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# Solution of Wave Optics Test 03 Class 12 Notes | EduRev

## Class 12 : Solution of Wave Optics Test 03 Class 12 Notes | EduRev

``` Page 1

CBSE TEST PAPER-03
CLASS - XII PHYSICS (Wave Optics)
Ans1: (1) They must emit waves continuously of same wavelengths.
(2) The phase difference between the waves must be zero or constant
Ans2:   '
2
D d
when d
d
?
ß = =
2
'
D
d
?
ß ? =
Ans3:  I = a
2
+ b
2
+ 2ab cos ?
Where a and b are amplitudes of two coherent waves having phase difference of? .
Here a
2
= I , b
2
= 4I
I = I + 4I + 2 4 cos I I ?
I = 5I + 4I cos ?
(i)
2
When
p
? =
5 4 cos
2
I I I
p
= +
(ii) Why ? p =
I = 5I + 4I cos ?
I = 5I – 4I
Ans4:  White light produces interference but due to different colour present in white light
interference pattern overlaps the central bright fringe for all the colours is at the
position, so its colour is white. The white central bright fringe is surrounded by few
coloured rings.
Ans5:
7
5000 5 10
o
A m ?
- = = ×
2
0.6 0.6 10 cm m
D
d
ß
?
ß
- = = ×
=
' 2 ß ß =
I = 5I
I = I
Page 2

CBSE TEST PAPER-03
CLASS - XII PHYSICS (Wave Optics)
Ans1: (1) They must emit waves continuously of same wavelengths.
(2) The phase difference between the waves must be zero or constant
Ans2:   '
2
D d
when d
d
?
ß = =
2
'
D
d
?
ß ? =
Ans3:  I = a
2
+ b
2
+ 2ab cos ?
Where a and b are amplitudes of two coherent waves having phase difference of? .
Here a
2
= I , b
2
= 4I
I = I + 4I + 2 4 cos I I ?
I = 5I + 4I cos ?
(i)
2
When
p
? =
5 4 cos
2
I I I
p
= +
(ii) Why ? p =
I = 5I + 4I cos ?
I = 5I – 4I
Ans4:  White light produces interference but due to different colour present in white light
interference pattern overlaps the central bright fringe for all the colours is at the
position, so its colour is white. The white central bright fringe is surrounded by few
coloured rings.
Ans5:
7
5000 5 10
o
A m ?
- = = ×
2
0.6 0.6 10 cm m
D
d
ß
?
ß
- = = ×
=
' 2 ß ß =
I = 5I
I = I
2
7
4
0.6 10
5 10
1.2 10 (1)
D D
d d
D
d
ß
?
- - ×
= ? =
×
= × - - - - - New Distance '
' '
New fringe width '
2
D
D
d
D D
d d
? ?
ß
=
= =
7 4
5 10 1.2 10
'
2
ß
- × × ×
=
Ans6:  The phenomenon of restricting the vibrations of a light vector in a particular direction in
a  plane perpendicular to the direction of propagation of light is called polarisation of
light.
Transverse waves show the property of polarisaiton.
Two methods to produce plane polarised light
(1) Polarisation by Reflection
(2) Polarization by scattering
Ans7:
Conditions for sustained interference of light
(1) Two sources must be coherent sources of light.
(2)Two sources should exist light waves continuously. Intensity mono
Ans8: Path difference between
S 1P and S 2P
2 1
2
( )

x S P S P A
In S BP
? = - - - - - - ?
( )
1
2 2 2
2
2
( ) ( ) S P S B PB ? ? = +
? ?
3
' 3 10 m ß
- = ×
Page 3

CBSE TEST PAPER-03
CLASS - XII PHYSICS (Wave Optics)
Ans1: (1) They must emit waves continuously of same wavelengths.
(2) The phase difference between the waves must be zero or constant
Ans2:   '
2
D d
when d
d
?
ß = =
2
'
D
d
?
ß ? =
Ans3:  I = a
2
+ b
2
+ 2ab cos ?
Where a and b are amplitudes of two coherent waves having phase difference of? .
Here a
2
= I , b
2
= 4I
I = I + 4I + 2 4 cos I I ?
I = 5I + 4I cos ?
(i)
2
When
p
? =
5 4 cos
2
I I I
p
= +
(ii) Why ? p =
I = 5I + 4I cos ?
I = 5I – 4I
Ans4:  White light produces interference but due to different colour present in white light
interference pattern overlaps the central bright fringe for all the colours is at the
position, so its colour is white. The white central bright fringe is surrounded by few
coloured rings.
Ans5:
7
5000 5 10
o
A m ?
- = = ×
2
0.6 0.6 10 cm m
D
d
ß
?
ß
- = = ×
=
' 2 ß ß =
I = 5I
I = I
2
7
4
0.6 10
5 10
1.2 10 (1)
D D
d d
D
d
ß
?
- - ×
= ? =
×
= × - - - - - New Distance '
' '
New fringe width '
2
D
D
d
D D
d d
? ?
ß
=
= =
7 4
5 10 1.2 10
'
2
ß
- × × ×
=
Ans6:  The phenomenon of restricting the vibrations of a light vector in a particular direction in
a  plane perpendicular to the direction of propagation of light is called polarisation of
light.
Transverse waves show the property of polarisaiton.
Two methods to produce plane polarised light
(1) Polarisation by Reflection
(2) Polarization by scattering
Ans7:
Conditions for sustained interference of light
(1) Two sources must be coherent sources of light.
(2)Two sources should exist light waves continuously. Intensity mono
Ans8: Path difference between
S 1P and S 2P
2 1
2
( )

x S P S P A
In S BP
? = - - - - - - ?
( )
1
2 2 2
2
2
( ) ( ) S P S B PB ? ? = +
? ?
3
' 3 10 m ß
- = ×
1
2
2 2
2
1 (1)
d
y
S P D
D
? ? ? ?
+
? ? ? ?
? ?
? ? = + - - - - - ? ?
? ?
? ?

Using Binomial theorem expand equation. (1) and neglect higher terms
2
2
2
2
d
y
S P D
D
? ?
+
? ?
? ?
= +
2
1
2
S P = D + (2)
2
Substituting equation (1) & (2) in equation (A)
d
y
Similarty
D
? ?
- ? ?
? ?
- - - - - 2 2
2 2
x =
2
d d
y y
D
? ? ? ?
+ - - ? ? ? ?
? ? ? ?
?
2 2
2 2
4 4
x =
2
d d
y yd y yd
D
+ + - - +
?
2
x =
2
yd
D
?
For bright fringes
Path difference = x?
.
yd
x
D
x D
i e y
d
?
?
=
=
1
2
?D
form =1   y =
d
?D
n = 2   y =
d
For fringe width
2
1
y y ß = - x =
yd
D
?
d
d
?
ß =
Page 4

CBSE TEST PAPER-03
CLASS - XII PHYSICS (Wave Optics)
Ans1: (1) They must emit waves continuously of same wavelengths.
(2) The phase difference between the waves must be zero or constant
Ans2:   '
2
D d
when d
d
?
ß = =
2
'
D
d
?
ß ? =
Ans3:  I = a
2
+ b
2
+ 2ab cos ?
Where a and b are amplitudes of two coherent waves having phase difference of? .
Here a
2
= I , b
2
= 4I
I = I + 4I + 2 4 cos I I ?
I = 5I + 4I cos ?
(i)
2
When
p
? =
5 4 cos
2
I I I
p
= +
(ii) Why ? p =
I = 5I + 4I cos ?
I = 5I – 4I
Ans4:  White light produces interference but due to different colour present in white light
interference pattern overlaps the central bright fringe for all the colours is at the
position, so its colour is white. The white central bright fringe is surrounded by few
coloured rings.
Ans5:
7
5000 5 10
o
A m ?
- = = ×
2
0.6 0.6 10 cm m
D
d
ß
?
ß
- = = ×
=
' 2 ß ß =
I = 5I
I = I
2
7
4
0.6 10
5 10
1.2 10 (1)
D D
d d
D
d
ß
?
- - ×
= ? =
×
= × - - - - - New Distance '
' '
New fringe width '
2
D
D
d
D D
d d
? ?
ß
=
= =
7 4
5 10 1.2 10
'
2
ß
- × × ×
=
Ans6:  The phenomenon of restricting the vibrations of a light vector in a particular direction in
a  plane perpendicular to the direction of propagation of light is called polarisation of
light.
Transverse waves show the property of polarisaiton.
Two methods to produce plane polarised light
(1) Polarisation by Reflection
(2) Polarization by scattering
Ans7:
Conditions for sustained interference of light
(1) Two sources must be coherent sources of light.
(2)Two sources should exist light waves continuously. Intensity mono
Ans8: Path difference between
S 1P and S 2P
2 1
2
( )

x S P S P A
In S BP
? = - - - - - - ?
( )
1
2 2 2
2
2
( ) ( ) S P S B PB ? ? = +
? ?
3
' 3 10 m ß
- = ×
1
2
2 2
2
1 (1)
d
y
S P D
D
? ? ? ?
+
? ? ? ?
? ?
? ? = + - - - - - ? ?
? ?
? ?

Using Binomial theorem expand equation. (1) and neglect higher terms
2
2
2
2
d
y
S P D
D
? ?
+
? ?
? ?
= +
2
1
2
S P = D + (2)
2
Substituting equation (1) & (2) in equation (A)
d
y
Similarty
D
? ?
- ? ?
? ?
- - - - - 2 2
2 2
x =
2
d d
y y
D
? ? ? ?
+ - - ? ? ? ?
? ? ? ?
?
2 2
2 2
4 4
x =
2
d d
y yd y yd
D
+ + - - +
?
2
x =
2
yd
D
?
For bright fringes
Path difference = x?
.
yd
x
D
x D
i e y
d
?
?
=
=
1
2
?D
form =1   y =
d
?D
n = 2   y =
d
For fringe width
2
1
y y ß = - x =
yd
D
?
d
d
?
ß =
(b)
2
1 1
2
2 2
4
1
a w
a w
= =
1
2
1 2
2
1
2
a
a
or a a
=
=
Using
( )
( )
2
1 2 max
2
min
1 2
I
I
a a
a a
+
=
- ( )
( )
2 2
1 2 max 2
2
min 2
1 2
2 I 3
I
2
a a a
a
a a
+ ? ?
= =
? ?
- ? ?
max
min
I 9
I 1
=
```
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