Chapter 28 - Interference and Diffraction of Light - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

DC Pandey (Questions & Solutions) of Physics: NEET

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NEET : Chapter 28 - Interference and Diffraction of Light - Physics, Solution by D C Pandey, NEET NEET Notes | EduRev

 Page 1


28 Interference and Diffraction
of Light
Introductory Exercise 28.1
1. Because they are incoherent ie, Df  does not
re main constant.
2. Since la ser is highly coherent and
mono chro matic source of light
3. I I =
0
2
2 cos / q
Þ 
3
4
2
0
0
2
I
I = cos / q
Þ cos / q 2
3
2
=
Þ 
q p
2 6
=
Þ q
p
=
3
Þ Dx = ´ = ´ =
l
l
f
l
p
p l
2 2 3 6
But Dx
yd
D
=
Þ y
D x
d
D
d
= = ´
D l
6
Þ y =
´ ´
´ ´
=
-
-
1.2 600 10
0 25 10 6
48
9
2
.
mm
4. 2
1
2
m l t n = -
æ
è
ç
ö
ø
÷
 for min i mum thick ness n = 1
Þ t = =
´
=
l
m 4
3
4 1.5
0.5cm
5. Here a a
1
3 = and a a
2
=
R a a a a
2 2 2
3 2 3 = + + ´ ´ ´ ( ) ( ) cos q
Þ I I I I = + + 9 6
0 0 0
cos q
Þ I I = + [ cos ] 10 6
0
q
I I = + ´ -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
10 6 2
2
1
2
0
cos
q
   = + -
é
ë
ê
ù
û
ú
10 12
2
6
2
0
cos
q
I
   = +
é
ë
ê
ù
û
ú
4 12
2
2
0
cos
q
I
    = +
é
ë
ê
ù
û
ú
4 1 3
2
0
2
I cos
q
Now, I
I
max
=
0
9
Þ          I I = +
é
ë
ê
ù
û
ú
4
9
1 3
2
2
max
cos
q
6. Dx
yd
D
t = - - ( ) m 1
if t =
-
l
m 2 1 ( )
Þ Dx
yd
D
= -
l
2
For maxima Dx n = l
Þ 
yd
D
n = +
æ
è
ç
ö
ø
÷
2
1
2
l
This become minima.
For minima Dx n = -
æ
è
ç
ö
ø
÷
1
2
l
          
yd
n
D
- = -
l
l
l
2 2
Þ 
yd
D
n - l this become maxima.
Hence maxima and minima are
interchanged.
Page 2


28 Interference and Diffraction
of Light
Introductory Exercise 28.1
1. Because they are incoherent ie, Df  does not
re main constant.
2. Since la ser is highly coherent and
mono chro matic source of light
3. I I =
0
2
2 cos / q
Þ 
3
4
2
0
0
2
I
I = cos / q
Þ cos / q 2
3
2
=
Þ 
q p
2 6
=
Þ q
p
=
3
Þ Dx = ´ = ´ =
l
l
f
l
p
p l
2 2 3 6
But Dx
yd
D
=
Þ y
D x
d
D
d
= = ´
D l
6
Þ y =
´ ´
´ ´
=
-
-
1.2 600 10
0 25 10 6
48
9
2
.
mm
4. 2
1
2
m l t n = -
æ
è
ç
ö
ø
÷
 for min i mum thick ness n = 1
Þ t = =
´
=
l
m 4
3
4 1.5
0.5cm
5. Here a a
1
3 = and a a
2
=
R a a a a
2 2 2
3 2 3 = + + ´ ´ ´ ( ) ( ) cos q
Þ I I I I = + + 9 6
0 0 0
cos q
Þ I I = + [ cos ] 10 6
0
q
I I = + ´ -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
10 6 2
2
1
2
0
cos
q
   = + -
é
ë
ê
ù
û
ú
10 12
2
6
2
0
cos
q
I
   = +
é
ë
ê
ù
û
ú
4 12
2
2
0
cos
q
I
    = +
é
ë
ê
ù
û
ú
4 1 3
2
0
2
I cos
q
Now, I
I
max
=
0
9
Þ          I I = +
é
ë
ê
ù
û
ú
4
9
1 3
2
2
max
cos
q
6. Dx
yd
D
t = - - ( ) m 1
if t =
-
l
m 2 1 ( )
Þ Dx
yd
D
= -
l
2
For maxima Dx n = l
Þ 
yd
D
n = +
æ
è
ç
ö
ø
÷
2
1
2
l
This become minima.
For minima Dx n = -
æ
è
ç
ö
ø
÷
1
2
l
          
yd
n
D
- = -
l
l
l
2 2
Þ 
yd
D
n - l this become maxima.
Hence maxima and minima are
interchanged.
7. For two slit experiment
d n sin q l =
Þ sin q
l
=
n
d
But sin q £ 1 Þ 
nl
d
£ 1
Þ       n
d
£
l
   Þ  n £
´
´
=
-
-
4 10
6 10
6
7
6.67
Þ         n = 6
8. Since am pli tude of each wave is equal. The
am pli tude of re sul tant wave is zero if waves
are equally dis placed in phase
ie, q =
°
= °
360
8
45
Hence phase difference must be 45°
AIEEE Corner
¢ Subjective Question (Level-1)
1. R a a a a
2
1
2
2
2
1 2
2 = + + f cos
(i) R a a a a = = = 2
1 2
,
4 2
2 2 2 2
a a a a = + + f cos
Þ  cos f = 1 Þ f = ° 0 
(ii) 2 2 2
2 2 2
a a a = + f cos
Þ f = ° 90
(iii) a a a
2 2 2
2 2 = + f cos
Þ cos f =
-1
2
 Þ f = ° 120
(iv) 0 2 2
2 2
= + f a a cos Þ f = ° 180
2.
I
I
a a
a a
a
a
a
a
max
min
( )
( )
=
+
-
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
1 2
2
1 2
2
1
2
2
1
2
1
1
ç
ö
ø
÷
÷
2
Þ 
I
I
max
min
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
ç
ö
ø
÷
÷
= =
5
3 1
5
3 1
8
2
16
2
2
2
2
    I I
max min
: : = 16 1
3. I I =
max
cos
2
2
q
Þ 
I
I
max
max
cos
2 2
2
=
q
Þ cos cos
q p
2 4
=
Þ       q
p
=
2
Now,        f = ´
2p
l
Dx
      
p p
l 2
2
= ´ Dx
Þ Dx =
l
4
But    Dx
yd
D
=
Þ y
D x
d
=
D
Þ y =
´
´
=
´ ´
´
-
-
1
4 1
1 500 10
4 10
9
3
m
mm
l
 
Þ y = ´
-
1.25 10
4
 m
4. I I =
max
cos
2
2
q
(a) 
I
I
max
max
cos
2 2
2
=
q
 Þ q
p
=
2
Now q
p
l
= ´
2
Dx
Þ Dx =
l
4
 and Dx
yd
D
= Þ =
D
d
l
4
(b) 
1
4 2
2
I I
max max
cos =
q
Þ cos cos
q p
2 3
= Þ q
p
=
2
3
f = ´
2p
l
Dx
 47
Page 3


28 Interference and Diffraction
of Light
Introductory Exercise 28.1
1. Because they are incoherent ie, Df  does not
re main constant.
2. Since la ser is highly coherent and
mono chro matic source of light
3. I I =
0
2
2 cos / q
Þ 
3
4
2
0
0
2
I
I = cos / q
Þ cos / q 2
3
2
=
Þ 
q p
2 6
=
Þ q
p
=
3
Þ Dx = ´ = ´ =
l
l
f
l
p
p l
2 2 3 6
But Dx
yd
D
=
Þ y
D x
d
D
d
= = ´
D l
6
Þ y =
´ ´
´ ´
=
-
-
1.2 600 10
0 25 10 6
48
9
2
.
mm
4. 2
1
2
m l t n = -
æ
è
ç
ö
ø
÷
 for min i mum thick ness n = 1
Þ t = =
´
=
l
m 4
3
4 1.5
0.5cm
5. Here a a
1
3 = and a a
2
=
R a a a a
2 2 2
3 2 3 = + + ´ ´ ´ ( ) ( ) cos q
Þ I I I I = + + 9 6
0 0 0
cos q
Þ I I = + [ cos ] 10 6
0
q
I I = + ´ -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
10 6 2
2
1
2
0
cos
q
   = + -
é
ë
ê
ù
û
ú
10 12
2
6
2
0
cos
q
I
   = +
é
ë
ê
ù
û
ú
4 12
2
2
0
cos
q
I
    = +
é
ë
ê
ù
û
ú
4 1 3
2
0
2
I cos
q
Now, I
I
max
=
0
9
Þ          I I = +
é
ë
ê
ù
û
ú
4
9
1 3
2
2
max
cos
q
6. Dx
yd
D
t = - - ( ) m 1
if t =
-
l
m 2 1 ( )
Þ Dx
yd
D
= -
l
2
For maxima Dx n = l
Þ 
yd
D
n = +
æ
è
ç
ö
ø
÷
2
1
2
l
This become minima.
For minima Dx n = -
æ
è
ç
ö
ø
÷
1
2
l
          
yd
n
D
- = -
l
l
l
2 2
Þ 
yd
D
n - l this become maxima.
Hence maxima and minima are
interchanged.
7. For two slit experiment
d n sin q l =
Þ sin q
l
=
n
d
But sin q £ 1 Þ 
nl
d
£ 1
Þ       n
d
£
l
   Þ  n £
´
´
=
-
-
4 10
6 10
6
7
6.67
Þ         n = 6
8. Since am pli tude of each wave is equal. The
am pli tude of re sul tant wave is zero if waves
are equally dis placed in phase
ie, q =
°
= °
360
8
45
Hence phase difference must be 45°
AIEEE Corner
¢ Subjective Question (Level-1)
1. R a a a a
2
1
2
2
2
1 2
2 = + + f cos
(i) R a a a a = = = 2
1 2
,
4 2
2 2 2 2
a a a a = + + f cos
Þ  cos f = 1 Þ f = ° 0 
(ii) 2 2 2
2 2 2
a a a = + f cos
Þ f = ° 90
(iii) a a a
2 2 2
2 2 = + f cos
Þ cos f =
-1
2
 Þ f = ° 120
(iv) 0 2 2
2 2
= + f a a cos Þ f = ° 180
2.
I
I
a a
a a
a
a
a
a
max
min
( )
( )
=
+
-
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
1 2
2
1 2
2
1
2
2
1
2
1
1
ç
ö
ø
÷
÷
2
Þ 
I
I
max
min
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
ç
ö
ø
÷
÷
= =
5
3 1
5
3 1
8
2
16
2
2
2
2
    I I
max min
: : = 16 1
3. I I =
max
cos
2
2
q
Þ 
I
I
max
max
cos
2 2
2
=
q
Þ cos cos
q p
2 4
=
Þ       q
p
=
2
Now,        f = ´
2p
l
Dx
      
p p
l 2
2
= ´ Dx
Þ Dx =
l
4
But    Dx
yd
D
=
Þ y
D x
d
=
D
Þ y =
´
´
=
´ ´
´
-
-
1
4 1
1 500 10
4 10
9
3
m
mm
l
 
Þ y = ´
-
1.25 10
4
 m
4. I I =
max
cos
2
2
q
(a) 
I
I
max
max
cos
2 2
2
=
q
 Þ q
p
=
2
Now q
p
l
= ´
2
Dx
Þ Dx =
l
4
 and Dx
yd
D
= Þ =
D
d
l
4
(b) 
1
4 2
2
I I
max max
cos =
q
Þ cos cos
q p
2 3
= Þ q
p
=
2
3
f = ´
2p
l
Dx
 47
Þ 
2
3
2 p p
l
= ´ Dx Þ Dx =
l
3
Now Dx
yd
D
= Þ y
D
d
=
l
3
5. (a) I I =
max
cos
2
2
q
 
I I
max
=
0
 and q = ° 60
Þ        I I I I = ° = =
0
2
0 0
30
3
4
cos 0.75
(b) q
p
l
= ´
2
Dx
Þ 
p p
l 3
2
= ´ Dx
Þ Dx = =
l
6
480
6
 nm
Þ Dx = 80 nm
6. l l
A B
= = 6m
For constructive interference difference
wavelength = 0 2 , , l l
Q l = > = 6 5 d
AB
Hence only constructive interference occur
at Dl = 0
Þ x = =
5
2
m 2.5 m
For destructive interference Dl l
l
= ,
3
2
Only possibility at Dl = 6
Which occur at x = 1 m and x = 4 m from A.
7. The wavelength l = =
´
´
=
c
n
3 10
120 10
8
6
2.5 m
For constructive interference
x x n - - = ( ) 9 l , where n = 0 1 2 , , K
x = 4.5, 5.75, 8.25 7,
The other points are 3.25, ,. 2 75
8. In Young’s dou ble slit experiment
w
D
d
=
l
Þ 2.82
2.2
´ =
´
´
-
-
10
460 10
3
3
m
l
.
Þ l = 590 nm
9. x
n
 for bright fringe is given by
x
nD
d
n
=
l
Þ x
D
d
1
=
l
 and   x
D
d
2
2
=
l
 
Dx
D
d
=
l
 the angular separation
sin ( ) D
D
q
l
= = =
´
´
= ´
-
-
-
x
D d
5 10
2 10
10
7
3
4
2.5
      ( ) sin [ ] Dq =
-1
0.00025
= ° 0.014
10. When whole appratus is im mersed in wa ter
l l = = ´ ´
-
n
w
4
3
700 10
9
m
w
D
d
= =
´ ´ ´ ´
´ ´
=
- -
-
l 48 10 4 7 10
3 25 10
2 7
5
0.90 mm
11. w
D
d
1
1
=
l
 and   w
D
d
2
2
=
l
 
Þ w w
D D
d
1 2
1 2
- =
- ( ) l
Þ 3 10
10
10
5
2
3
´ =
´
´
-
-
-
1.5
l
Þ l m = ´ = ´ =
-
-
-
3 10
10
2 10 2
8
2
6
1.5
m m
12. For bright fringe x
D n
d
n
=
l
For first light ( ) l = 480 nm the third order
Bright fringe is x
3
9
3
1 3 480 10
5 10
=
´ ´ ´
´
-
-
For second light ( ) l = 600 nm
x
3
9
3
1 3 600 10
5 10
¢ =
´ ´ ´
´
-
-
48
A B
5 m
A B
x
(9 – x)
Page 4


28 Interference and Diffraction
of Light
Introductory Exercise 28.1
1. Because they are incoherent ie, Df  does not
re main constant.
2. Since la ser is highly coherent and
mono chro matic source of light
3. I I =
0
2
2 cos / q
Þ 
3
4
2
0
0
2
I
I = cos / q
Þ cos / q 2
3
2
=
Þ 
q p
2 6
=
Þ q
p
=
3
Þ Dx = ´ = ´ =
l
l
f
l
p
p l
2 2 3 6
But Dx
yd
D
=
Þ y
D x
d
D
d
= = ´
D l
6
Þ y =
´ ´
´ ´
=
-
-
1.2 600 10
0 25 10 6
48
9
2
.
mm
4. 2
1
2
m l t n = -
æ
è
ç
ö
ø
÷
 for min i mum thick ness n = 1
Þ t = =
´
=
l
m 4
3
4 1.5
0.5cm
5. Here a a
1
3 = and a a
2
=
R a a a a
2 2 2
3 2 3 = + + ´ ´ ´ ( ) ( ) cos q
Þ I I I I = + + 9 6
0 0 0
cos q
Þ I I = + [ cos ] 10 6
0
q
I I = + ´ -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
10 6 2
2
1
2
0
cos
q
   = + -
é
ë
ê
ù
û
ú
10 12
2
6
2
0
cos
q
I
   = +
é
ë
ê
ù
û
ú
4 12
2
2
0
cos
q
I
    = +
é
ë
ê
ù
û
ú
4 1 3
2
0
2
I cos
q
Now, I
I
max
=
0
9
Þ          I I = +
é
ë
ê
ù
û
ú
4
9
1 3
2
2
max
cos
q
6. Dx
yd
D
t = - - ( ) m 1
if t =
-
l
m 2 1 ( )
Þ Dx
yd
D
= -
l
2
For maxima Dx n = l
Þ 
yd
D
n = +
æ
è
ç
ö
ø
÷
2
1
2
l
This become minima.
For minima Dx n = -
æ
è
ç
ö
ø
÷
1
2
l
          
yd
n
D
- = -
l
l
l
2 2
Þ 
yd
D
n - l this become maxima.
Hence maxima and minima are
interchanged.
7. For two slit experiment
d n sin q l =
Þ sin q
l
=
n
d
But sin q £ 1 Þ 
nl
d
£ 1
Þ       n
d
£
l
   Þ  n £
´
´
=
-
-
4 10
6 10
6
7
6.67
Þ         n = 6
8. Since am pli tude of each wave is equal. The
am pli tude of re sul tant wave is zero if waves
are equally dis placed in phase
ie, q =
°
= °
360
8
45
Hence phase difference must be 45°
AIEEE Corner
¢ Subjective Question (Level-1)
1. R a a a a
2
1
2
2
2
1 2
2 = + + f cos
(i) R a a a a = = = 2
1 2
,
4 2
2 2 2 2
a a a a = + + f cos
Þ  cos f = 1 Þ f = ° 0 
(ii) 2 2 2
2 2 2
a a a = + f cos
Þ f = ° 90
(iii) a a a
2 2 2
2 2 = + f cos
Þ cos f =
-1
2
 Þ f = ° 120
(iv) 0 2 2
2 2
= + f a a cos Þ f = ° 180
2.
I
I
a a
a a
a
a
a
a
max
min
( )
( )
=
+
-
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
1 2
2
1 2
2
1
2
2
1
2
1
1
ç
ö
ø
÷
÷
2
Þ 
I
I
max
min
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
ç
ö
ø
÷
÷
= =
5
3 1
5
3 1
8
2
16
2
2
2
2
    I I
max min
: : = 16 1
3. I I =
max
cos
2
2
q
Þ 
I
I
max
max
cos
2 2
2
=
q
Þ cos cos
q p
2 4
=
Þ       q
p
=
2
Now,        f = ´
2p
l
Dx
      
p p
l 2
2
= ´ Dx
Þ Dx =
l
4
But    Dx
yd
D
=
Þ y
D x
d
=
D
Þ y =
´
´
=
´ ´
´
-
-
1
4 1
1 500 10
4 10
9
3
m
mm
l
 
Þ y = ´
-
1.25 10
4
 m
4. I I =
max
cos
2
2
q
(a) 
I
I
max
max
cos
2 2
2
=
q
 Þ q
p
=
2
Now q
p
l
= ´
2
Dx
Þ Dx =
l
4
 and Dx
yd
D
= Þ =
D
d
l
4
(b) 
1
4 2
2
I I
max max
cos =
q
Þ cos cos
q p
2 3
= Þ q
p
=
2
3
f = ´
2p
l
Dx
 47
Þ 
2
3
2 p p
l
= ´ Dx Þ Dx =
l
3
Now Dx
yd
D
= Þ y
D
d
=
l
3
5. (a) I I =
max
cos
2
2
q
 
I I
max
=
0
 and q = ° 60
Þ        I I I I = ° = =
0
2
0 0
30
3
4
cos 0.75
(b) q
p
l
= ´
2
Dx
Þ 
p p
l 3
2
= ´ Dx
Þ Dx = =
l
6
480
6
 nm
Þ Dx = 80 nm
6. l l
A B
= = 6m
For constructive interference difference
wavelength = 0 2 , , l l
Q l = > = 6 5 d
AB
Hence only constructive interference occur
at Dl = 0
Þ x = =
5
2
m 2.5 m
For destructive interference Dl l
l
= ,
3
2
Only possibility at Dl = 6
Which occur at x = 1 m and x = 4 m from A.
7. The wavelength l = =
´
´
=
c
n
3 10
120 10
8
6
2.5 m
For constructive interference
x x n - - = ( ) 9 l , where n = 0 1 2 , , K
x = 4.5, 5.75, 8.25 7,
The other points are 3.25, ,. 2 75
8. In Young’s dou ble slit experiment
w
D
d
=
l
Þ 2.82
2.2
´ =
´
´
-
-
10
460 10
3
3
m
l
.
Þ l = 590 nm
9. x
n
 for bright fringe is given by
x
nD
d
n
=
l
Þ x
D
d
1
=
l
 and   x
D
d
2
2
=
l
 
Dx
D
d
=
l
 the angular separation
sin ( ) D
D
q
l
= = =
´
´
= ´
-
-
-
x
D d
5 10
2 10
10
7
3
4
2.5
      ( ) sin [ ] Dq =
-1
0.00025
= ° 0.014
10. When whole appratus is im mersed in wa ter
l l = = ´ ´
-
n
w
4
3
700 10
9
m
w
D
d
= =
´ ´ ´ ´
´ ´
=
- -
-
l 48 10 4 7 10
3 25 10
2 7
5
0.90 mm
11. w
D
d
1
1
=
l
 and   w
D
d
2
2
=
l
 
Þ w w
D D
d
1 2
1 2
- =
- ( ) l
Þ 3 10
10
10
5
2
3
´ =
´
´
-
-
-
1.5
l
Þ l m = ´ = ´ =
-
-
-
3 10
10
2 10 2
8
2
6
1.5
m m
12. For bright fringe x
D n
d
n
=
l
For first light ( ) l = 480 nm the third order
Bright fringe is x
3
9
3
1 3 480 10
5 10
=
´ ´ ´
´
-
-
For second light ( ) l = 600 nm
x
3
9
3
1 3 600 10
5 10
¢ =
´ ´ ´
´
-
-
48
A B
5 m
A B
x
(9 – x)
Dx x x = ¢ - =
´ ´ -
´
-
-
3 3
9
3
3 10 600 480
5 10
( )
        =
´
´
-
3 120
5
10
6
  Dx = ´
-
72 10
6
, Dx = 72 mm
13. Fringe width :
w
l
=
D
d
=
´ ´
´
- -
-
( ) ( )
( )
500 10 75 10
10
9 2
3
0.45
m
= ´
-
0.83 10
3
 m = 0.83 mm
Distance between second and third dark
line = one fringe width = 0.83 mm.
14. For first or der bright fringe
         x
D
d
=
l
Þ   4.94 ´ =
´ ´
-
-
10
3 600 10
3
9
d
Þ    d =
´
´
= ´
-
-
-
18 10
10
18
10
7
3
4
4.94 5.94
m
Let for wavelength l first dark fringe is
obtained at this point for first dark fringe
x
D
d
=
l
2
 Þ 4.94
4.94
´ =
´ ´
´ ´
-
-
10
3
2 18 10
3
4
l
Þ 2 18 10 3
7
´ ´ =
-
l
Þ       l = ´
-
12 10
7
m l = 1200 nm
15.
For dark fringe
d n sin ( ) q
l
= - 2 1
2
For first dark fringe n = 1 and for 2nd, n = 2
dsin q
l
=
2
dsin q
l
¢ =
3
2
  Þ  sin q
l
=
2d
Þ      
y
D y
d
1
2
1
2
2
+
=
l
and  
y
D y
d
2
2
1
2
3
2
+
=
l
Þ 
y
y
1
2 2
1
2
9
6
35 10
550 10
2 10
( ) ´ +
=
´
´ ´
-
-
-
1.8
…(i)
and 
y
y
2
2 2
2
2
9
6
35 10
3 550 10
2 10
( ) ´ +
=
´ ´
´ ´
-
-
-
1.8
On solving y y
2 1
- = 12.6 cm
16. Wavelength of source l
l
=
2 meV
Þ l =
´
´ ´ ´ ´ ´
-
- -
6.62
9.1 1.6
10
2 10 100 10
34
31 19
Þ l = ´
-
1.24 m 10
10
w
D
d
= =
´ ´
´
=
-
-
l 3 10
10 10
10
10
1.24
m 36.6 cm
17. Given l = = ´
-
546 10
7
nm 5.46 m, D = 1m
d = = ´
-
0 mm 0.3 m .3 10
3
(a) At distance y = = ´
-
10 10 10
3
mm m from
the central fringe path difference will be
Dy
y d
D
= =
´ ´ ´
- -
. 10 10 10
1
3 3
0.3
      = ´
-
3 10
6
m
Corresponding phase difference
f
l
l
p
= ´ =
´
´ ´
-
-
2 2
10
3 10
7
6
Dx
5.46
 rad
   = ° 1978 
 49
q
q¢
S
1
S
2
y
2
y
1
screen
I
Page 5


28 Interference and Diffraction
of Light
Introductory Exercise 28.1
1. Because they are incoherent ie, Df  does not
re main constant.
2. Since la ser is highly coherent and
mono chro matic source of light
3. I I =
0
2
2 cos / q
Þ 
3
4
2
0
0
2
I
I = cos / q
Þ cos / q 2
3
2
=
Þ 
q p
2 6
=
Þ q
p
=
3
Þ Dx = ´ = ´ =
l
l
f
l
p
p l
2 2 3 6
But Dx
yd
D
=
Þ y
D x
d
D
d
= = ´
D l
6
Þ y =
´ ´
´ ´
=
-
-
1.2 600 10
0 25 10 6
48
9
2
.
mm
4. 2
1
2
m l t n = -
æ
è
ç
ö
ø
÷
 for min i mum thick ness n = 1
Þ t = =
´
=
l
m 4
3
4 1.5
0.5cm
5. Here a a
1
3 = and a a
2
=
R a a a a
2 2 2
3 2 3 = + + ´ ´ ´ ( ) ( ) cos q
Þ I I I I = + + 9 6
0 0 0
cos q
Þ I I = + [ cos ] 10 6
0
q
I I = + ´ -
é
ë
ê
ù
û
ú
é
ë
ê
ù
û
ú
10 6 2
2
1
2
0
cos
q
   = + -
é
ë
ê
ù
û
ú
10 12
2
6
2
0
cos
q
I
   = +
é
ë
ê
ù
û
ú
4 12
2
2
0
cos
q
I
    = +
é
ë
ê
ù
û
ú
4 1 3
2
0
2
I cos
q
Now, I
I
max
=
0
9
Þ          I I = +
é
ë
ê
ù
û
ú
4
9
1 3
2
2
max
cos
q
6. Dx
yd
D
t = - - ( ) m 1
if t =
-
l
m 2 1 ( )
Þ Dx
yd
D
= -
l
2
For maxima Dx n = l
Þ 
yd
D
n = +
æ
è
ç
ö
ø
÷
2
1
2
l
This become minima.
For minima Dx n = -
æ
è
ç
ö
ø
÷
1
2
l
          
yd
n
D
- = -
l
l
l
2 2
Þ 
yd
D
n - l this become maxima.
Hence maxima and minima are
interchanged.
7. For two slit experiment
d n sin q l =
Þ sin q
l
=
n
d
But sin q £ 1 Þ 
nl
d
£ 1
Þ       n
d
£
l
   Þ  n £
´
´
=
-
-
4 10
6 10
6
7
6.67
Þ         n = 6
8. Since am pli tude of each wave is equal. The
am pli tude of re sul tant wave is zero if waves
are equally dis placed in phase
ie, q =
°
= °
360
8
45
Hence phase difference must be 45°
AIEEE Corner
¢ Subjective Question (Level-1)
1. R a a a a
2
1
2
2
2
1 2
2 = + + f cos
(i) R a a a a = = = 2
1 2
,
4 2
2 2 2 2
a a a a = + + f cos
Þ  cos f = 1 Þ f = ° 0 
(ii) 2 2 2
2 2 2
a a a = + f cos
Þ f = ° 90
(iii) a a a
2 2 2
2 2 = + f cos
Þ cos f =
-1
2
 Þ f = ° 120
(iv) 0 2 2
2 2
= + f a a cos Þ f = ° 180
2.
I
I
a a
a a
a
a
a
a
max
min
( )
( )
=
+
-
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
1 2
2
1 2
2
1
2
2
1
2
1
1
ç
ö
ø
÷
÷
2
Þ 
I
I
max
min
=
+
æ
è
ç
ç
ö
ø
÷
÷
-
æ
è
ç
ç
ö
ø
÷
÷
= =
5
3 1
5
3 1
8
2
16
2
2
2
2
    I I
max min
: : = 16 1
3. I I =
max
cos
2
2
q
Þ 
I
I
max
max
cos
2 2
2
=
q
Þ cos cos
q p
2 4
=
Þ       q
p
=
2
Now,        f = ´
2p
l
Dx
      
p p
l 2
2
= ´ Dx
Þ Dx =
l
4
But    Dx
yd
D
=
Þ y
D x
d
=
D
Þ y =
´
´
=
´ ´
´
-
-
1
4 1
1 500 10
4 10
9
3
m
mm
l
 
Þ y = ´
-
1.25 10
4
 m
4. I I =
max
cos
2
2
q
(a) 
I
I
max
max
cos
2 2
2
=
q
 Þ q
p
=
2
Now q
p
l
= ´
2
Dx
Þ Dx =
l
4
 and Dx
yd
D
= Þ =
D
d
l
4
(b) 
1
4 2
2
I I
max max
cos =
q
Þ cos cos
q p
2 3
= Þ q
p
=
2
3
f = ´
2p
l
Dx
 47
Þ 
2
3
2 p p
l
= ´ Dx Þ Dx =
l
3
Now Dx
yd
D
= Þ y
D
d
=
l
3
5. (a) I I =
max
cos
2
2
q
 
I I
max
=
0
 and q = ° 60
Þ        I I I I = ° = =
0
2
0 0
30
3
4
cos 0.75
(b) q
p
l
= ´
2
Dx
Þ 
p p
l 3
2
= ´ Dx
Þ Dx = =
l
6
480
6
 nm
Þ Dx = 80 nm
6. l l
A B
= = 6m
For constructive interference difference
wavelength = 0 2 , , l l
Q l = > = 6 5 d
AB
Hence only constructive interference occur
at Dl = 0
Þ x = =
5
2
m 2.5 m
For destructive interference Dl l
l
= ,
3
2
Only possibility at Dl = 6
Which occur at x = 1 m and x = 4 m from A.
7. The wavelength l = =
´
´
=
c
n
3 10
120 10
8
6
2.5 m
For constructive interference
x x n - - = ( ) 9 l , where n = 0 1 2 , , K
x = 4.5, 5.75, 8.25 7,
The other points are 3.25, ,. 2 75
8. In Young’s dou ble slit experiment
w
D
d
=
l
Þ 2.82
2.2
´ =
´
´
-
-
10
460 10
3
3
m
l
.
Þ l = 590 nm
9. x
n
 for bright fringe is given by
x
nD
d
n
=
l
Þ x
D
d
1
=
l
 and   x
D
d
2
2
=
l
 
Dx
D
d
=
l
 the angular separation
sin ( ) D
D
q
l
= = =
´
´
= ´
-
-
-
x
D d
5 10
2 10
10
7
3
4
2.5
      ( ) sin [ ] Dq =
-1
0.00025
= ° 0.014
10. When whole appratus is im mersed in wa ter
l l = = ´ ´
-
n
w
4
3
700 10
9
m
w
D
d
= =
´ ´ ´ ´
´ ´
=
- -
-
l 48 10 4 7 10
3 25 10
2 7
5
0.90 mm
11. w
D
d
1
1
=
l
 and   w
D
d
2
2
=
l
 
Þ w w
D D
d
1 2
1 2
- =
- ( ) l
Þ 3 10
10
10
5
2
3
´ =
´
´
-
-
-
1.5
l
Þ l m = ´ = ´ =
-
-
-
3 10
10
2 10 2
8
2
6
1.5
m m
12. For bright fringe x
D n
d
n
=
l
For first light ( ) l = 480 nm the third order
Bright fringe is x
3
9
3
1 3 480 10
5 10
=
´ ´ ´
´
-
-
For second light ( ) l = 600 nm
x
3
9
3
1 3 600 10
5 10
¢ =
´ ´ ´
´
-
-
48
A B
5 m
A B
x
(9 – x)
Dx x x = ¢ - =
´ ´ -
´
-
-
3 3
9
3
3 10 600 480
5 10
( )
        =
´
´
-
3 120
5
10
6
  Dx = ´
-
72 10
6
, Dx = 72 mm
13. Fringe width :
w
l
=
D
d
=
´ ´
´
- -
-
( ) ( )
( )
500 10 75 10
10
9 2
3
0.45
m
= ´
-
0.83 10
3
 m = 0.83 mm
Distance between second and third dark
line = one fringe width = 0.83 mm.
14. For first or der bright fringe
         x
D
d
=
l
Þ   4.94 ´ =
´ ´
-
-
10
3 600 10
3
9
d
Þ    d =
´
´
= ´
-
-
-
18 10
10
18
10
7
3
4
4.94 5.94
m
Let for wavelength l first dark fringe is
obtained at this point for first dark fringe
x
D
d
=
l
2
 Þ 4.94
4.94
´ =
´ ´
´ ´
-
-
10
3
2 18 10
3
4
l
Þ 2 18 10 3
7
´ ´ =
-
l
Þ       l = ´
-
12 10
7
m l = 1200 nm
15.
For dark fringe
d n sin ( ) q
l
= - 2 1
2
For first dark fringe n = 1 and for 2nd, n = 2
dsin q
l
=
2
dsin q
l
¢ =
3
2
  Þ  sin q
l
=
2d
Þ      
y
D y
d
1
2
1
2
2
+
=
l
and  
y
D y
d
2
2
1
2
3
2
+
=
l
Þ 
y
y
1
2 2
1
2
9
6
35 10
550 10
2 10
( ) ´ +
=
´
´ ´
-
-
-
1.8
…(i)
and 
y
y
2
2 2
2
2
9
6
35 10
3 550 10
2 10
( ) ´ +
=
´ ´
´ ´
-
-
-
1.8
On solving y y
2 1
- = 12.6 cm
16. Wavelength of source l
l
=
2 meV
Þ l =
´
´ ´ ´ ´ ´
-
- -
6.62
9.1 1.6
10
2 10 100 10
34
31 19
Þ l = ´
-
1.24 m 10
10
w
D
d
= =
´ ´
´
=
-
-
l 3 10
10 10
10
10
1.24
m 36.6 cm
17. Given l = = ´
-
546 10
7
nm 5.46 m, D = 1m
d = = ´
-
0 mm 0.3 m .3 10
3
(a) At distance y = = ´
-
10 10 10
3
mm m from
the central fringe path difference will be
Dy
y d
D
= =
´ ´ ´
- -
. 10 10 10
1
3 3
0.3
      = ´
-
3 10
6
m
Corresponding phase difference
f
l
l
p
= ´ =
´
´ ´
-
-
2 2
10
3 10
7
6
Dx
5.46
 rad
   = ° 1978 
 49
q
q¢
S
1
S
2
y
2
y
1
screen
I
I I =
0
2
2
cos
q
    =
°
æ
è
ç
ö
ø
÷
= ° I I
0
2
0
2
1978
2
989 cos cos
I I = ´
-
3 10
4
0
(b) Fringe width w
D
d
=
l
w =
´ ´
´
=
-
-
5.46
0.3
1.82 mm
10 1
10
7
3
Number of fringes = =
10 mm
1.82 mm
5.49
Hence the number of fringe is five.
18. Shift due to sheet of thick ness 10m and
refractive index 1.6 is
Dx
tD
d
1
1
=
- ( ) m
Þ Dx
1
6
3
1
10 10
10
= - ´
´ ´
´
-
-
( ) 1.6
1.5
1.5
Þ Dx
1
3 3
10 10 6 10 = ´ ´ = ´
- -
0.6 m
Shift due to sheet of thickness 15m and
refractive index 1.2 is
Dx
2
6
3
1 15 10
10
=
- ´ ´ ´
´
-
-
( ) 1.2 1.5
1.5
Dx
2
3
3 10 = ´
-
m
Since these shifts are in opposite direction
of central maxima hence net shift
D D D x x x = - = ´ - ´
- -
1 2
3 3
6 10 3 10 m m
= ´
-
3 10
3
m = 3 mm
19. Let l is the wavelength of light D is screen
dis tance from source and d is the sep a ra tion
between slits (all are in me tres)
         Shift = =
-
Dx
tD
d
( ) m 1
Þ Dx
D
d
=
- ´ ´ ´
-
( ) 1.6 1.964 1 10
6
Þ Dx
D
d
= ´
-
1.17
m
84
10
6
Now when t is removed and D is doubled the
distance between successive maximum (or
minima) i e . ., fringe width
w
D
d
=
2 l
but according to question Dx w =
Þ   
1.178 ´
-
10
6
D
d
 =
2 D
d
l
Þ   l = ´
-
0.589 m 10
6
 = 589 nm
20. Let n bright fringe (l = 5500 Å) concide with
10th 
bright fringe of 6000 Å
Þ n ´ = ´ 5500 6000 10 Å Å
Þ n
~
- 11
Similarly first bright fringe concide with 1st
fringe. Now fringe width
w =
-
=
14.74 12.5
0.224 mm
10
Hence position of 10th bright fringe
= - 14.74 0 224 . 
~
- 14.55 mm
Position of zero order bright fringe
= - 12.75 0.224 
~
- 12.25 mm
21. Here d
~
- 1 cm
D = 100 m
l = 500 nm
50
1 cm
q
P
y
S
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