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5. DETERMINATION OF COEFFICIENT OF VISCOSITY OF THE GIVEN
LIQUID OF USING POISEUILLE’S FLOW METHOD
AIM
To determine the coefficient of viscosity of the given liquid by Poiseuille’s flow method
APPARATUS REQUIRED:
Burette
Burette stand
Capillary tube
Beaker
Stop clock
Travelling microscope
Liquid
Rubber tube
3
DETERMINATION OF COEFFICIENT OF VISCOSITY OF THE GIVEN LIQUID OF
USING POISEUILLE’S FLOW METHOD
FORMULA
DESCRIPTION OF FORMULA:
4
2
The coefficient of viscosity of
ga ht
Ns
the liqu
v
id
m
8


 

2
2
2
2
densityof the liquid used.
g acceleration due to gravity.(ms )
a radius of capillary tube.( 10 m)
h mean press
t time taken for 5 cc flow of water.(seco
ure head
nd)
length of
.( 1
the capillary tube
0 m)
( 10 m)
.






 

 
 


6
v v olume of the liquid flow ( 10 m)

 
4
MEASUREMENT OF TIME OF LIQUID FLOW
Height of Capillary tube from Table h0
S.No Water
level in the
Burette
(cc)
Time
(t)
(second)
Range
(cc)
Time for
flow of 10
cc liquid
(second)
Height of
initial water
level from
table (h1) x
10-2 m
Height of
initial water
level from
table (h2) x
10-2 m
Pressure head
ht x 10-2 m
1 0 0 0-5 13 76.2 71.2 64.7 841.1
2 5 13 5-10 13 71.2 66.2 59.7 716.1
3 10 26 10-15 16 66.2 61.2 54.7 875.2
4 15 42 15-20 20 61.2 56.1 49.7 994
5 20 62 20-25 23 56.2 51.2 44.7 1028.1
6 25 85 25-30 25 51.2 46.2 39.7 992.5
7 30 110 30-35 32 46.2 41.2 34.7 1110.4
8 35 142 35-40 46 41.2 36.2 29.7 1366.2
9 40 188 40-45
990.5X10-2 m
OBSERVATION:
1 2
3
2

 
h h
h h
5
DETERMINATION OF COEFFICIENT OF VISCOSITY OF THE
GIVEN LIQUID OF USING POISEUILLE’S FLOW METHOD
CALCULATION:
RESULT
The Coefficient of viscosity of the liquid ( )= 0.0022 Nsm-2
 
 
 
 
 








 
 
 

   

 

2 2
2 6
3 14 2
4
2
4
13
8
2
6
5
2
3.14 x 1000 x9.8x 0.05x10 x 990.5 x 10
8x21.4x10 x5x10
3.14 x 10 x9
ga ht
Nsm
8 v
( ( )
( (6 )
190497 10
222.54 10
85
.8x .25x10 x 990.95 x 10
8x21.
6 10
0.00
4x10 x5x10
22 Nsm
2

6
4. DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD
CONDUCTOR USING LEE’S DISC METHOD
AIM
To determine the thermal conductivity of a bad conductor using Lee’s disc method
APPARATUS REQUIRED
Lee’s disc apparatus
Steam boiler
Thermometers
Bad conductor
Screw gauge
Stop clock
7
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR
USING LEE’S DISC METHOD
1 1
2
1 2
MSRd(r 2h)
K Wm K
r ( )(2r 2h)
 


    
 
 
3
1 1
2
M Mass of metallic brass disc 10 kg
S Specific heat capacity of the metallic disc J Kg K
d
R Rate of cooling at stady state temperature second calculated from graph .
dt
d
( )
( )
( )
Change in tempera

 
 


  
  o
3
2
o
1
2
ture C
dt Change in time second
d Thickness of the bad conductor disc 10 m
r Radius of the metallic disc 10 m
Steady state temperature of the stream chamber C
Stea
( )
( )
dy s
( )
(
tate temperatur
)
( )
e



 
 
 
  o
3
of the metallic disc C
h Thickness of the metallic disc 10
)
( )
m
(

 
8
To find the thickness of the brass disc (h) using screw gauge
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR
USING LEE’S DISC METHOD
ZE = -2 division
LC = 0.01 mm ZC = 0.02 mm
S.No PSR
mm
HSC
division
1. 10 89 10.89 10.91
2. 10 92 10.92 10.94
3. 10 88 10.88 10.90
4. 10 90 10.90 10.92
5. 10 91 10.91 10.93
Mean 10.92X10-3m
9
To find the thickness of the cardboard (h) using screw gauge
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR
USING LEE’S DISC METHOD
ZE = -2 division
LC = 0.01 mm ZC = +0.02 mm
S.No PSR
mm
HSC
division
1. 1 53 1.53 1.55
2. 1 54 1.54 1.56
3. 1 55 1.55 1.57
4. 1 54 1.54 1.56
5. 1 53 1.53 1.55
Mean 1.558X10-3m
10
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING
LEE’S DISC METHOD
Temperature
(oC)
Time
(Second)
67 27
66 74
65 113
64 148
63 193
62 242
61 292
60 332
59 383
58 434
OBSERVATION:


     

  

1 2
d
To find the rate of cooling R
dt
98 C 63 C
64 63 1
Slope 0.02
193 148 45
11
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR
USING LEE’S DISC METHOD
3
3
3
2
Mass of the brass disc M) 810 10 Kg
Thickness of the brass disc (h) 10.92 10 m
Thickness of the bad conductor disc (d) 1.558 10 m
Radius of the bad conductor disc (r) 5.5 10 m
Specific heat capaci




 
 
 
 
1 1
o
1
o
2
ty of the material of the brass disc (S) 372 J Kg K
Steady state temperature of the steam chamber ( ) 98 C
Steady state temperature of the brass disc( ) 63
Rate of cooling at the steady state temp
 

 
 
o
erature R 0.021 C / Sec

12
CALCULATION:
RESULT
DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC
METHOD
Thermal conductivity of bad conductor using Lee's disc method
 
     
 
   
3 3 2 3
2 2 3
3 3 3 3
2
2
2
810x 10 x 0.021x 372 x1.558x10 x 5.5 x10 21.84x10
K
3.14 x 5.5x10 98 63 x 11 x10 21.84x 10
810x 10 x 0.021x 372 x1.558x10 x 55 x10 21.84x10
K
3.14 x 5.5x10 9
(
(
8 63 x 1
   
  
   

 


 




 
 
  
3 3
9
7
2 1 1
10 x10 21.84x 10
810x0.021x 372 x1.558 x76 .84x10
K
3.14 x 30.25 35x131 .84x10
K 1.728 x 10 W m K
 


  



 
 
 

2 1 1
K 1.728 x 10 W K
  
 m
13
3. ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN
A LIQUID AND COMPRESSIBILITY OF THE LIQUID
AIM:
(i)To determine the velocity of Ultrasonic waves in the given liquid by using
ultrasonic Interferometer.
(ii)To determine the compressibility of the liquid.
APPARATUS REQUIRED:
Ultrasonic Interferometer
Measuring cell
Frequency generator
Liquid
14
3.ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN
A LIQUID AND COMPRESSIBILITY OF THE LIQUID
15
FORMULA:
DESCRIPTION OF FORMULA:
ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC
WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID
 
 
 
 

 
 


2 1
2
m
Velocity of Ultrasonic waves in the liquid (v) n
s
2d
where, Wavelength of ultrasonic wave (m)
x
1
Compressibility of the given liquid K m N
v
 
 
3
n Frequency of the generator Hertz
Wavelength of the ultrasonic wave metre
Density of the liquid (kg m )
d Distance moved by the micrometer screw metre
x Number of oscillatio
(
ns No un t
)
)
i
(


 
 


16
S.No.
No. of
Oscillations
(x)
(No Unit)
Readings for "x" Oscillations
Distance
Wavelength Velocity
Initial Reading
PSR
x 10-3m
HSC
Div
TR
x 10-3m
1 n 21 0 21
2 n+5 19 47 19.47 1.53 0.612 1.22
3 n+10 17 27 17.27 2.2 0.88 1.76
4 n+15 15 25 15.35 1.92 0.768 1.536
5 n+20 13.5 46 13.96 1.39 0.556 1.12
6 n+25 11.5 6 11.56 2.4 0.96 1.92
Mean 1.51X103 m/s
ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC
WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID
OBSERVATION:
3
2d
x
10 m




3
v n
m
10
s






1 2
3
d d
10 m
17
CALCULATION:
RESULT
The velocity of Ultrasonic waves in the liquid =
Compressibility of the liquid =
ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC
WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID
 
 
 

 

 
 
 
 
 
   


  

1
6 3 3 1
2 1
2
2
10 2 1
9
3
Velocity of Ultrasonic waves in the liquid (v) n ms
v n 2 x 10 x 0.612 x 10 1.224x10 ms
1
Compressibility of the given liquid K m N
v
1 1
K 4.385x10 m N
2.28 10
1.51x10 x 1000

3 1
1.51x10 ms
 
10 2 1
4.385x10 m N
AIM:
To determine the wavelength of the given laser source using diffraction grating.
APPARATUS REQUIRED
Laser Source
Scale
Diffraction grating
Screen
1. DETERMINATION OF LASER PARAMETERS – WAVELENGTH
19
FORMULA:
DESCRIPTION OF FORMULA:
DETERMINATION OF LASER PARAMETERS – WAVELENGTH
1
sin
Wavelength of the laser m
Nm
d
tan
D
d
tan Degree
D


 
 
 
   
 
 
 
th
th th
N number of lines per meter on the grating lines / m
m order of diffraction No unit
angle ofm ^ thorder spectrum from zero order degrees
d distance of m order spectrum from zero order.
D distance be
( )




 2
tween grating and the screen. )
10 m
( 

20
'
DETERMINATION OF LASER PARAMETERS – WAVELENGTH
OBSERVATION:
S. No. Order of
diffraction
‘m’
(No unit) Left Right Mean
1 18 1 1.3 1.3 1.3 4.1308 7318
2 2.4 2.4 2.4 7.594 6503
3 3.6 3.6 3.6 11.309 6433
2 22 1 1.5 1.5 1.5 3.9004 6911
2 3 3 3 7.765 6649
3 4.5 4.5 4.5 11.560 6574
3 26 1 1.7 1.7 1.7 4.4186 7827
2 3.7 3.7 3.7 8.099 6933
3 5.5 5.5 5.5 11.944 6789
Mean λ 6881 x 10-10 m
21
CALCULATION:
DETERMINATION OF LASER PARAMETERS – WAVELENGTH
RESULT
The wavelength of the laser source = 6881 x 10-10 m




 
 
 
   
 
 
  
 
 
   

1
1
10
sin
Wavelength of the laser m
Nm
d
tan
D
d
tan
D
1.3
tan 4.130 Degree
18
sin 4.130
7318 10 m
1 98425
AIM:
To determine
(i) The acceptance angle of the optical fibre.
(ii) The numerical aperture.
APPARATUS REQUIRED
Semiconductor laser
optical fibre
NA jig
2. DETERMINATION OF ACCEPTANCE ANGLE OF
AN OPTICAL FIBRE AND NUMERICAL APERTURE
23
FORMULA:
DESCRIPTION OF FORMULA:
DETERMINATION OF ACCEPTANCE ANGLE OF AN
OPTICAL FIBRE AND NUMERICAL APERTURE
 
1 r
i Acceptance angle of optical fibre tan degree
d
(ii Numerical Aperture NA sin No u
) nit
( )   
   
 
 
2
2
r radius of the circular opening in NA jig 10 m
d Distance between the tip of the optical fibre and aperture of the j
( )
(
ig 10 m)


 
 
24
'
DETERMINATION OF ACCEPTANCE ANGLE OF AN OPTICAL
FIBRE AND NUMERICAL APERTURE
OBSERVATION:
S.No
1. 0.4 0.2 26°56’
2. 0.8 0.5 32°00’
3. 1.1 0.6 29°01’
4. 1.6 1.1 34°50’
5. 2 1.3 33°02’
Mean 31°09’
25
CALCULATION:
DETERMINATION OF ACCEPTANCE ANGLE OF AN OPTICAL
FIBRE AND NUMERICAL APERTURE
RESULT (i) The acceptance angle of the optical fibre = 31°09’
(ii) Numerical Aperture = 0.51 (No unit)
1
1
'
'
r
tan degree
d
NA Sin no unit
0.3
tan degree
0.5
31 36
NA sin No unit)
NA sin31 36
NA 0.52 no unit
(


 
   
 
 
 
   
 
  
 
 


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physics lab manual.pdf

  • 1.
  • 2. 2 5. DETERMINATION OF COEFFICIENT OF VISCOSITY OF THE GIVEN LIQUID OF USING POISEUILLE’S FLOW METHOD AIM To determine the coefficient of viscosity of the given liquid by Poiseuille’s flow method APPARATUS REQUIRED: Burette Burette stand Capillary tube Beaker Stop clock Travelling microscope Liquid Rubber tube
  • 3. 3 DETERMINATION OF COEFFICIENT OF VISCOSITY OF THE GIVEN LIQUID OF USING POISEUILLE’S FLOW METHOD FORMULA DESCRIPTION OF FORMULA: 4 2 The coefficient of viscosity of ga ht Ns the liqu v id m 8      2 2 2 2 densityof the liquid used. g acceleration due to gravity.(ms ) a radius of capillary tube.( 10 m) h mean press t time taken for 5 cc flow of water.(seco ure head nd) length of .( 1 the capillary tube 0 m) ( 10 m) .                6 v v olume of the liquid flow ( 10 m)   
  • 4. 4 MEASUREMENT OF TIME OF LIQUID FLOW Height of Capillary tube from Table h0 S.No Water level in the Burette (cc) Time (t) (second) Range (cc) Time for flow of 10 cc liquid (second) Height of initial water level from table (h1) x 10-2 m Height of initial water level from table (h2) x 10-2 m Pressure head ht x 10-2 m 1 0 0 0-5 13 76.2 71.2 64.7 841.1 2 5 13 5-10 13 71.2 66.2 59.7 716.1 3 10 26 10-15 16 66.2 61.2 54.7 875.2 4 15 42 15-20 20 61.2 56.1 49.7 994 5 20 62 20-25 23 56.2 51.2 44.7 1028.1 6 25 85 25-30 25 51.2 46.2 39.7 992.5 7 30 110 30-35 32 46.2 41.2 34.7 1110.4 8 35 142 35-40 46 41.2 36.2 29.7 1366.2 9 40 188 40-45 990.5X10-2 m OBSERVATION: 1 2 3 2    h h h h
  • 5. 5 DETERMINATION OF COEFFICIENT OF VISCOSITY OF THE GIVEN LIQUID OF USING POISEUILLE’S FLOW METHOD CALCULATION: RESULT The Coefficient of viscosity of the liquid ( )= 0.0022 Nsm-2                                  2 2 2 6 3 14 2 4 2 4 13 8 2 6 5 2 3.14 x 1000 x9.8x 0.05x10 x 990.5 x 10 8x21.4x10 x5x10 3.14 x 10 x9 ga ht Nsm 8 v ( ( ) ( (6 ) 190497 10 222.54 10 85 .8x .25x10 x 990.95 x 10 8x21. 6 10 0.00 4x10 x5x10 22 Nsm 2 
  • 6. 6 4. DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD AIM To determine the thermal conductivity of a bad conductor using Lee’s disc method APPARATUS REQUIRED Lee’s disc apparatus Steam boiler Thermometers Bad conductor Screw gauge Stop clock
  • 7. 7 DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD 1 1 2 1 2 MSRd(r 2h) K Wm K r ( )(2r 2h)              3 1 1 2 M Mass of metallic brass disc 10 kg S Specific heat capacity of the metallic disc J Kg K d R Rate of cooling at stady state temperature second calculated from graph . dt d ( ) ( ) ( ) Change in tempera             o 3 2 o 1 2 ture C dt Change in time second d Thickness of the bad conductor disc 10 m r Radius of the metallic disc 10 m Steady state temperature of the stream chamber C Stea ( ) ( ) dy s ( ) ( tate temperatur ) ( ) e            o 3 of the metallic disc C h Thickness of the metallic disc 10 ) ( ) m (   
  • 8. 8 To find the thickness of the brass disc (h) using screw gauge DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD ZE = -2 division LC = 0.01 mm ZC = 0.02 mm S.No PSR mm HSC division 1. 10 89 10.89 10.91 2. 10 92 10.92 10.94 3. 10 88 10.88 10.90 4. 10 90 10.90 10.92 5. 10 91 10.91 10.93 Mean 10.92X10-3m
  • 9. 9 To find the thickness of the cardboard (h) using screw gauge DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD ZE = -2 division LC = 0.01 mm ZC = +0.02 mm S.No PSR mm HSC division 1. 1 53 1.53 1.55 2. 1 54 1.54 1.56 3. 1 55 1.55 1.57 4. 1 54 1.54 1.56 5. 1 53 1.53 1.55 Mean 1.558X10-3m
  • 10. 10 DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD Temperature (oC) Time (Second) 67 27 66 74 65 113 64 148 63 193 62 242 61 292 60 332 59 383 58 434 OBSERVATION:              1 2 d To find the rate of cooling R dt 98 C 63 C 64 63 1 Slope 0.02 193 148 45
  • 11. 11 DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD 3 3 3 2 Mass of the brass disc M) 810 10 Kg Thickness of the brass disc (h) 10.92 10 m Thickness of the bad conductor disc (d) 1.558 10 m Radius of the bad conductor disc (r) 5.5 10 m Specific heat capaci             1 1 o 1 o 2 ty of the material of the brass disc (S) 372 J Kg K Steady state temperature of the steam chamber ( ) 98 C Steady state temperature of the brass disc( ) 63 Rate of cooling at the steady state temp        o erature R 0.021 C / Sec 
  • 12. 12 CALCULATION: RESULT DETERMINATION OF THERMAL CONDUCTIVITY OF A BAD CONDUCTOR USING LEE’S DISC METHOD Thermal conductivity of bad conductor using Lee's disc method               3 3 2 3 2 2 3 3 3 3 3 2 2 2 810x 10 x 0.021x 372 x1.558x10 x 5.5 x10 21.84x10 K 3.14 x 5.5x10 98 63 x 11 x10 21.84x 10 810x 10 x 0.021x 372 x1.558x10 x 55 x10 21.84x10 K 3.14 x 5.5x10 9 ( ( 8 63 x 1                              3 3 9 7 2 1 1 10 x10 21.84x 10 810x0.021x 372 x1.558 x76 .84x10 K 3.14 x 30.25 35x131 .84x10 K 1.728 x 10 W m K                  2 1 1 K 1.728 x 10 W K     m
  • 13. 13 3. ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID AIM: (i)To determine the velocity of Ultrasonic waves in the given liquid by using ultrasonic Interferometer. (ii)To determine the compressibility of the liquid. APPARATUS REQUIRED: Ultrasonic Interferometer Measuring cell Frequency generator Liquid
  • 14. 14 3.ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID
  • 15. 15 FORMULA: DESCRIPTION OF FORMULA: ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID                2 1 2 m Velocity of Ultrasonic waves in the liquid (v) n s 2d where, Wavelength of ultrasonic wave (m) x 1 Compressibility of the given liquid K m N v     3 n Frequency of the generator Hertz Wavelength of the ultrasonic wave metre Density of the liquid (kg m ) d Distance moved by the micrometer screw metre x Number of oscillatio ( ns No un t ) ) i (        
  • 16. 16 S.No. No. of Oscillations (x) (No Unit) Readings for "x" Oscillations Distance Wavelength Velocity Initial Reading PSR x 10-3m HSC Div TR x 10-3m 1 n 21 0 21 2 n+5 19 47 19.47 1.53 0.612 1.22 3 n+10 17 27 17.27 2.2 0.88 1.76 4 n+15 15 25 15.35 1.92 0.768 1.536 5 n+20 13.5 46 13.96 1.39 0.556 1.12 6 n+25 11.5 6 11.56 2.4 0.96 1.92 Mean 1.51X103 m/s ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID OBSERVATION: 3 2d x 10 m     3 v n m 10 s       1 2 3 d d 10 m
  • 17. 17 CALCULATION: RESULT The velocity of Ultrasonic waves in the liquid = Compressibility of the liquid = ULTRASONIC INTERFEROMETER – VELOCITY OF ULTRASONIC WAVES IN A LIQUID AND COMPRESSIBILITY OF THE LIQUID                               1 6 3 3 1 2 1 2 2 10 2 1 9 3 Velocity of Ultrasonic waves in the liquid (v) n ms v n 2 x 10 x 0.612 x 10 1.224x10 ms 1 Compressibility of the given liquid K m N v 1 1 K 4.385x10 m N 2.28 10 1.51x10 x 1000  3 1 1.51x10 ms   10 2 1 4.385x10 m N
  • 18. AIM: To determine the wavelength of the given laser source using diffraction grating. APPARATUS REQUIRED Laser Source Scale Diffraction grating Screen 1. DETERMINATION OF LASER PARAMETERS – WAVELENGTH
  • 19. 19 FORMULA: DESCRIPTION OF FORMULA: DETERMINATION OF LASER PARAMETERS – WAVELENGTH 1 sin Wavelength of the laser m Nm d tan D d tan Degree D                   th th th N number of lines per meter on the grating lines / m m order of diffraction No unit angle ofm ^ thorder spectrum from zero order degrees d distance of m order spectrum from zero order. D distance be ( )      2 tween grating and the screen. ) 10 m (  
  • 20. 20 ' DETERMINATION OF LASER PARAMETERS – WAVELENGTH OBSERVATION: S. No. Order of diffraction ‘m’ (No unit) Left Right Mean 1 18 1 1.3 1.3 1.3 4.1308 7318 2 2.4 2.4 2.4 7.594 6503 3 3.6 3.6 3.6 11.309 6433 2 22 1 1.5 1.5 1.5 3.9004 6911 2 3 3 3 7.765 6649 3 4.5 4.5 4.5 11.560 6574 3 26 1 1.7 1.7 1.7 4.4186 7827 2 3.7 3.7 3.7 8.099 6933 3 5.5 5.5 5.5 11.944 6789 Mean λ 6881 x 10-10 m
  • 21. 21 CALCULATION: DETERMINATION OF LASER PARAMETERS – WAVELENGTH RESULT The wavelength of the laser source = 6881 x 10-10 m                               1 1 10 sin Wavelength of the laser m Nm d tan D d tan D 1.3 tan 4.130 Degree 18 sin 4.130 7318 10 m 1 98425
  • 22. AIM: To determine (i) The acceptance angle of the optical fibre. (ii) The numerical aperture. APPARATUS REQUIRED Semiconductor laser optical fibre NA jig 2. DETERMINATION OF ACCEPTANCE ANGLE OF AN OPTICAL FIBRE AND NUMERICAL APERTURE
  • 23. 23 FORMULA: DESCRIPTION OF FORMULA: DETERMINATION OF ACCEPTANCE ANGLE OF AN OPTICAL FIBRE AND NUMERICAL APERTURE   1 r i Acceptance angle of optical fibre tan degree d (ii Numerical Aperture NA sin No u ) nit ( )            2 2 r radius of the circular opening in NA jig 10 m d Distance between the tip of the optical fibre and aperture of the j ( ) ( ig 10 m)      
  • 24. 24 ' DETERMINATION OF ACCEPTANCE ANGLE OF AN OPTICAL FIBRE AND NUMERICAL APERTURE OBSERVATION: S.No 1. 0.4 0.2 26°56’ 2. 0.8 0.5 32°00’ 3. 1.1 0.6 29°01’ 4. 1.6 1.1 34°50’ 5. 2 1.3 33°02’ Mean 31°09’
  • 25. 25 CALCULATION: DETERMINATION OF ACCEPTANCE ANGLE OF AN OPTICAL FIBRE AND NUMERICAL APERTURE RESULT (i) The acceptance angle of the optical fibre = 31°09’ (ii) Numerical Aperture = 0.51 (No unit) 1 1 ' ' r tan degree d NA Sin no unit 0.3 tan degree 0.5 31 36 NA sin No unit) NA sin31 36 NA 0.52 no unit (                            