……. (3)
Let ‘R’ be the radius of curvature of the plano-convex lens. Consider a point ‘P’ where the
thickness of the film is ‘t’ and radius of the ring through ‘P’ is ‘r’. Then from the property of the
circle, (refer fig.b),
PAAQ = BAAO
i.e. r2 = 2R t t = 2Rt t2
26 Optics & Electricity Practical II
20th ring
25th ring
25th ring
20th ring
Precautions
The glass plate G1 should be 45 with the light (horizontal) from the sodium lamp.
The glass plate G1 should be oriented such that the light from the sodium lamp incident at
the inner side of it. This helps the incident light to reflect towards the Newton’s rings
arrangement.
Focus the microscope to see the interference bands clearly. The objective lens of the
microscope must be at a certain distance from the air wedge. This can be achieved by
adjusting the main clamping screw and the rack and pinion arrangement of the
microscope.
In order to avoid the backlash error of the travelling microscope, initially the tangential
screw of the microscope is rotated in a direction and the cross wire is moved from back
of the 20th ring (say, from 25th ring) and is then made to coincide with the 20th ring. The
tangential screw should not be rotated in the opposite direction (or to and fro) while
coinciding with the 20th ring on one side and throughout the experiment. By mistake, if
you have moved the microscope in the opposite direction while taking the reading
abandon all the readings taken and do the experiment from the beginning itself.
Before starting to take readings ensure that we can move the microscope from the 25 th
ring on one side to more than 20th ring on the other side. If it is not so loose the main
screw of the vernier and loosen or tighten sufficiently the tangential screw and then
tighten the main screw.
Distance of the lens form the object when side by side image is formed with dark screen
behind the lens, d = ……… cm
Distance of the lens form the object when side by side image is formed with plane mirror
behind the lens, f = ……… cm
fd
Radius of curvature of the lens, R = = …….. m
f d
D2n+k Dn2
Wavelength of sodium light = = ………..
4kR
= ……….. nm
Result
Wavelength of sodium light = ……….. nm
Standard data*
Wavelength of sodium light = 589.3 nm
Practical-II MCT 29
Exp.No.2.9
Laser-Slit width from diffraction pattern
Aim: To find the width of a narrow slit by producing single slit diffraction pattern by laser
source.
Apparatus: A laser source, single slit, a screen with a graph paper pasted on it, scale etc.
Theory: Parallel beam of light from a laser source is passed through a slit of width AB = b. Let
XY be a screen at a distance D X
from the slit. The diffraction
pattern of a single slit consists of a L2
L1 P
number of bright and dark spots as A
shown in fig.b. The theory of
Fraunhofer diffraction by a single N
S O C
slit (since the theory is too long we
avoid it here) shows that the BM
condition for minimum intensity D
(or directions of minimum
intensity) is Fig.a: Diffraction by a single slit Y
Let x1, x2, x3, …… are the distances of the dark Screen
spots from the centre of the central maximum.
(Refer fig.c and fig.e). Then, y2
Slit x2
x
sin θ n = n
D
nλD λD x1
Thus, b = = (2) y1
xn xn
n D
xn Fig.c
Since and D are constants is a constant. Thus
n xn
xn-n graph will be a straight line as shown in fig.d. The slope
x
of the graph gives the mean value of n . Thus,
n
λD
b = (3)
slope of x n - n graph
n
Fig.d
30 Optics & Electricity Practical II
The directions of the secondary maxima (condition for secondary maxima) approximately
are given by,
λ
b sinn = ± 2n+1 where, n = 1, 2, 3, ….. (4)
2
Let y1, y2, y3, …… are the distances of the centres of the secondary maxima (bright spots)
from the centre of the central maximum. (Refer fig.c and fig.e).
Then, yn
y
sin θ n = n
D
Thus, b =
2n+1 λD = λD (5)
2y n yn
2
2n+1 Fig.d 2n+1
yn
Since and D are constants, is a constant. Thus the graph
2n+1
yn
between yn and 2n+1 is a straight line. Its slope gives average . Then,
2n+1
λD
b = (6)
Twice the slope of yn -(2n+1) graph
Procedure: The slit is mounted on the stand. Its width is made narrow. The laser beam is
allowed to pass through the slit. Ensure that the slit is at the middle of the laser beam. A screen
fixed with a graph paper is arranged at a large distance ‘D’. The plane of the screen must be
parallel to the plane of the slit. Adjust the width of the slit and its orientation to get well defined
diffraction pattern consisting of a number of dark and bright spots on the screen. Since the length
of the slit is large the diffraction effect (the dark and bright spots in the pattern) occurs only in
the direction of the width of the slit. Using a pencil draw the outline of all bright spots on the
graph paper and mark the centre of the spots. Also mark the midpoint of the dark region in
between the bright spots. Measure the distance D
between the slit and the screen. Determine the distances y4
xn from the centre of central bright spot to the midpoint y3
Theory of potentiometer
Let a steady current I be passed through the
wire AB with the help of a cell of e m f E. Let
ρ be the resistance per unit length of the E
potentiometer wire and J is a sliding contact. L
A J B
Let AB = L and AJ = l. Then, l
Potential difference across AB = ILρ E G
Potential difference across AJ = Ilρ HR
PD across AB ILρ L
= = (1)
PD across AJ Ilρ l
PD across AB
PD across AJ = l = PD per unit lengthlength of the wire (2)
L
Thus, when a steady current is flowing through the potentiometer wire AB, the PD across
any length of the wire is proportional to the length of the wire.
If a DC voltmeter is connected between A and the variable point J it can be seen that the
voltmeter registers greater values as the contact maker J moves from A to B.
If another cell of e m f equal to PD across AJ is connected between A and J as shown in the
figure, no current will flow in the secondary circuit and the galvanometer will show no
deflection.
Exp.No.2.10
Potentiometer- Calibration of ammeter
Aim: To calibrate the given ammeter using a potentiometer.
Apparatus: A potentiometer, the given ammeter, rheostats, two accumulators (or power
sources), a Daniel cell (or a power source of standard voltage), a standard resistance, six terminal
key or a three terminal key, etc.
Theory
E K1 Rh1 E K1 Rh1
A J B A J B
E
E 1
6 1 G
5 2 HR 2 HR
4 3 3 G
R + R +
A A
K2 Rh2 K2 Rh2
Fig.a: Six terminal key is used Fig.b: Three terminal key is used
Practical-II MCT 33
Let L be the balancing length when a cell of standard e m f E (for Daniel cell E = 1.08 V) is
connected in the secondary circuit. Then by the theory of potentiometer,
E L (1)
The potential difference developed across the standard resistance when a current I flows
through it is,
V = IR (2)
Let l be the balancing length when the potential difference across R is applied to the
potentiometer. Then,
V l
IR l (3)
Dividing eqn.3 by eqn.1, we get,
IR l
=
E L
E
I = l (4)
LR I0
If E = 1.08 volt,
1.08l
I = (4a) Fig.c
LR
E
Correction to the ammeter reading = I I0 = l I0 (5)
LR
The graph between the measured current I0 in the X axis and the correction I I0 in the Y axis is
called the calibration graph. A model of it is shown in the fig.c.
Procedure: The connections are made as shown in the fig.a or b. A steady current is allowed to
flow through the wire AB by connecting the terminals of it to the cell of e m f E through the
rheostat Rh1 and key K1. E is a standard cell (may be a Daniel cell). The ammeter to be
calibrated is connected in the secondary circuit in series with the battery, key K2, rheostat Rh2
and a standard resistance R (1 or 2 ohms).
Now connect terminals 1 and 2 in fig.b (for fig.a insert keys in between 1 and 2 and 5 and
6). Adjust the sliding contact J till the galvanometer shows zero deflection and the balancing
length L corresponding to e m f E is measured from the end A. Next disconnect 1 and 2 and
connect terminals 2 and 3 (for fig.a unplug the keys in between 1 and 2 and 5 and 6 and insert in
between 2 and 3 and 4 and 5) . Adjust the rheostat Rh2 so that the ammeter reads a value I0, say
0.1 A. Let I be the actual current flowing through the circuit. This current produces a potential
difference IR across the resistance R. Again adjust the contact maker and find the balancing
length l corresponding to this potential difference. Next the rheostat Rh2 is adjusted successively
for currents 0.2A, 0.3A, 0.4A,………., 1A and the corresponding balancing lengths are
determined in each case. The current I and I I0 are calculated. Then the calibration graph is
plotted with ammeter reading I0 in the X axis and the correction I I0 in the Y axis. The
different points obtained are joined by straight lines.
34 Optics & Electricity Practical II
Precautions
Clean the ends of the wires before the connecting.
Ensure that the wires are not broken.
In all the electricity experiments, it is advised to do the series connections first and then
the parallel connections.
If there is no deflection, check the voltages of the cells or power supplies used and also
check the continuity of the circuit with a multimeter.
Ensure that the secondary voltage applied to the potentiometer (in this case P D across
the standard resistance R) should not exceed the P D across A B of the potentiometer
wire. If the balancing length for 1.08 volt is about 920 cm and the standard resistance
used is 1 the current greater than 1A is not suitable.
Ensure that all the positive potential sides are connected to the terminals 4, 5 and 6 and
negative sides are connected to the terminals 1, 2 and 3.
You can check the circuit by a method as follows. Unplug the key K1 in the primary
circuit. By inserting keys in the secondary circuit and the six terminal keys apply the
secondary voltage to the potentiometer. Press the sliding contact J at the ends A and B.
Make sure that the deflections in the galvanometer are in the same direction. If there is no
deflection check the voltage and continuity of secondary circuit. Now insert the key K1 in
the primary circuit and check the deflections at A and B. If the deflections are in the
opposite directions connections are correct. Otherwise, check the voltage and continuity
of the primary circuit. This checking for opposite deflections must be done separately
with standard voltage and P D across R.
Ensure that the key of the high resistance is to be inserted during the determination of
final balance point.
Keep the potential difference in the primary circuit (p d across AB) undisturbed
throughout the experiment.
Observation and tabulation
Standard resistance R = …………
Standard voltage E = …………volts
Balancing length for standard e m f E = L = ……… cm
Ammeter Balancing length Calculated current Correction I I0
reading for PD across R E
I0 A l cm I l ampere ampere
LR
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1
Result
The given ammeter is calibrated and the calibration graph is drawn.
Practical-II MCT 35
Exp.No.2.11
Potentiometer-Reduction factor of TG
Aim: To determine the reduction factor of a tangent galvanometer and hence to find out the
horizontal component of earth’s magnetic field.
Apparatus: A potentiometer, tangent galvanometer, rheostats, two accumulators (or power
sources), a Daniel cell (or a power source of standard voltage), a standard resistance, six terminal
key or a three terminal key, etc.
Theory
E K1 Rh1 E K1 Rh1
A J B A J B
E
E 1
6 1 G
5 2 HR 2 HR
4 3 3 G
R R
TG TG
K2 Rh2 K2 Rh2
Fig.a: Six terminal key is used Fig.b: Three terminal key is used
A tangent galvanometer consists of a circular coil of radius ‘a’ and a compass box placed at
the centre of the coil. The magnetic field produced at the centre of the coil when a current I flows
through the coil is,
μ nI
B= 0 (1)
2a
If the plane of the coil is set along the north-south direction, the magnetic field will be in
the east-west direction. Hence the resultant magnetic field at the centre of the coil makes an
angle with the direction of earth’s magnetic field and the magnetic needle in the compass box
will be aligned itself in the direction of the resultant field. Then,
B E
= tan
Bh
Or, B = Bh tan (2) B
μ 0 nI
i.e. = Bh tan
2a
2aBh Bh
N
I = tan θ = K tan (3)
μ0n
2aBh
where, K is called the reduction factor of the T G. (4)
μ0n
36 Optics & Electricity Practical II
Let L be the balancing length when a cell of standard e m f E (for Daniel cell E = 1.08 V) is
connected in the secondary circuit. Then by the theory of potentiometer,
E L (5)
The potential difference developed across the standard resistance when a current I flows
through it is,
V = IR (6)
Let l be the balancing length when the potential difference across R is applied to the
potentiometer. Then,
V l
IR l (7)
Dividing eqn.7 by eqn.5, we get,
IR l
=
E L
E
I = l (8)
LR
E
Using eqn.3, K tan = l
LR
El
K = (9)
LR tan θ
1.08l
If E = 1.08 volt, K = (9a)
LR tan θ
μ nK
By eqn.4, Bh = 0 (10)
2a
Procedure: The connections are made as shown in the fig.a or b. Before inserting the different
keys do the initial adjustments of the TG. To set the TG in the magnetic meridian, rotate the
compass box alone till the 90-90 mark line coincides with the vertical plane of the coil. Then the
TG as a whole is rotated till the aluminum pointer reads 0-0. For leveling, if necessary, rotate the
leveling screws provided at the base.
To standardize the potentiometer for a particular p d per unit length, connect the standard
cell (Daniel cell or power source) to the potentiometer by inserting keys in the gap between 1
and 2 and 5 and 6. Also close key K1. Keep the sliding contact J at some point, say 920 cm, and
adjust rheostat Rh1 for no deflection in the galvanometer. Then close the high resistance key and
the exact balancing length L for the standard cell is determined.
Then unplug the keys in between 1 & 2 and 5 & 6 and close keys in between 2 & 3 and 4 &
5. Also close key K2. Now the p d across R is applied to the potentiometer. The rheostat Rh2 is
adjusted for a suitable deflection in between 30 and 60. The readings at the ends of the
aluminum pointer are noted. Find the balancing length l. Repeat the experiment by interchanging
the commutator keys. Repeat the experiment for different deflections and in each case the
deflection and the corresponding balancing length are determined.
Using a piece of twine the circumference of the T G coil is determined and from which its
radius is calculated. Finally, K and Bh are calculated using eqns.9 and 10.
Practical-II MCT 37
Precautions
Precautions given in exp.No.10 are applicable in this case also.
Ensure that the 90-90 line is at the centre of the coil and parallel to plane of the coil.
Place a small plastic scale close to the coil with its plane parallel to the coil plane and
look through the gap in between the coil and the scale. While looking lean a little forward
and close one of the eyes.
Avoid the parallax error during measurement. For that set the TG at a convenient place.
Keep TG away from galvanometer, ammeter, rheostat etc., since the magnetic fields due
to them may affect the reading.
Ensure that the current through the TG is such that the deflection in the TG is in between
30 and 60.
Before taking reading tap on the frame of TG.
Checking for opposite deflection must be done for each current through R. This is to
ensure that the p d across R is less than p d between A and B due to primary voltage.
If an ammeter is connected in series with the T G circuit you can check whether there is
any change in the current during the determination of balancing point and also after the
interchange of commutator keys.
Observation and tabulation
Standard resistance, R = ………
Standard voltage, E = ……… volt
Balancing length for standard voltage, L = …….. cm
Reading against the pointer of Mean Balancing length in cm
the T G in degrees K
Sl.No Commutator Commutator Degree Position Position Mean ampere
position 1 position 2 1 2 l cm
1 2 3 4
Mean K
Number of turns of the coil n = ……..
Circumference of the T G coil C = ………. cm
C
Radius of the coil, a = = ……… m
2π
μ 0 nK
Horizontal component of earth’s magnetic field Bh = = ……… tesla
2a
Result
Reduction factor of T G, K = ………. ampere
Horizontal component of earth’s magnetic field Bh = ……… tesla
Standard data*: Horizontal component of earth’s magnetic field Bh = 0.38 104 tesla
38 Optics & Electricity Practical II
Exp.No.2.12
Potentiometer-Calibration of high range voltmeter
Aim: To calibrate the given high range voltmeter using a potentiometer.
Apparatus: A potentiometer, the given high range voltmeter, rheostats, two accumulators (or
power sources), a Daniel cell (or a power source of standard voltage), two resistance boxes, six
terminal key or a three terminal key, etc.
Theory
E K1 Rh1 E K1 Rh1
A J B A J B
E
E 1
6 1 G G
5 2 HR 2 HR
4 3 3
P Q P Q
+V +V
Rh2 Rh2
K2 K2
Fig.a: Six terminal key is used Fig.b: Three terminal key is used
Let L be the balancing length when a cell of standard e m f E (for Daniel cell E = 1.08 V) is
connected in the secondary circuit. Then by the theory of potentiometer,
E L (1)
Let V be the actual potential difference across P and Q. Then the current I through P and Q is
given by,
V
I =
P+Q
VP
Potential difference across P, V1 = IP = (2)
P+Q
If l is the balancing length for the p d developed across the resistance P, we can write,
V1 l
VP
i.e. l (3)
P+Q
VP l
Dividing eqn.3 by eqn.1, =
P+Q E L
Practical-II MCT 39
E P+Q
V = l (4)
L P
1.08 P+Q
= l (4a)
L P
Voltmeter correction = V V0
E P+Q
l V0
L P
= (5)
The key K1 is closed and check for opposite deflections with the standard voltage and the
voltage across P. By inserting suitable keys, the standard voltage source is applied to the
potentiometer and the balancing length L is determined. Then, take suitable resistances in P and
Q. The keys in the six terminal key or three terminal key are changed such that the standard
voltage is removed from the potentiometer circuit and the p d across P is applied to the
potentiometer. The voltmeter reading is adjusted to, say 1 volt by adjusting the rheostat Rh2. The
balancing length l is determined. Then the rheostat Rh2 is adjusted for voltmeter reading 2V, 3V,
………. etc. and in each case the balancing length is determined.
The voltages and corrections are calculated using eqns. 4 and 5. A calibration graph is
plotted with the voltmeter readings V0 in the X axis and the correction VV0 in the Y axis.
Precautions:
Same as given in exp.No.10.
Be careful that the voltage across P is less that the primary voltage across AB. That is,
resistance P << Q.
Tight all the plugged keys in all resistance boxes, since the loose keys create unwanted
resistance in the circuit.
40 Optics & Electricity Practical II
Result
The given high range voltmeter is calibrated and the calibration graph is drawn.
Exp.No.2.13
Circular coil- Determination of Bh and m
Aim: To determine the
value of the horizontal North
component of earth’s
magnetic field and the West East
dipole moment of a bar
magnet using a circular coil. x d
South S N
Apparatus: Circular coil
apparatus (fig.a), power
supply, compass box, bar
magnet, ammeter, rheostat,
commutator key and another
key.
Theory: The magnetic Fig.a
field at a point on the axis at
a distance ‘x’ from the centre of a current carrying circular coil of ‘n’ turns is given by,
μ 0 na 2 I
B = (1)
2 a2 x2
3
2
where, ‘a’ is the radius of the coil and ‘I’ is the current through it. Since, while doing the
experiment the axis of the coil is set perpendicular to the magnetic meridian, the field due to the
coil and the earth’s horizontal field (directed from geographic south to geographic north) are
mutually perpendicular, the resultant field is at an angle with the magnetic meridian. Then
applying the tangent law, we get,
B = Bh tan (2)
μ 0 na 2 I
From eqns.1 and 2, Bh tan =
2 a2 x2
3
2
μ 0 na 2 I
Bh = (3)
2 a 2 x 2 2 tan θ
3
If a magnet of half the length ‘l’ and dipole moment ‘m’ is placed at a distance ‘d’ from the
centre of the compass box such that the field due to the coil is exactly cancelled by the field due
to the magnet, we can write,
μ 0 na 2 I μ 2md
= 0
4π d 2 l 2 2
3
2 a2 x2 2
πna 2 I d 2 l 2
2
m = (4)
d a2 x2
3
2
42 Optics & Electricity Practical II
Procedure: Connections are made as shown in the fig.b. Usually the coil consists of a set of
coils of different number of turns. Select any one of them, say 5 turns. The initial adjustments to
set the plane of the coil parallel to the magnetic meridian are made as follows. The compass box
is placed at the centre of the coil. Rotate the
compass box alone till the 90-90 mark line
coincides with the vertical plane of the coil. Then
the apparatus as a whole is rotated till the
aluminum pointer reads 0-0. Now the plane of the
coil is in the magnetic meridian and the platform
Fig.b
is in the east-west direction as shown in the fig.a.
To find B0: Now the compass box is kept at a
distance ‘x’ from the centre of the coil, (say, 5
cm) on one side. The circuit is closed and the
rheostat is adjusted for a current ‘I’. The current A+
is such that the deflection in the compass box is K Rh
in between 30 and 60. The ammeter reading
and the readings corresponding to both ends of the pointer are noted. Then the current is reversed
by changing the commutator keys and the pointer readings are again taken. Now the compass
box is placed at the same distance on the other side of the coil and four more readings are taken.
The average of these eight deflections is calculated. Let it be . The circumference of the coil is
determined with a twine and from it the radius ‘a’ of the coil is calculated. Bh is calculated using
eqn.3. The experiment is repeated for different values of distance x and current I. The average
value of Bh is determined.
To find m: After making connections and the initial adjustments of the apparatus as discussed
above, the compass box is placed at distance ‘x’ from the centre of the coil. The current is
adjusted to get a deflection in between 30 and 60. The given magnet is placed along the axis on
one side of the compass box as shown in fig.a. The distance between the magnet and the
compass box is adjusted so that the deflection in the compass box is reduced to zero. The current
through the coil and the distance d1 between the magnet and the compass box are measured. The
current is reversed by changing the commutator keys. The magnet also must be reversed. The
distance is adjusted for null deflection. The distance d2 from the centre of the magnet and the
centre of the compass box is again measured. Now the compass box and the magnet are placed
on the other side of the coil and two more readings, d3 and d4, for null deflection are determined.
The average distance ‘d’ of four distances is found.
The number of turns of the coil is noted. The radius of the coil ‘a’ is determined by
measuring its circumference. The length of the magnet ‘2l’ is also measured. Then using eqn.4
the dipole moment ‘m’ is calculated. The experiment is repeated for different values of distance
x and current I. The average value of m is determined.
Precautions
Ensure that the plane of the coil is set parallel to the magnetic meridian.
The axis of the bar magnet should pass through the centre of the compass needle and the
centre of the circular coil.
The distances and the currents are such that the deflection is in between 30 and 60.
To get zero deflection with magnet, the field due to the coil and the field due to the
magnet are in the opposite directions. If you are not getting a zero deflection reverse the
magnet for changing the direction of field due to the magnet.
Practical-II MCT 43
Result
Horizontal component of earth’s magnetic field, Bh = ……….. tesla
Dipole moment of the given bar magnet, m = ……….. ampere metre2
Standard data*
Permeability of the free space, 0 = 4π 107 henry/metre
44 Optics & Electricity Practical II
Exp.No.2.14
Carey Fosters’ Bridge-Temperature coefficient of resistance
Aim: To determine the temperature coefficient of the resistance of the material of the coil by
measuring its resistance at different temperatures using Carey Foster’s bridge.
Apparatus: Carey Foster’s bridge, given coil of wire, a cell (power supply), standard
resistances, a resistance box, key, galvanometer, high resistance, heating arrangement and a
thermometer.
Theory: The temperature coefficient of resistance is defined by the change in resistance to its
resistance at zero degree Celsius per unit resistance per unit degree rise of temperature.
X t X0
Temperature coefficient of resistance = (1)
X0 t
Xt = X0 1 αt
Let X1 and X2 be the resistances at temperatures t1C and t2C respectively. Then,
X1 = X0 1 αt1 (2)
X2 = X0 1 αt 2 (3) E K
X P Q R
Dividing eqn.2 by eqn.3,
X1
=
1 αt1 G1 G2
D
G3 G4
X2 1 αt 2 C
G
F
J
Cross multiplying and rearranging,
A B
X 2 X1 l1 Fig.a 100l1
= (4)
X1t 2 X 2 t1
In this experiment we use
D D
Carey Foster Bridge to
determine the unknown
P Q P Q
resistance X1 and X2. The basic
principle of it is Wheatstone’s
principle. Carey Foster Bridge C G F C G F
consists of a uniform wire AB X R R X
of length 1 m stretched on a
wooden board. Five metallic
strips are fixed on the wooden ρl1 ρ(100l1) ρl2 ρ(100l2)
J J
board as shown the figure. G1,
G2, G3 and G4 are gaps between
the metal strips. Two equal
Fig.b Fig.c
resistances P and Q are
connected in the gaps G2 and G3 respectively. The unknown resistance X is connected in the gap
G1. A standard resistance R is connected in the gap G4. A standard cell is connected across the
terminals C and F. A galvanometer G is connected between D and the contact maker J, that is
able to slide along AB.
Practical-II MCT 45
Theory*: The contact maker J is moved along the wire AB until the galvanometer shows no
deflection. Then the bridge is said to be balanced. Let l1 be the balancing length as measured
from the end A. Let and , respectively, be the end resistances at A and B. Let ρ be the
resistance per unit length of the wire AB. The above bridge is equivalent to a Wheatstone’s
bridge as shown fig.b.
Applying Wheatstone’s principle we get,
P X+α+ρl1
= (5)
Q R+β+ρ 100 l1
The resistances R and X are interchanged and the bridge is again balanced. The balancing
length l2 is measured from the same end A. Then,
P R+α+ρl2
= (6)
Q X+β+ρ 100 l2
Since the numerators are equal, we can equate the denominators. Thus we get,
R+β+ρ 100 l1 = X+β+ρ 100 l2
Procedure
To find E K
X P Q R
The connections are
made as shown in the D
G1 G2 G3 G4
fig.d. Suitable standard C F
resistances P and Q are G J
connected in the gaps G2 A B
l1 Fig.d 100l1
and G3. Instead of X
connect a thick copper
strip in the gap G1 and a resistance box with fractional resistance in the gap G4. Take a resistance
R = 0.2 in the box. Find the balancing length l3. It is measured from the end A. Then
interchange the copper strip and the resistance box. The balancing length l4 is determined. It is
again measured from the end A. Calculate ‘’ using eqn.8. The experiment is repeated for R =
0.3 , 0.4 , ……….. The average of ‘’ is calculated.
To find X1 and X2
Sl.No. Resistance Balancing length with
Temperature R R in l2 l1 X R ρ l2 l1
ohms Right gap Left gap cm ohm
l1 (cm) l2 (cm)
1
2
t1 = …….C 3
4
5
Mean X1
1
2
t2 = …….C 3
4
5
Mean X2
X 2 X1
= = ……… = ………. Per C
X1t 2 X 2 t1
Result
Temperature coefficient of resistance of the material of the coil, = ………per C
Standard data*
Temperature coefficient of resistance of copper, = 0.00400.0002 per celsius
The accepted value is 0.0039 per celsius.
48 Optics & Electricity Practical II
Exp.No.2.15
Conversion of a Galvanometer into voltmeter- calibration using
potentiometer
Aim: To convert the given galvanometer into a voltmeter and calibrate it with a potentiometer.
Apparatus: The given pointer type galvanometer, potentiometer, resistance boxes,
commutator key, accumulator, Daniel cell (or power supplies), etc.
Theory: A galvanometer can be converted into a
voltmeter by connecting a high resistance in series V
I
with it. The range of the voltmeter (converted
galvanometer) depends on this series resistance. Let
the galvanometer is converted to a voltmeter to Ig R G
measure a voltage range 0-V. Let Ig be the full scale
deflection current (current flowing through the V
galvanometer when it shows full scale deflection) of Fig.a
the galvanometer. For example, the given
galvanometer is graduated such that the division at the centre is zero and on each side of the zero
line there are 30 divisions. Then the full scale deflection current means the current required to
produce a deflection of 30 divisions in the galvanometer. Suppose this galvanometer is converted
into a voltmeter of range 0 to V volt. When the converted galvanometer reads a voltage V, the
current trough it is Ig ampere. Let G be the resistance of
the galvanometer. Then, from fig.a,
I E K
Ig R G = V
P Q
V
R = G (1)
Ig V
Let ‘k’ be the figure of merit (current sensitivity) of the
galvanometer. It is the current required to produce a R
deflection of one division. If there are ‘n’ divisions on
each side of the zero division, G
Fig.b
Ig = kn (2)
To find out k and G we consider a circuit as shown in fig.b. Using Ohm’s law, the current
through P and Q is (since P << R + G),
E
I =
P+Q
EP
Potential difference across P is, V =
P+Q
V EP
Current through the galvanometer, I1 = =
R G P+Q R G
EP
If R = 0, I1 = (3)
P+Q G
Practical-II MCT 49
Due to this current the galvanometer shows a deflection of ‘d’ divisions. Then current sensitivity,
I1 EP E P
P+Q G d
k = = = (4)
d P+Q Gd
When a resistance is introduced in R, current I1 decreases and hence the deflection in the
galvanometer decreases. The current and hence the galvanometer deflection reduces to half when
R G + P. (5)
(See the appendix at the end). Using eqns.2, 4 and 5 in eqn.1 we get the resistance needed to
convert a galvanometer to a voltmeter of range 0 to V volt. Its volt per division is V/n. Then
convert the galvanometer into
voltmeter and calibrate it. VV0
Draw the calibration graph as
shown in the model.
Procedure
To find G and k: Connections
are made as shown in the fig.b.
Suitable resistances are
introduced in P and Q with P
<< Q. For example, P = 10 V0
and Q = 990 so that P + Q =
1000 . The voltage across P
(much smaller voltage) is
applied to the galvanometer through the resistance box R. All the keys in R are plugged and
tightened so that R = 0. The deflection ‘d’ in the galvanometer is noted. Now increase the
resistance in R and find out the resistance needed to reduce the deflection to half of its original
value. This value of resistance is noted. The experiment is repeated after reversing the
commutator. The entire experiment is repeated for different values of P and Q. In each case ‘d’
and ‘G’ are determined.
To convert galvanometer into voltmeter: Using eqns.2, 4 and 5 in eqn.1calculate the resistance
R needed for the conversion of galvanometer to the voltmeter for the selected voltage range. For
example, If there are 30 divisions in the galvanometer we can conveniently choose a range of 0
to 0.3, 0 to 3 or 0 to 30 volt. Connect the calculated resistance R in series with the galvanometer.
Thus the galvanometer is converted to a voltmeter.
E K1 Rh1 E K1 Rh1
L l
A J B A J B
E G G
HR + R
Fig.c
Voltmeter
Fig.d
50 Optics & Electricity Practical II
Calibration of the voltmeter: Depending upon the range of the voltmeter constructed, we can
follow the method of calibration of low range voltmeter (Exp.No.1.20 of practical-I) or the
method for high range voltmeter (Exp.No.2.12 of practical-II).
Here we use another simple method as follows. Standardise the potentiometer with a
standard voltage using the diagram as shown in fig.c. The balancing length ‘L’ for the standard
voltage (Daniel cell 1.08 V) is determined. Then the potential difference per unit length of the
E
potentiometer wire is found out. Connect the newly constructed voltmeter as shown in fig.d.
L
Move the contact maker J along the wire such that the voltmeter reads a particular value V0, say,
0.1 volt. Measure the length ‘l’ corresponding to this voltage and calculate the voltage
El 1.08l
V . Repeat the experiment for 0.2 volt, 0.3 volt, …………. A calibration graph is
L L
plotted with V0 along the X axis and the correction along the Y axis.
To make calculations easy choose P + Q as a multiple of 10 and keep it constant. For
example If P = 10 and Q = 90, P +Q = 100 or, if P = 10, Q = 990, so that P + Q = 1000. If
deflection is too large either reduce the voltage E or increase P + Q. Ensure P << Q. Also
P << R + G.
Tight all the plugged keys in all resistance boxes, since the loose keys create unwanted
resistance in the circuit.
It can be shown that the resistance needed for half deflection in the galvanometer is, (see
the appendix at the end),
P Q r
R = G , where ‘r’ is the internal resistance of the cell E.
PQr
P Q r
G = R
PQr
Since, r and P are much less than Q,
G R P
Observation and tabulation
To find G, k and R
Voltage applied across P and Q = E = ……… volt
P + Q = ………. ohm
P Q Galvanometer deflection P Resistance for half deflection GRP
Ohm Ohm ‘d’ d Ohm
left Right Mean Left Right Mean
Mean Mean G
Practical-II MCT 51
Result
The given galvanometer is converted to a voltmeter to read a voltage range 0 to ….. volt.
Its readings are checked with a potentiometer and a calibration graph is plotted.
Standard data*
E m f of the Daniel cell, E = 1.08 volt
52 Optics & Electricity Practical II
Appendix*
For the closed circuits we can write,
E = I1P I1 I2 Q I1 I2 r (1)
I1P = I2 R G (2)
When R = 0, I1 = I1 and I2 = Ig, then the equations I r I=I1+I2
E
reduce to
E = I1P I1 Ig Q I1 Ig r
I1 I=I1+I2
(3) P Q
I2
I1P = Ig G (4) V
Ig
I1 = G R
P
Ig Ig G
E = Ig G G Ig Q G Ig r (5)
P P
Ig
When R = R, the deflection in the galvanometer is half, i.e. the I2 = and I1 = I1 . Then the
2
eqns. 1 and 2 become,
Ig Ig
E = I1P I1 Q I1 r (6)
2 2
Ig
I1P = R G
2
Ig
I1 = R G
2P
Ig I I I I
E = R G g R G g Q g R G g r (7)
2 2P 2 2P 2
Equating the R H Ss of eqns.5 and 7,
I I I I I I I
Ig G g G Ig Q g G Ig r = g R G g R G g Q g R G g r
P P 2 2P 2 2P 2
2P
Multiplying throughout by and rearranging,
Ig
R P Q r = G P Q r P Q r
P Q r
R = G
P Q r
P Q r
G = R (8)
P Q r
When P and r are much less than Q,
G R P
If P << R , G R
Practical-II MCT 53
Exp.No.2.16
Conversion of Galvanometer into ammeter- calibration using
potentiometer
Aim: To convert the given galvanometer into an ammeter and calibrate it using a potentiometer.
Apparatus: The given pointer type galvanometer, potentiometer, resistance boxes, standard
resistance wire, commutator key, accumulator, Daniel cell (or power supplies), etc.
Theory: A galvanometer can be converted into an ammeter by connecting a small resistance
(shunt resistance) in parallel with it. The range of the ammeter (converted galvanometer)
depends on this parallel resistance. Let the galvanometer is converted to an ammeter to measure
a current range 0 to I ampere. Let Ig be the full scale deflection current (current flowing through
the galvanometer when it shows full scale deflection) of the
galvanometer. Suppose this galvanometer is converted into an I Ig G I
ammeter of range 0 to I ampere. When the converted galvanometer Is
reads a current I ampere, the current trough it is only Ig ampere and S
the remaining current passes through the shunt resistance S. Let G Fig.a
be the resistance of the galvanometer. Then from fig.a,
I g G = IsS = I I Sg I E K
Ig G P Q
S = (1)
I Ig
V
Let ‘k’ be the figure of merit (current sensitivity) of the
galvanometer. It is the current required to produce a R
deflection of one division. If there are ‘n’ divisions on
each side of the zero division,
G
Ig = kn (2) Fig.b
To find out k and G we consider a circuit as shown in fig.b. Using Ohm’s law, the current
through P and Q is,
E
I =
P+Q
EP
Potential difference across P is, V =
P+Q
V EP
Current through the galvanometer, I1 = =
R G P+Q R G
EP
If R = 0, I1 = (3)
P+Q G
Due to this current the galvanometer shows a deflection of ‘d’ divisions. Then current sensitivity,
54 Optics & Electricity Practical II
I1 EP E P
P+Q G d
k = = = (4)
d P+Q Gd
When a resistance is introduced in R, current I1 decreases and hence the deflection in the
galvanometer decreases. The current and hence the galvanometer deflection reduces to half when
R G + P (5)
Using eqns.2, 4 and 5 in eqn.1
we get the resistance needed to
II0
convert a galvanometer into an
ammeter of range 0 to I
ampere. Its current per division
is I/n.
Let be the resistance
per unit length of a standard
resistance wire. Length of the
wire needed for a resistance S
is S/. I0
Then convert the
galvanometer into ammeter
and calibrate it. Draw the calibration graph as shown in the model.
Procedure
To find G and k: Connections are made as shown in the fig.b. Suitable resistances are
introduced in P and Q with P << Q. For example, P = 10 and Q = 990 so that P + Q = 1000
. The voltage across P (much smaller voltage) is applied to the galvanometer through the
resistance box R. All the keys in R are plugged and
tightened so that R = 0. The deflection ‘d’ in the
galvanometer is noted. Now increase the resistance in Rh1
E K1
R and find out the resistance needed to reduce the
deflection to half of its original value. This value of
A J B
resistance is noted. The experiment is repeated after E
reversing the commutator. The entire experiment is
repeated for different values of P and Q. In each case 6 1 G
5 2 HR
‘d’ and ‘G’ are determined. 4 3
To convert galvanometer into ammeter: Using G
eqns.2, 4 and 5 in eqn.1, calculate the resistance S +
needed for the conversion of galvanometer to the R
ammeter for the selected current range. For example, I S I
If there are 30 divisions in the galvanometer we can
conveniently choose a range of 0 to 0.3, 0 to 3
ampere. Connect the calculated resistance S in parallel K2 Rh2
with the galvanometer. Thus the galvanometer is Fig.c
converted to an ammeter.
Calibration of the ammeter: To calibrate the ammeter we follow the method described in
exp.No.2.10. The balancing length L, corresponding to the standard voltage E and the balancing
Practical-II MCT 55
length l corresponding to the voltage across the standard resistance R for various currents are
determined. The current and the error in the measurement of it are calculated as follows.
E
I = l
LR
If E = 1.08 volt,
1.08l
I =
LR
E
Correction to the ammeter reading = I I0 = l I0
LR
The calibration graph between the measured current I0 in the X axis and the correction I I0 in
the Y axis is plotted.
Precautions
All precautions given in exp.No.15 applicable in this experiment.
Take an excess length ( 2 cm) of standard resistance wire required to connect at the two
terminals of the galvanometer.
Observation and tabulation
To find G, k and R
Voltage applied across P and Q = E = ……… volt
P + Q = ………. ohm
P Q Galvanometer deflection P Resistance for half deflection GRP
ohm Ohm ‘d’ d Ohm
left Right Mean Left Right Mean
Mean Mean G
Resistance per unit length of the standard resistance wire, = ……. ohm/metre
S
Length of the wire needed for the resistance S is L = metre.
ρ
(Read the precautions given)
To check the correctness of the ammeter (converted voltmeter) readings
Standard voltage E = ……. volt
Balancing length with standard voltage, L = …….. cm
I
Current per division of the constructed ammeter, i = = ……ampere/division
n
Deflection in Measured current Length of the Calculated El
the converted I0 = id potentiometer wire current = Error = I0
RL
galvanometer, ampere corresponding to the El ampere
‘d’ measured current, l cm ampere
LR
Result
The given galvanometer is converted to an ammeter to read a current range 0 to ….. A.
Its readings are checked with a potentiometer and a calibration graph is plotted.
Standard data*
E m f of the Daniel cell, E = 1.08 volt
*Resistance per metre of copper wire of
Gauge number Diameter in mm Resistance per metre /m
20 0.9144 0.0263
22 0.7112 0.0434
24 0.5588 0.0703
26 0.4572 0.105
28 0.3759 0.155
30 0.3150 0.221
Practical-II MCT 57
Exp.No.2.17
Verification of Thevenin’s and Norton’s theorems
Aim: To verify the Thevenin’s and Norton’s network theorems.
Apparatus: Power supply, resistance box or standard resistances, voltmeter, ammeter etc.
Theory r R1 R2 C
Thevenin’s theorem: It states that a
I IIL IL
two terminal network containing
resistances (linear impedances) and
voltage sources can be replaced by a E R3 RL VL
single voltage source VTh, in series
with a single resistance RTh, where,
VTh, called Thevenin voltage, is the
open circuit voltage between the Fig.a: Network of resistances and a voltage source D
terminals and RTh, called the Thevenin
resistance, is the resistance that would be measured between RTh C
the terminals with all the voltage sources are replaced by their IL
internal resistances.
Fig.a represents a circuit containing simple linear
network of three resistances R1, R2 and R3 with two terminals VTh RL VL
C and D. RL is a load resistance connected between the two
terminals C and D. E is the e m f of the voltage source and r is
its internal resistance. Fig.b represents the Thevenin D
equivalent. Fig.b: Thevenin equivalent
Theoretical calculation of Thevenin voltage and Thevenin
resistance: From fig.a,
I R1 r I L R 2 R L = E (1)
IL R 2 R L I IL R 3 = 0 (2)
IR 3 = IL R 2 R 3 R L
IL
I = R2 R3 R L (3)
R3
Using eqn.3 in eqn.1,
IL
R 2 R 3 R L R1 r IL R 2 R L = E
R3
ER 3
IL = (4)
R 2 R 3 R L R1 r R 3 R 2 R L
ER 3R L
VL = (5)
R 2 R 3 R L R1 r R 3 R 2 R L
58 Optics & Electricity Practical II
RTh = R 2
R1 r R 3 (7)
r R1 R 3 R3 RTh
From fig.b,
VTh
IL =
R Th R L
Fig.d: Calculation of RTh D
Using eqns.6 and 7
ER 3
= (4)
R 2 R 3 R L R1 r R 3 R 2 R L
ER 3R L
VL = ILRL = (5)
R 2 R 3 R L R1 r R 3 R 2 R L
Experimental verification: From the Thevenin equivalent circuit given in fig.b we can write,
VTh = ILRTh + VL VL
Or, VL = ILRTh + VTh (8)
Eqn.8 represents a straight line. Thus the graph
between IL and VL is a straight line with slope equal
to RTh and the Y intercept VTh. VTh
To verify the Thevenin’s theorem we show that
the value of VTh and RTh obtained by the calculation
method and the graphical method are same as those
obtained by the direct measurement.
Fig.e IL
Measurement of VTh: Thevenin voltage VTh can be
measured directly from the r R1 R2 C R I
open circuit shown in fig.c.
Measurement of RTh: We
V
use the circuit as shown in r
fig.f for the measurement of R3 V
RTh. The same power
supply used in fig.a is used E
here. Send a current I
through the circuit. Measure Fig.f: Determination of RTh D
Practical-II MCT 59
RN = RTh = R 2
R1 r R 3 Fig.j VL
r R1 R 3
Calculation of IN: Let I be the main
current in the circuit with the terminals r R1 R2 C
short circuited as shown in fig.k. From the I IIN
fig.k, IN
I IN R 3 = IN R 2 E R3 A
IN R 2 R 3
I =
R3
Also, E = I r R1 I N R 2 Fig.k: Determination of IN D
I N R 2 R 3 r R1
= IN R 2
R3
60 Optics & Electricity Practical II
ER 3
IN = (11)
R1 r R 2 R 3 R 2 R 3
To verify the Norton’s theorem we show that the value of IN and RN obtained by the
calculation method and the graphical method are same as those obtained by the direct
measurement. (Remember RN is same as RTh)
Procedure:
Step 1: Determination of internal resistance ‘r’ of power supply
Measure the open circuit voltage E (=1015 volt) of the power r
supply. Connect a suitable resistance R, say 10 as shown in fig.m
and measure the terminal voltage Vt. I
+
E V E R V
I = = t
Rr R
r =
E Vt R
Vt Fig.m
Power supply voltage E must be kept constant throughout the experiment. So use
regulated power supply
Ensure that all the unplugged keys of the resistance boxes are tightened.
Formula for theoretical calculation is applicable only for the ‘T’ type network given in
the figure. It is different for different networks.
Observation and Tabulation
Measurement of internal resistance of the power supply
Open circuit voltage of the power supply, E = ……..volt
Value of resistance connected to the power supply, R = ……. Ohm
Terminal voltage (voltage across R), Vt = ……. Volt
Internal resistance, r =
E Vt R = …….. ohm
Vt
Theoretical calculation of VTh, RTh = RN and IN
VTh
ER 3
RTh = R 2
R1 r R 3 IN =
ER 3
R1 R2 R3
r R1 R 3 r R1 R 3 R1 r R 2 R 3 R 2 R 3
volt
Ampere
Graphical verification
Thevenin voltage, VTh = Y intercept of VL-IL graph = ……… volt
Thevenin resistance, RTh = Slope of the VL-IL graph = ……… ohm
Norton’s current, IN = Y intercept of IL-VL graph = ……… ampere
Experimental verification - by direct measurement method
Thevenin voltage, VTh = Open circuit voltage = ……… volt
Norton’s current, IN = Short circuited current = ……… ampere
Determination of RTh
Sl.No. R Voltage across the Voltage across R VR
RTh =
terminals C D, V volt V volt V
1
2
3
4
5
Result
Thevenin voltage, VTh
By calculation = ……… volt
By graphical method = ……… volt
By direct measurement = ………. volt
Thevenin resistance, RTh
By calculation = ……… ohm
By graphical method = ……… ohm
By direct measurement = ……… ohm
Norton current, IN
By calculation = ……… ampere
By graphical method = ……… ampere
By direct measurement = ……… ampere
Since the values obtained for VTh, RTh and IN by different methods are nearly equal
Thevenin’s and Norton’s theorem are verified.