PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
1
MEASURING INSTRUMENTS
1.
MEASUREMENT OF LENGTH
For the measurement of length
usually meter scales are used with an accuracy upto
millimeters. To measure length say 1/100
th
of a centimeter (or) 1/100
th
of
a millimeter
, it is not
possible to measure accurately using meter scale. Hence the following instruments are used for
more accuracy.
1.
Vernier Calipers (Accuracy upto 1/100
th
of cm)
2
.
Screw gauge (Accuracy upto 1/100
th
of mm)
1.
VERNIER CALIPERS
Design of the Instrument
It consists of two scales viz, main scale and vernier scale. The main scale is a long steel bar
graduated in inches on the top and i
n centimeters on the bottom. 10 main scale divisions are equal
to 1cm. The vernier scale has 10 divisions which are equal to 9 main scales divisions. There are
two metal jaws A & B. A is fixed and B is movable which can be fixed at any position using the
s
crew(s) as shown in figure
1
.
Fig
1. Vernier Caliper
The
projections P1 and P2 of the jaws in the upward direction are
used to measure the
internal diameter of the calorimeter, cylinder, etc.
Least Count
Least count of vernier calipers is
defined as the smallest length that can be accurately
measured and is equal to the difference between the main scale division (MSD) and the vernier
scale division (VSD).
Derivation
LC
=
1 MSD
–
VSD
10 MSD
=
1 cm
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1 MSD
=
1 /10 cm
10 VSD
=
9 MSD
1 VSD
=
9/10 MSD (or)
{
9/10
} X
{1/10}
=
9/100 cm
LC
=
{1/10

9/100} cm
=
1/100 cm =
0.01 cm
No Zero Error
To find the zero error of the vernier calipers, the two
jaws A
and B brought in contact with each other. If the zero of the vernier
scale coincides with the zero of the main scale then there is no
zero
error (Z.E = 0) (fig. 1.1).
Hence
there is no need to apply zero
correction
(Z.C = 0)
.
Fig. 1.1 No zero error.
Positive Zero Error
If the zero of the vernier scale lies on the right side the zero
of the main scale, then the
instrument has
an error called
positive
zero
error
(fig. 1.2).
This error should be subtracted from the final
reading. Thus, the zero correction is negative.
Example
If the 4
th
vernier division coincides with any of the
main
scale
division, then
zero error
=
Vernier Scale coincidence x Least count
Fig. 1.2
positive
zero error.
=
VSC x LC
=
4 x 0.01 = 0.04 cm
Zero Correction
=

0.04 cm
Negative Zero Error
If the zero
of
the vernier
scale
lies on the left side of
the zero of main scale, then the instrument is said to have an
error called negative zero
error (fig.
3
)
.
This error should be
added to the final reading. Thus, the zero correction is positive.
Fig. 1.3
negative
zero error.
Example
If the 6
th
division of the vernier coincides with any of the main scale division, then
Zero Error
=

(no. of vernier scale divisions
–
Vernier scale coincidence) x Least count
=

(10

6) x 0.01 cm
=

4 x 0.01 =

0.04 cm
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Zero Correction
=
0.04 cm
To find the length of the specimen
The specimen whose length (or) outer diameter is to be determined is held between the two
jaws. The position of zero of the vernier on the main scale gives the main scale reading
(MSR).
The vernier division which coincides with any one of the main
scale division gives the vernier scale coincidence (VSC), then
Observed reading (OR)
= MSR
+
(VSC X LC) in cms
Actual (or) correct Reading
= OR ±Zero
Correcti0n
in cms
Example
OR = 1.2 + (3 x 0.01) cm
(
Zero
Error
= Nil
)
CR = 1.23 cm
2
.
SCREW GAUGE
Fig
2. Screw gauge
Design of the Instrument
It consists of two scales namely pitch scale and head scale. Pitch scale is a millimeter scale
engraved on the cylinder (C), which is rigidly attached with the frame (f). The head scale carrier
100 equal divisions. The specimen can be held in

between the t
wo edges A (fixed) and B
(movable) shown in fig.
2
.
Derivation of Least count
LC
=
Pitch/No. of head scale divisions
Pitch
=
Distance
moved on the pitch scale / No. of head scale rotations
To find pitch the head is given say, two rotations and the distance
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moved by the head on the pitch scale is noted
Pitch
=
2mm/2
=
1 mm
Number of head scale division
=
100
LC
=
1/100 mm
LC
=
0.01 mm
To find zero
error
To find zero error
the stud A and the tip B are kept in contact. If the zero of the head scale coincides with zero of the
pitch scale on the reference line, the instrument is said to have no zero error
(Fig. 2.1)
.
Fig. 2.1
No zero e
rror.
Positive zero error
If the zero of the head scale lies below the reference line of the pitch scale, the zero error i
s
positive and the correction is
negative
(Fig. 2.2)
.
Example
If the
2
nd
division of the head scale coincides with the reference line of
the pitch scale then
Zero error
=
head scale coincidence x Least count
=
2
x 0.01 mm
=
0.0
2
mm
Zero Correction
=

0.0
2
mm
Fig.
2.
2
positive
zero error.
Negative zero error
If the zero of the head scale lies above the reference line of the pitch
scale the
zero error is
negative
and the correction is positive
(Fig. 2.3)
.
Example
If the 9
6
th
division of the head scale coincides with reference line of the
pitch scale then,
Zero error
=

(100
–
head scale coincidence) x least count
=

(100
–
96
) x 0.01 mm
=

0.04
mm
Zero correction
=
0.04
mm
Fig.
2.
3
negative
zero error.
To find the diameter of the given wire
The wire is gripped gently between the faces A and B. The number of pitch scale divisions
just in front of the head scale gives the pitch scale reading. The
division on the head scale that
coincides with the reference line gives the head scale coincidence. The readings are noted and are
tabulated.
Example
Suppose if the PSR
=
5mm
H.S.C
=
35 div.
HSR
=
HSC x L.C.
=
35 x0.01 = 0.35 mm
Observed reading
=
PSR + HSR = 5 + 0.35 = 5.35 mm
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If the zero
error is
0.04
mm
(Positive zero error)
, then the zero correction is
–
0.04 mm,
Correct reading
=
OR±ZC
=
(5.35

0.04) mm
=
5.31 mm
3
. TRAVELLING MICROSCOPE
Design of the Instru
ment
Travelling Microscope consists of an
ordinary
compound microscope which slides
along a graduated vertical pillar, attached to
horizontal base resting on the leveling
screws. The
main scales divisions along
with vernier scale divisions are marked on
the horizontal base and the vertical pillar as
shown in fig
.3
.
The microscope can be moved up
and down in the vertical pillar and can be
moved in to and fro direction over the
hori
zontal base. Thus the microscope can
be moved both in the vertical and horizontal
directions. Two fine
adjustment screws
are
provided for the horizontal and vertical
movements respectively.
The image of the object
can be focused by adjusting the side
screw
(S) attached to the microscope. The eye
piece of the microscope is provided with a
cross wire.
Fig. 3 Travelling
Microscope.
The main scale is divided into millimeter and half a millimeter. Therefore
the
value of one
main scale division (MSD) is 0.5 mm. The vernier scale of the travelling microscope is divided into
50 divisions which are equivalent to 49 main scale divisions.
Least count Derivation
LC
=
1 MSD
–
1 VSD
20 MSD
=
1 cm
1 MSD
=
1/20 cm
Here
50 VSD
=
49 MSD
1 VSD
=
49/50 MSD
=
(
49/50
) x
(
1/20) cm
1
VSD
=
49/1000 cm
LC
=
(1/20
–
49/1000) cm
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=
1/1000 cm
LC
=
0.001 cm
4. MEASUREMENT OF ANGLES
Fig. 4
Spectrometer.
Spectrometer
–
Design of the instrument
Spectrometer consists of collimator (c) telescope (T), prism table (p) and vernier table as
shown
in
fig
4
. The collimator consists of a convex lens fixed at one end and a slit of adjustable
widt
h on the other end.
The telescope consists of an objective (O) at one end and an eye piece fixed with the
cross wires on the other end. The prism table consists of two circular disc connected by three
leveling screws. The vernier
tables have
two verniers V
A
and V
B
each having main scale and
vernier scale. Here one main scale division is equal to half a degree. Each vernier scale has 30
divisions, which is equal to 29 main scale divisions.
Initial adjustments
(i)
Eye piece adjustment:
The eye piece (E) is adjusted until the cross wires are
clearly seen when viewed through the telescope.
(ii)
Distant object method:
A clear, well defined inverted and diminished image of a
distance object is seen by adjusting the telescope.
(iii)
Slit
adjustment:
Th
e slit is made narrow with the help of the screw, provided
aside of
the
slit.
(iv)
Collimator adjustment:
The telescope is brought on line with the collimator. The
slit is
illuminated by a source of light. If the image of the slit appears blurred, then
the
screw of the collimator is adjusted until a clear image is seen when viewed
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through the telescope. Now the rays of light
emerging from the collimator will be
rendered parallel.
(v)
Prism table adjustment:
The spirit level is placed on the prism table, parallel
to
the line joining any two leveling screws. The air bubble in the spirit level is brought
to the centre by adjusting the two screws. It
is then placed in a perpendicular
direction and the air bubble is brought at the centre by adjusting the third screw.
Now the prism table will be horizontal.
(vi)
Spectrometer base:
The base of the spectrometer is adjusted to the horizontal
with the help of the three leveling screws.
Least count
LC
=
1 MSD
–
1 VSD
1 MSD
=
½ degree
30 VSD
=
29
MSD
1 VSD
=
29/30 MSD
1 VSD
=
29/30 x ½ degree
1 VSD
=
29/60 degree
LC
=
(1/2

29/60) degree
=
(30

29)/60 degree
=
1/60 degree
Since 60’
=
1°
1’
=
1°/60
LC
=
1’
Measu
rement
Before doing any experiment using spectrometer, the above mentioned initial adjustments
have to be made. While performing the experiment, the main scale and vernier scale readings are
noted from both the verniers in V
A
and V
B
in degree and minutes.
Fig. 5
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Fig.
6
.
Different order of spectrum
Expt. No.:
Date:
1. a. PARTICLE SIZE DETERMINAION USING DIODE LASER
Aim
To determine the size of the given particle using the laser source.
Apparatus requir
ed
A laser source, laser grating, given particle, screen, scale, etc.
Formula
Explanation of symbols
Symbol
Explanation
Unit
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
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λ
Wavelength of the laser source
Å
m
Order of spectrum
Unit
Y
m
Distance of m
th
order from Zeroth order
met
r
e
D
1
Distance between the particle and the
screen
metre
Theory
`
A device for producing spectra by diffraction is known as diffraction grating. While producing
diffraction spectra using the particle, the size of
the particle should be comparably equal to the
wavelength of the source. The diffracted wave undergoes constructive and destructive interference
effect. The intensities of the spectra depend up to the diffraction angle. By measuring the diffraction
angle i
nterms of orders of spectra. The wavelength of the given laser source and the size of the
particle can be determined.
To fine the size of the given particle
Wavelength of the given laser source = ………………………
Å
S.No.
Order (m)
Y
m
Mean
Y
m
Y
2
m
D
2
1
√
d
L䡓
剈R
啮楴
乯N
X10

2
m
X10

2
m
X10

2
m
X10

4
m
2
X10

4
m
2
X10

2
m
Metre
1
2
3
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4
5
6
7
8
9
10
Procedure
To find the size of the given particle
Now the laser grating is removed and the size of the particle of be found is introduced. The
laser source is switched ON and the light is made of
fa
l
l on the particle. The screen is moved back
and forth until the clear image of the spectrum is seen and the
distance between the screen and
the particle (D
1
) is noted. Due to diffraction of laser light by the particle, different orders of spectrum
are obtained as shown in fig.
5
.
The positions Y
1
, Y
2
, Y
3
…… of the spots belonging to the first
order, second order, third
orders etc. on either side of the central maximum are
noted in similar
way as noted above.
Then
by using the given formula the size of the particle can be determined.
Result:
The size of
the given particle = ………………………….. meter.
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VIVA
–
VOCE
1.
Define the term “Diffraction”, with its conditions.
2.
What is meant by grating element?
3.
Is the diffraction possible if laser light is replaced by ordinary light for the same particle?
Explain.
4.
What
happens to the order of spectrum, if the distance between the particle and the screen
is increased?
5.
Will the laser undergo diffraction through ordinary grating? Explain.
6.
What will happen to the order of spectrum if the size of the particle is decreased?
WO
RK SHEET
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WORK SHEET
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Fig.
7
Diffraction Pattern
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Fig.
8
Angle of divergence.
Expt. No.:
Date:
1. b. DETERMINATION OF WAVELENGTH OF LASER USING GRATING
Aim
To
determine Wavelength of the given laser source, using a laser grating.
Apparatus requi
r
ed
A laser source, laser grating, screen, scale, etc.
Formula
Explanation of symbols
Symbol
Explanation
Unit
λ
Wavelength of the
laser source
Å
m
Order of spectrum
Unit
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
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X
m
Distance of m
th
order from Zeroth order
metre
D
Distance between the particle and the
screen
metre
N
Number of rulings in the grating
Lines/metre
Theory
A device for producing spectra by
diffraction is known as diffraction grating. While producing
diffraction spectra using the particle, the size of the particle should be comparably equal to the
wavelength of the source. The diffracted wave undergoes constructive and destructive interferenc
e
effect. The intensities of the spectra depend upto the diffraction angle. By measuring the diffraction
angle interms of orders of spectra. The wavelength of the given laser source and the size of the
particle can be determined.
(i) To find the wavelen
gth of the laser source
Number of rulings in the grating = ………………………
S.No
.
Order
(m)
X
m
Mean
X
m
X
2
m
D
2
√
λ
=
i䡓
=
剈o
=
啮楴
=
乯k
=
uNM

2
m
X10

2
m
X10

2
m
X10

4
m
2
X10

4
m
2
X10

2
m
Aº
1
2
3
4
5
6
7
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8
9
10
(ii) To find the Angle of Divergence (
Ф
) using Laser source
S.No.
Distance between laser
and screen
(X 10

2
m )
Radius of circular image
(X 10

2
m )
Angle of divergence
Ф
= r
2

r
1
/d
2

d
1
(degree)
1
d
1
=
r
1
=
2
d
2
=
r
2
=
Procedure
(i)
To find wavelength of the laser source
The laser source and the laser grating are mounted on separate stands as shown
in fig.7. A
fixed distance (D) is kept between the laser grating and the screen. The laser source is
switched ON and the beam of laser is allowed to fall on the laser grating. The diffracted beams
are collected on the screen. The diffracted beams are in the form of s
pots as shown in fig.
7
.
In the figure 7
, the intensity of the irradiance is found to decrease, from zeroth order to
higher orders, i.e. the first order is brighter than the second order and so on. The positions
X
1
,
X
2
,
X
3
… of the spots belonging to the
first order, second order, third order etc., on either side of the
central maximum are marked on the screen and is noted.
The experiment is repeated for various values of D and the positions of the spots are noted.
Then by using the given formula the wave
length of the source can be calculated and the mean is
taken.
(ii)
To find the angle of a divergence (
Ф
)
Angle of divergence given the angular spread of the laser beam. A simple
diagrammatic
explanation of finding the angle of divergence show in fig.
8
.
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WORK SHEET
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Result
(a)
The wavelength of the given laser source =………………………
Å
(b)
Angle of Divergence of the Laser beam= ………………….. Degrees.
VIVA
–
VOCE
1.
Define the term “Diffraction”, with its conditions.
2.
What is meant
by grating element?
3.
Is the diffraction possible if laser light is replaced by ordinary light for the same particle?
Explain.
4.
What happens to the order of spectrum, if the distance between the particle and the screen
is increased?
5.
Will the laser undergo dif
fraction through ordinary grating? Explain.
PHYSICS PRACTICAL MANUAL CUM RECORD
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6.
What will happen to the order of spectrum if the size of the particle is decreased?
7.
What is meant by “Numerical aperture” and “Acceptance angle”?
8.
What is the principle used in the propagation of light through opt
ical fibres?
9.
What are the parts of an optical fibre?
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Fig
9. Numerical Aperture
Expt. No.:
Date:
1. c. DETERMINATION OF ACCEPTANCE ANGLE OF OPTICAL FIBRE
Aim
To measure the numerical aperture and acceptance angle of the
given optical fibre.
Apparatus requir
ed
O
ptical fibre, numerical aperture measurement Jig.
,
scale,
etc.
Formula
Numerical apertur
e of the given optical fibre
NA
=
√
(No unit)
Acceptance angle
‘
θ
’
m
ax
= sin

1
(NA)
degrees
Explanation of symbols
Symbol
Explanation
Unit
d
Distance between the tip of the optical fibre and the aperture of
the Numerical Aperture (NA) Jig.
metre
r
Radius of the circular opening in NA jig
metre
Theory
Numerical Aperture (NA) and Acceptance
Angle:
It is the light collecting efficiency of the fibre and is the measure of the
maximum amount of
light
that can be accepted by the fibre. Using
Snell’s law
mathematically we can say
NA
=
sin
θ
max
=
√n
1
2

n
2
2
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Where
,
n
1
=
refractive Index of core,
n
2
=
refractive index of cladding
θ
max
=
Acceptance angle of the fibre
=
sin

1
(Numerical aperture)
i.e.
θ
max
=
sin

1
(NA)
Measurement of Numerical aperture
S.No.
Length of
the given
fibre
Distance between NA
Jig opening and the
fibre (d)
Radius of the circular
opening in Numerical
aperture Jig (r)
Numerical
aperture=
√
Unit
meter
X10

3
m
X10

2
m
(No Unit)
1.
2.
3.
4.
5.
6.
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Acceptance angle (
θ
)
θ
max
= sin

1
(NA)
θ
max
= ……………………. degrees
Procedure
Measurement of Numerical aperture and Acceptance angle
A known length of fibre is taken. One end of the fibre is connected to the laser source and
the other end is
connected to the numerical a
perture (NA) Jig as shown in fig
.
9
. The source is
switched ON. The opening in the NA jig is completely opened so that a circular red patch of laser
light is observed on the screen. Now the opening in the NA Jig is slowly closed
with the knob
provided, so that at a particular points the circular light patch in the screen just cuts. The radius of
the circular opening (r) of NA Jig at which the circular patch of light just cuts is measured.
The distance between the NA jig opening
and the fibre can be measured directly with the
help
of the calibration in NA jig. By
substituting the values in the given formula the numerical
aperture can be calculated.
The same procedure can be adopted for various distances between the fibre and the
opening of NA jig. The same procedure can be adopted for various length of fibre cable.
By finding the mean of numerical aperture (NA) and substituting it in the given formula the
acceptance angle can be determined.
Result
(i) The Numerical aperture of
the given optical fibre = …………. (No unit)
(ii) The acceptance angle of the given optical fibre = …………… Degrees
VIVA
–
VOCE
1.
What is meant by “Numerical aperture” and “Acceptance angle”?
2.
What is the principle used in the propagation of light through
optical fibres?
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3.
What are the parts of an optical fibre?
4.
What is the type of the laser beam used in the experiment?
WORK SHEET
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WORK SHEET
CALCULATIONS:
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Fig.
10
. Air Wedge
Fig.
11
. Fringe pattern.
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Expt. No.:
Date:
2. DETERMINATION OF THICKNESS OF A THIN WIRE

AIR WEDGE METHOD
Aim
To determine the thickness of a given thin wire (or) thin sheet of paper by forming interference
fringes due to an air

wedge.
Apparatus requi
r
ed
Traveling microscope, two optically plane glass plates,
45° angle glass plate,
given wire (or) thin
paper, sodium
vapour
lamp
, Transformer
etc.
Formula
Thickness of the given wire (or) thin sheet of paper is given by
t =
Explanation of symbols
Symbol
Explanation
Unit
l
Distance between edge of contact and the wire (or) paper
metre
λ
Wavelength of the monochromatic source of light (5893
Å
)
metre
β
Mean fringe width
metre
Theory
Two plane glass plates are
inclined at an angle by introducing thin materials. (e.g.. hair), forming a
wedge shaped air film. This film is illuminated by sodium light, interference occurs between the two rays,
one reflected from the front surface and the other obtained by internal r
eflection at the back surface. Since
in the case of a wedge shaped film, thickness of the material remains constant only in direction parallel to
the thin edge of the wedge, straight line fringes parallel to the edge of the wedge are obtained.
(i) To
find the band width (
β
)
LC = 0.001 cm
TR = MSR + (VS
C
X LC)
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Order of the
fringes
Microscope reading
Width of 15
fringes
Fringes
width (
β
)
MSR
VSC
TR
X10

2
m
(Div)
X10

2
m
X10

2
m
X10

2
m
n
n+5
n+10
n+15
n+20
n+25
n+30
n+35
n+40
n+45
n+50
Mean = ……………. X 10

2
m
(ii)
To find the distance between the edge of contact and the material of wire (or)
paper.
LC = 0.001 cm
TR = MSR + (VSC x
LC)
Position of the Microscope
Microscope reading
l = R
1~R2
x
10

2
m
MSR
x 10

2
m
VSC
(Div)
TR
x 10

2
m
At the edge of contact
(R1)
At the edge of material of wire (or) paper
(R2)
Procedure
Two optically plane glass plates are placed
one over the other and tied by means of a
rubber band at one end. The given material of wire (or) paper is introduced on the other end, so
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that an airwedge is formed between the plates as shown in fig.1
0
. This set up is placed on the
horizontal bed plate
of the traveling microscope.
Light from the sodium vapour lamp (S) is rendered parallel by means of a condensing lens
(L). The parallel beam of light is incident on a plane glass plate (G) inclined at an angle of 45
º
and
gets reflected. The reflected ligh
t is incident normally on the glass plates in contact Pc. Interference
taken place between the light reflected from the top and bottom surfaces of the glass plates and is
viewed through the traveling microscope (M). Hence large number of equally spaced dar
k and
bright fringes are formed which are parallel to the edge of contact (fig.
11
).
The microscope is adjusted so that the bright (or) dark fringe near the edge of contact is
made to coincide with the vertical cross wire and this is taken as the nth frin
ge. The reading from
the horizontal scale of the traveling microscope is noted. The microscope is moved across the
fringes using the horizontal traverse screw and the readings are taken when the vertical cross wire
coincides with every successive 5 fringes
(5, 10, 15….). The width of every 15 fringes is calculated
and the width of one fringe is calculated. The mean of this gives the fringe width (
β
).
The cross wire is fixed at the inner edge of the rubber band and the reading from the
microscope is noted.
Similarly reading from the microscope is noted keeping the cross wire at the
edge of the material. The difference between these two values gives the value of ‘l’. Substituting ‘
β
’
and
‘
l
’
in the given formula, the thickness of the given material can be determined.
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
29
Result
The thickness of the given material = ………………….. meter.
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VIVA
–
VOCE
1.
What is the principle behind the
formation of fringes in Air wedge?
2.
What is meant by an “Air Wedge” and explain how it can be formed?
3.
What is meant by fringe width?
4.
What is the theory of formation of thin films? Give examples.
5.
What is the use of inserting 45˚ angled glass plate?
6.
Why do we
get straight line fringes in an airwedge?
7.
Explain the reason for color formation in soap bubbles, when white light falls on it.
8.
What happens to the fringe width, if the thickness of the material is increased?
9.
What is the condition for the formation of bri
ght fringes?
10.
Why do we get bright and dark fringes alternatively?
PHYSICS PRACTICAL MANUAL CUM RECORD
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31
Fig.12 Ultrasonic Interferometer
Expt. No.:
Date:
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3. DETERMINATION
OF
VELOCITY OF
SOUND
AND COMPRESSIBILITY OF THE LIQUID

ULTRASONIC INTERFEROMETER
Aim
(i) To determine the velocity of Ultrasonic waves in the given liquid using Ultrasonic Interferometer.
(ii) To determine the compressibility of the given liquid.
Apparatus requi
r
ed
Ultrasonic Interferometer, measuring cell, frequenc
y generator, given liquid, etc.
Formula
(i
) Velocity
of Ultrasonic
waves in the given liquid v = n
λ
ms

1
Where, Wavelength
met
r
e
(or)
Å
(ii)Compressibility of the given liquid
m
2
/ N
Explanation of symbols
Symbol
Explanation
Unit
n
Frequency of the generator which excites the crystal
Hertz
λ
Wavelength of the Ultrasonic
metre
ρ
Density of the given liquid
Kg/m
2
d
Distance moved by the micrometer screw
metre
x
Number of oscillations
Unit
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
33
Fig. 13
Distance moved by the reflector Vs Crystal current
Theory
An Ultrasonic Interferometer is a simple and direct to determine the velocity of Ultrasonic waves in
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34
liquid with a high degree of accuracy. Here the high frequency
genera
tor generates
variable frequency,
which excited the Quartz Crystal placed at the bottom of the measuring cell (Fig. 1
2
). The excited Quartz
Crystal generates Ultrasonic waves in the experimental liquid. The liquid will now serve as an acoustical
grating
element. Hence when Ultrasonic waves passes through the rulings of grating, successive maxima
and minima occurs, satisfying the condition for diffraction.
Initial adjustments:
In high frequency generator two knobs are provided for initial adj
ustments. One is marked with ‘Adj’
(set
)
and the other with ‘Gain’ (Sensitivity). With knob marked ‘Adj’ the position of the needle on the
ammeter is adjusted and with the knob marked ‘Gain’, the sensitivity of the instrument can be increased for
greater d
eflection, if desired.
Procedure
The measuring cell is connected to the output terminal of the high frequency generator through a
shielded cable. The cell is filled with the experimental liquid before switching ON the generator is switched
ON,
the Ultrasonic waves move normal form the Quartz Crystal till
they
are reflected back by the movable
reflector plate. Hence, standing waves are formed in the liquid in

between the reflector plate and the quartz
Crystal.
The distance between t
he reflector and crystal is varied using the micrometer screw such that the
anode current of the generator increases to a maximum and then decreases to a minimum and again
increases to a maximum in the anode current is equal to half the wave length of th
e Ultrasonic waves in the
liquid (Fig.
13
.). Therefore, by nothing the initial and final position of the micrometer screw for one complete
oscillation (maxima
–
minima

maxima) the distance moved by the reflector can be determined.
To minimize the erro
r, the distance (d) moved by the micrometer screw i
s
noted for ‘x’
number of oscillations (successive maxima), by noting the initial and final reading in the micrometer screw
and is tabulated. From the total distance (d) moved by the micrometer screw and t
he number of
oscillations(x), the wavelength of ultrasonic waves can be determined using the formula
λ
= 2d/x.
From the
value of
λ
and by not
ing the frequency of the generator (n), the velocity of the Ultrasonic waves can be
calculated using the given
formula.
After determining the velocity of the Ultrasonic waves in liquids. The
compressibility of the liquid is calculated using the given formula.
PHYSICS PRACTICAL MANUAL CUM RECORD
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Result
(i)The velocity of Ultrasonic waves in the given liquid
=……………. ms

1
(ii)Compressibility of the given liquid
= ……………. m
2
N

1
VIVA
–
VOCE
1.
What is meant by Compressibility?
2.
Explain the terms stress and strain.
3.
What is the frequency range of Ultrasonics?
4.
What is mean by acoustical grating?
5.
Explain the principle of
determining the compressibility of liquids.
6.
What is meant by node
s
and
Antinodes
?
7.
What do
you
understand by the term “Over tones”?
8.
Is Ultrasonic wave, an Electromagnetic wave? Explain.
9.
What are the various liquids that can be used for finding the compressi
bility using Ultrasonic
interferometer?
10.
What type crystal is thrown into vibrations in the case of Ultrasonic interferometer?
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
37
WORK SHEET
CALCULATIONS:
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38
(i) To find the number of
lines
per metre on the
grating (N)
Order of the spectrum (n) =
Total Reading = MSR +
(VSCXLC)
N =
sin
θ
/n
λ
Lines / m
Mean
θ
degrees
Mean
θ
= 2
θ
/2
V
B
degrees
PHYSICS PRACTICAL MANUAL CUM RECORD
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39
V
A
degrees
2
θ
= R1

R2
V
B
degrees
V
A
degrees
Vernier B (V
B
)
TR
degrees
VSC
divisions
MSR
degrees
Vernier A (V
A
)
TR
degrees
VSC
divisions
LC = 1´
λ
= 5893 A°
MSR
degrees
Diffracted ray
readings
Left (R
1
)
Right (R2)
Expt. No.:
Date:
4. DETERMINATION OF WAVELENGTH OF
MERCURY SPECTRUM

SPECTROMETER
GRATING
Aim
To determine the wavelength of the mercury (Hg) spectrum by standardizing the plane
transmission grating.
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Apparatus requi
r
ed
Spectrometer, plane transmis
sion grating, mercury vapour lamp, sodium vapour lamp, spirit
level
, lens, torch light
etc.
Formula
(i
) The
number of lines drawn on the grating per meter is given by
lines /meter
(ii)The wavelength of the prominent lines of the mercury spectrum is given by
Å
Explanation of symbols
Symbol
Explanation
Unit
n
Order of spectrum
Unit
λ
Waveleng
th of the Sodium vapour lamp (58
93
Å)
Met
re
(or)
Å
θ
Angle of
diffraction
Degree
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
41
Fig. 1
4.
Position of the Grating
Fig.15.
Spectrometer Grating First
to Get
Reflected Ray
order Diffraction
Procedure
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42
(i)
Adjustment of grating for normal incidence
Preliminary adjustments of the spectrometer are made. The grating (G) is mounted on the
grating table with its ruled surface facing the collimator. The slit is
i
lluminated by a source of light
(either sodium (or) mercury vapour lamp) and is made to coincid
e with the vertical cross wire. (The
vernier scales are adjusted to read 0˚ and 180˚ for the direct
ray.
The telescope is rotated through
an angle 90˚ and is fixed. The grating table is adjusted until the image coincides with the vertical
cross wire. Both
the
grating
table and the
telescope
are
f
ixed at this position (fig.
1
4
). Now rotate
the
vernier tables
through 45˚in the same direction in which the telescope has been previously
rotated. The light from the collimator incidents normally (perpendicularly
)
o
n the grating. The
telescope is released and is brought on line with the direct image of the slit. Now the grating is said
to b
e
in the normal incidence position (fig.
15
).
(ii)
Standardization of grating (To find the number of lines drawn in the grating per
me
ter)
The
slit is illuminated by sodium vapour lamp. The telescope is released to get the diffracted
image of the first order on the left side of the central direct image. The readings are tabulated from
the two verniers V
A
and V
B
. Similarly readings are taken for the right side of the central direct
image (fig. 15
). The difference between the two readings gives 2
θ
, where
θ
is the angle of first
order diffraction. The number of lines per meter (N) of the grating is calculated using
the given
formula. The experiment is repeated for the second order and the readings are tabulated.
(iii)
Determination of wavelength of the mercury spectrum
The sodium vapour lamp is replaced by mercury vapour lamp. The diffracted images
of the first
order are
seen on either side of the central direct image (fig.
15
). As before the readings
ar
e
tabulated by coinciding the vertical cross wire with the prominent lines namely violet, blue, blueish
green, green, yellow, red of the mercury spectrum. The difference bet
ween the readings give 2
θ
,
from this
θ
can be found. The wavelength of each spectral line is calculated using the given
formula.
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Result
(i)
Number of lines drawn in the grating per meter = …………..
Lines
/ meter.
(ii)
Wavelength of various spectral lines of the
mercury spectrum are
λ
V
=
……….………….
Aº
λ
B
=
……….………….
Aº
λ
BG
=
……….………….
Aº
λ
G
=
……….………….
Aº
λ
Y
=
……….………….
Aº
λ
R
=
……….………….
Aº
VIVA

VOCE
1.
How many rulings are there in the grating element given?
2.
What is the condition for diffraction?
3.
What
is the difference between reflection and scattering?
4.
What do you understand by the term least count?
5.
What is the diff
erence between transmission and reflection grating?
6.
Why the
skies
appear red during sunset?
7.
What is meant by dispersion?
8.
Define wave packet
.
9.
What is the use of collimator and
telescope?
10.
How the astronomical
telescopes
differ from ordinary telescope?
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
45
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
2010

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46
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
47
Fig. 16 Lee’s Disc apparatus arrangement
(i)
To find
(d
θ
/dt)
θ
2
θ
1 = ………..°C
θ
2 = ………..°C
S.No
Temperature
Time Taken
°C
Kelvin
Sec.
1
θ
2
+
5
2
θ
2
+
4
3
θ
2
+
3
4
θ
2
+
2
5
θ
2
+
1
6
θ
2
7
θ
2

1
8
θ
2

2
9
θ
2

3
10
θ
2

4
11
θ
2

5
Expt. No.:
Date:
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5.
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD
CONDUCTOR

LEE’S DISC METHOD.
Aim
To determine the thermal conductivity of a bad conductor using Lee’s disc apparatus.
Apparatus requi
r
ed
Lee’s disc
apparatus, two thermometers, bad conductor, stop watch, steam boiler, vernier
caliper, screw gauge, biscuit balance etc.
Formula
Thermal
conductivity of the given band conductor
(
)
Explanation
of symbols
Symbol
Explanation
Unit
M
Mass of the metallic disc
Kg
S
Specific heat
capacity
of the material of the
disc
J/Kg/K
(d
θ
/dt)
θ
2
Rate of cooling at steady temperature
θ
2
Kelvin
θ
1
Steady temperature of the steam chamber
Kelvin
θ
2
Steady
temperature of the metallic disc
Kelvin
R
Radius of the metallic disc
metre
h
Thickness of the metallic disc
metre
x
Thickness of the bad conductor
metre
PHYSICS PRACTICAL MANUAL CUM RECORD
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49
Fig. 17 Model Graph.
(ii)
To find the diameter of the metallic disc
LC = 0.01 cm
ZE = ………….
ZC = ………….
x
10

2
m
S.No.
Main Scale
Reading
(MSR)
Vernier Scale
Coincidence (VSC)
Observed Reading
OR=MSR+(VSCXLC)
Correct Reading
CR
= OR±ZC
Unit
X 10

2
m
Div.
X 10

2
m
X 10

2
m
1
2
3
4
5
6
Mean = ……….. x 10

2
m
Description
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Lee’s disc apparatus consists of a brass metal disc (
B
) suspended horizontally by three
strings form a stand. A hollow steam chamber (
A
) with inlet and outlet for steam is placed above.
The given bad conductor is placed between them. Two
thermometers T
1
&
T
2
are
inserted
to
measure the temperatures of A
and
B
respectively.
Procedure
The experimental arrangement is as shown in fig
16
. Steam is allowed to pass through the
steam chamber. The temperature indicated by two thermomet
ers start rising. After 30 minutes a
steady state is reached (i.e) the temperature of the lower disc will no longer rises. At this stage,
steady temperatures
θ
1
and
θ
2
are recorded from the thermometers T
1
and T
2
.
Now the cardboard is removed and the
lower disc is heated directly by keeping it in contact
with the steam chamber. When the temperature of the lower disc attains a value of about 10°C
more than its steady state temperature, the chamber is removed and the lower disc is allowed to
cool down on
its own accord.
When the temperature of the disc reaches 5°C above the steady temperature (
θ
2
) of
the disc
i.e., (
θ
2
+5°C), a stop watch is started and the time is noted for every 1°C fall of temperature until
the metallic disc attains temperature
(
θ
2

5°C).
The thickness and radius of the metallic disc is found using screw gauge and vernier caliper
respectively. The thickness of the bad conductor is found using screw gauge. The mass of the
metallic disc is found by using biscuit balance. The
readings are tabulated in the tabular column.
Graph
A graph is drawn by taking time along the x

axis and the temperature along y

axis (fig.
17
)
Cooling curve is drawn. From the cooling curve d
θ
/dt is calculated by drawing a triangle by
taking
0.5°C above and 0.5°C below the steady temperature
θ
2
.
Substituting this d
θ
/dt i
n the
given formula, thermal conductivity of the card board can be calculated.
(iii)
To find the thickness of the metallic disc
LC = 0.01 mm
ZE = ………….
ZC = …………. x 10

3
m
S.No.
Pitch Scale
Reading (PSR)
Head Scale
Coincidence (HSC)
Observed Reading
OR=PSR+(HSCXLC)
Correct Reading
CR = OR±ZC
PHYSICS PRACTICAL MANUAL CUM RECORD
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51
Unit
X 10

3
m
Div.
X 10

3
m
X 10

3
m
1
2
3
4
5
Mean = ……….. x 10

3
m
(i)
To find the
thickness of the Bad Conductor
LC = 0.01 mm
ZE = ………….
ZC = …………. x 10

3
m
S.No.
Pitch Scale
Reading (PSR)
Head Scale
Coincidence (HSC)
Observed Reading
OR=PSR+(HSCXLC)
Correct Reading
CR = OR±ZC
Unit
X 10

3
m
Div.
X 10

3
m
X 10

3
m
1
2
3
4
5
6
Mean = ……….. x 10

3
m
Result
Thermal Conductivity of given
Bad Conductor = ……………………. W/m/K
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VIVA

VOCE
1.
What do you understand by the term conduction, convection and radiation?
2.
What is meant by thermal
conductivity?
3.
Name any four bad conductors?
4.
What do you understand by the term steady state?
5.
What
is meant by “Rate of Cooling”?
6.
What is the use of cooling the slab and noting the time?
7.
What happen to the thermal conductivity if the thickness of the given
bad conductor is
increased?
8.
Is the diameter of the bad conductor should match diameter of the disc (D) and the steam
chamber (S)? Why?
9.
Will there be any change in thermal conductivity if the area of cross section of bad conductor
is decreased: Justify.
10.
St
ate
some of the applications of bad conductors.
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
53
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
2010

2011
54
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
55
Fig 18 a. Top view of the B

H Curve Unit
Fig 18b. Hysteresis loop
Expt. No.:
Date:
6
. DETERMINATION OF
HYSTERESIS LOSS IN A FERROMAGNETIC
MATERIAL
.
Aim
To determine the hysteresis loss in the transformer core using B

H curve unit.
Apparatus requi
r
ed
CRO, B

H curve unit, patch cords
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Formula
Energy loss =
Explanation of symbols
Symbol
Explanation
Unit
Number of turns in the primary coil

Number of turns in the secondary coil

A
Area of cross section of
core
m
2
L
Length of the core
m
S
V
Steady temperature of the metallic disc

S
H
Horizontal sensitivity of the CRO

DESCRIPTION
The experimental arrangement is shown in the figure 18
a
. One of the specimens
used in the unit is made using
transformer stampings. There are two windings on the specimen
(primary and secondary). The primary is fed to low A.C voltage (50Hz). This produces a magnetic
field H in the specimen. The voltage across R
1
(resistance connected in series with primary) is
pr
oportional to the magnetic field.
It is
given to the imput of the CRO. The A.C magnetic field
induces a voltage in the secondary coil. The voltage induced is proportional to dB/dt. This voltage
is applied to passive integrating circuit. The output of
the i
ntegrator is proportional to B and fed to
the vertical input of the CRO. As a result of the application of voltage proportional to H the
horizontal axis and a voltage proportional to B is the vertical axis, the loop is formed as shown in fig
18b
.
S.No.
N
1
N
2
R
1
ohm
R
2
ohm
S
V
S
H
Area of the
loop
Energy
loss
1
2
3
4
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
57
5
6
7
8
9
10
Observations
Number of turns in the primary
N
1
=
Number of terms in the secondary
N
2
=
R
1
=
ohm
R
2
=
ohm
C
2
=
µF
S
V
=
Vm

1
S
H
=
Vm

1
Calculation
Area o f the loop
=
m
2
Energy loss =
PROCEDURE
The unit one to force the B

H loop (hysteresis) of a ferromagnetic specimen using a CRO is
shown in figure. A measurement of the area of the loop leads to the evaluation of the energy loss in
the specimen. The top view of the unit is shown in the figure. T
here are 12 terminals on the
panel;
sin patch cords are supplied with the kit.
The value of R1 can be selected connecting terminal D to
B or C (A

D = 50 ohm; B

D = 150
ohm; C

D = 50 ohm)
A is connected to D. The primary terminals of the specimen is connec
ted to p, p secondary to s, s
terminals. The CRO is calibrated as per the instructions given the Instruction Manual of CRO. CRO
is adjusted to work on external mode (the time base is switched off). The horizontal and vertical
position controls are adjusted
such that the spot is at the centre of
the CRO screen.
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The terminal marked GND is connected to the ground of the CRO. The H is
connected to
the Horizontal input
to
the ground
of the CRO. The H is connected to the Horizontal input of the
CRO. The
terminals V are
switched on. The hysteresis loop is formal. The horizontal and vertical
gains are adjusted such that the loop occupies maximum area of the screen of the CRO. Once this
adjustment is made, the gain controls should not
be disturbed
. The loop
is traced on a translucent
graph paper. The area of the loop is estimated.
The
connections from CRO are
removed without disturbing the horizontal and vertical gain
controls. The vertical sensitivity of the CRO is determined by applying a known A.C voltage
say 1
volt (peak to peak).
If the spot deflects by
X
cms for 1 volt, the vertical sensitivity
is 1
/
X x
10

2
(volt/m). Let it be
dV. The horizontal sensibility of CRO is determined by
applying a known A.C. voltage s
a
y
1 volt
(peak to peak). Let
the horizo
ntal sensitivity be S
H
(volt/m).
The transformer core may be replaced by ferrite ring and hysteresis loss in ferrite core can
be found.
WORK SHEET
CALCULATIONS:
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
59
Result:
Energy loss = ……………………………. Joules cycle

1
m

3
.
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Fig.
19
Non
–
Uniform Bending
Model Graph
S.No
Distance
between
Knife
edges (l)
Load
(M)
Microscope Reading
Mean
Depress
ion (y)
M/y
Increasing Load
Decreasing Load
MSR
VSC
TR
MSR
VSC
TR
Unit
X 10

2
m
X 10

3
Kg
X 10

2
m
Div
X 10

2
m
X 10

2
m
Div
X 10

2
m
X 10

2
m
X 10

2
m
X 10

1
Kg m

1
1
W
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
61
To find M/y
LC = 0.001 cm
TR = MSR +(VSCXLC)
Mean
M/y
=
……..Kg/m
Expt. No.:
Date:
7
.
DETERNINATION OF YOUNG’S MODULUS OF THE MATERIAL
–
NON UNIFORM BENDING
Aim
To determine the young’s modulus of the given material of the beam by Non Uniform
bending
.
Apparatus requited
A long uniform beam usually a meter scale, travelling microscope, pin, weight hanger
with slotted
weights, vernier calipers, Screw gauge, knife edges etc.,
Formula
The Young’s Modulus of
the given material of the beam
By Calculation method
Nm

2
By Graphical method
Nm

2
Explanation of symbols
Symbol
Explanation
Unit
g
Acceleration due to gravity
m/s
2
l
Distance between Knife
edge
m
eter
b
Breadth of the beam
meter
2
W+50
3
W+100
4
W+150
5
W+200
6
W+250
PHYSICS PRACTICAL MANUAL CUM RECORD
2010

2011
62
d
Thickness of the beam
meter
y
Depression produced for ‘M’ kg of load
meter
M
Load
appl
i
ed
Kg
k
Slope 1/k from graph
Kg m

1
By Graphical method
t
o find h/M
AXIS
W

(W+50)
W

(W+100)
W

(W+150)
W

(W+200)
W

(W+250)
X (M)
Y ( h)
To find Breadth (b) of the beam using vernier
calipers
LC
=
0.01cm
ZE = ………….
ZC
= …………. x 10

2
m
S.No.
Main Scale
Reading (MSR)
Vernier Scale
Coincidence (VSC)
Observed Reading
OR=MSR+(VSCXLC)
Correct Reading CR
= OR±ZC
Unit
X 10

2
m
Div.
X 10

2
m
X 10

2
m
1
2
3
4
5
Mean
‘b’ =
……….. x 10

2
m
To find the
thickness (
d
) of
the beam us
ing screw g
a
u
ge
.
LC = 0.01 mm
ZE = ………….
ZC = …………. x 10

3
m
S.No.
Pitch Scale
Reading (PSR)
Head Scale
Coincidence (HSC)
Observed Reading
OR=PSR+(HSCXLC)
Correct Reading CR
= OR±ZC
PHYSICS PRACTICAL MANUAL CUM RECORD
I & II Sem
63
Unit
X 10

3
m
Div.
X 10

3
m
X 10

3
m
1
2
3
4
5
Mean‘d’
= ……….. x 10

3
m
Procedure
The given beam is placed on the two knife edges (A and B) at a distance say 70 cm or 80
cm. A weight hanger is suspended at the centre
(C
) of the beam and a pin is fixed
vertically on
the
frame
of the hanger as shown in fig.19
. Taking the weigh
t
hanger alone as the dead load the tip o
the pin is focused by the microscope, and is adjusted in such a way that the tip of the pin just
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