HC Verma Solutions: Chapter 17 - Light Waves | Physics Class 11 - NEET PDF Download

 Page 1


17.1
SOLUTIONS TO CONCEPTS 
CHAPTER 17
1. Given that, 400 m < ? < 700 nm.
1 1 1
700nm 400nm
? ?
?
?
8 8
7 7 7 7
1 1 1 3 10 c 3 10
7 10 4 10 7 10 4 10
? ? ? ?
? ?
? ? ? ? ?
? ? ? ? ? ?
(Where, c = speed of light = 3 ? 10
8
m/s)
? 4.3 ? 10
14
< c/ ? < 7.5 ? 10
14
? 4.3 ? 10
14
Hz < f < 7.5 ? 10
14
Hz. ?
2. Given that, for sodium light, ? = 589 nm = 589 ? 10
–9
m
a) f
a
= 
8
9
3 10
589 10
?
?
?
= 5.09 ? 10
14 1
c
sec f
?
? ?
?
? ?
? ? ?
?
b)
a w w
w
9
w a
1
1.33 589 10
?
? ? ?
? ? ? ? ?
? ? ?
= 443 nm
c) f
w
= f
a
= 5.09 ? 10
14
sec
–1
[Frequency does not change]
d)
8
a a a w
w
w a w
v v 3 10
v
v 1.33
? ? ?
? ? ? ?
? ?
= 2.25 ? 10
8
m/sec.
3. We know that, 
2 1
1 2
v
v
?
?
?
So, 
8
8
400
400
1472 3 10
v 2.04 10 m/ sec.
1 v
?
? ? ? ?
[because, for air, ? = 1 and v = 3 ? 10
8
m/s]
Again, 
8
8
760
760
1452 3 10
v 2.07 10 m/ sec.
1 v
?
? ? ? ? ?
4.
8
t
8
1 3 10
1.25
(2.4) 10
? ?
? ? ?
?
velocity of light in vaccum
since,  = 
velocity of light in the given medium
? ?
?
? ?
? ?
5. Given that, d = 1 cm = 10
–2
m, ? = 5 ? 10
–7
m and D = 1 m
a) Separation between two consecutive maxima is equal to fringe width.
So, ? = 
7
2
D 5 10 1
d 10
?
?
? ? ?
? m = 5 ? 10
–5
m = 0.05 mm.
b) When, ? = 1 mm = 10
–3
m
10
–3
m = 
7
5 10 1
D
?
? ?
? D = 5 ? 10
–4
m = 0.50 mm. ?
6. Given that, ? = 1 mm = 10
–3
m, D = 2.t m and d = 1 mm = 10
–3
m
So, 10
–3
m = 
3
25
10
?
? ?
? ? = 4 ? 10
–7
m = 400 nm. ?
7. Given that, d = 1 mm = 10
–3
m, D = 1 m.
So, fringe with = 
D
d
?
= 0.5 mm.
a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm
b) No. of fringes = 10 / 0.5 = 20.
8. Given that, d = 0.8 mm = 0.8 ? 10
–3
m, ? = 589 nm = 589 ? 10
–9
m and D = 2 m.
So, ? = 
D
d
?
= 
9
3
589 10 2
0.8 10
?
?
? ?
?
= 1.47 ? 10
–3
m = 147 mm. ?
Page 2


17.1
SOLUTIONS TO CONCEPTS 
CHAPTER 17
1. Given that, 400 m < ? < 700 nm.
1 1 1
700nm 400nm
? ?
?
?
8 8
7 7 7 7
1 1 1 3 10 c 3 10
7 10 4 10 7 10 4 10
? ? ? ?
? ?
? ? ? ? ?
? ? ? ? ? ?
(Where, c = speed of light = 3 ? 10
8
m/s)
? 4.3 ? 10
14
< c/ ? < 7.5 ? 10
14
? 4.3 ? 10
14
Hz < f < 7.5 ? 10
14
Hz. ?
2. Given that, for sodium light, ? = 589 nm = 589 ? 10
–9
m
a) f
a
= 
8
9
3 10
589 10
?
?
?
= 5.09 ? 10
14 1
c
sec f
?
? ?
?
? ?
? ? ?
?
b)
a w w
w
9
w a
1
1.33 589 10
?
? ? ?
? ? ? ? ?
? ? ?
= 443 nm
c) f
w
= f
a
= 5.09 ? 10
14
sec
–1
[Frequency does not change]
d)
8
a a a w
w
w a w
v v 3 10
v
v 1.33
? ? ?
? ? ? ?
? ?
= 2.25 ? 10
8
m/sec.
3. We know that, 
2 1
1 2
v
v
?
?
?
So, 
8
8
400
400
1472 3 10
v 2.04 10 m/ sec.
1 v
?
? ? ? ?
[because, for air, ? = 1 and v = 3 ? 10
8
m/s]
Again, 
8
8
760
760
1452 3 10
v 2.07 10 m/ sec.
1 v
?
? ? ? ? ?
4.
8
t
8
1 3 10
1.25
(2.4) 10
? ?
? ? ?
?
velocity of light in vaccum
since,  = 
velocity of light in the given medium
? ?
?
? ?
? ?
5. Given that, d = 1 cm = 10
–2
m, ? = 5 ? 10
–7
m and D = 1 m
a) Separation between two consecutive maxima is equal to fringe width.
So, ? = 
7
2
D 5 10 1
d 10
?
?
? ? ?
? m = 5 ? 10
–5
m = 0.05 mm.
b) When, ? = 1 mm = 10
–3
m
10
–3
m = 
7
5 10 1
D
?
? ?
? D = 5 ? 10
–4
m = 0.50 mm. ?
6. Given that, ? = 1 mm = 10
–3
m, D = 2.t m and d = 1 mm = 10
–3
m
So, 10
–3
m = 
3
25
10
?
? ?
? ? = 4 ? 10
–7
m = 400 nm. ?
7. Given that, d = 1 mm = 10
–3
m, D = 1 m.
So, fringe with = 
D
d
?
= 0.5 mm.
a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm
b) No. of fringes = 10 / 0.5 = 20.
8. Given that, d = 0.8 mm = 0.8 ? 10
–3
m, ? = 589 nm = 589 ? 10
–9
m and D = 2 m.
So, ? = 
D
d
?
= 
9
3
589 10 2
0.8 10
?
?
? ?
?
= 1.47 ? 10
–3
m = 147 mm. ?
Chapter 17
17.2
9. Given that, ? = 500 nm = 500 ? 10
–9
m and d = 2 ? 10
–3
m
As shown in the figure, angular separation ? = 
D
D dD d
? ? ?
? ?
So, ? = 
9
3
500 10
D d 2 10
?
?
? ? ?
? ?
?
= 250 ? 10
–6
= 25 ? 10
–5
radian = 0.014 degree. ?
10. We know that, the first maximum (next to central maximum) occurs at y = 
D
d
?
Given that, ?
1
= 480 nm, ?
2
= 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10
–3
m
So, y
1
= 
9
1
3
D 1.5 480 10
d 0.25 10
?
?
? ? ?
?
?
= 2.88 mm
y
2 
= 
9
3
1.5 600 10
0.25 10
?
?
? ?
?
= 3.6 mm.
So, the separation between these two bright fringes is given by,
? separation = y
2
– y
1
= 3.60 – 2.88 = 0.72 mm.
11. Let m
th
bright fringe of violet light overlaps with n
th
bright fringe of red light.
?
m 400nm D n 700nm D m 7
d d n 4
? ? ? ?
? ? ?
? 7
th
bright fringe of violet light overlaps with 4
th
bright fringe of red light (minimum). Also, it can be 
seen that 14
th
violet fringe will overlap 8
th
red fringe.
Because, m/n = 7/4 = 14/8.
12. Let, t = thickness of the plate
Given, optical path difference = ( ? – 1)t = ?/2
? t = 
2( 1)
?
? ?
?
13. a) Change in the optical path = ?t – t = ( ? – 1)t
b) To have a dark fringe at the centre the pattern should shift by one half of a fringe.
? ( ? – 1)t = t
2 2( 1)
? ?
? ?
? ?
. ?
14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10
–3
m and ? = 620 nm = 620 ? 10
–9
m
We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t.
Again, for shift of one fringe, the optical path should be changed by ?.
So, no. of fringes crossing through the centre is given by,
n = 
3
9
( 1)t 0.45 0.02 10
620 10
?
?
? ? ? ?
?
? ?
= 14.5 ?
15. In the given Young’s double slit experiment, 
? = 1.6, t = 1.964 micron = 1.964 ? 10
–6
m
We know, number of fringes shifted = 
( 1)t ? ?
?
So, the corresponding shift = No.of fringes shifted ? fringe width
= 
( 1)t D ( 1)tD
d d
? ? ? ? ?
? ?
?
… (1)
Again, when the distance between the screen and the slits is doubled,
Fringe width = 
(2D)
d
?
…(2)
From (1) and (2), 
( 1)tD
d
? ?
= 
(2D)
d
?
? ? = 
( 1)t ? ?
?
= 
6
(1.6 1) (1.964) 10
2
?
? ? ?
= 589.2 ? 10
–9
= 589.2 nm. ?
B
S 1
S 2
? ?
D
Page 3


17.1
SOLUTIONS TO CONCEPTS 
CHAPTER 17
1. Given that, 400 m < ? < 700 nm.
1 1 1
700nm 400nm
? ?
?
?
8 8
7 7 7 7
1 1 1 3 10 c 3 10
7 10 4 10 7 10 4 10
? ? ? ?
? ?
? ? ? ? ?
? ? ? ? ? ?
(Where, c = speed of light = 3 ? 10
8
m/s)
? 4.3 ? 10
14
< c/ ? < 7.5 ? 10
14
? 4.3 ? 10
14
Hz < f < 7.5 ? 10
14
Hz. ?
2. Given that, for sodium light, ? = 589 nm = 589 ? 10
–9
m
a) f
a
= 
8
9
3 10
589 10
?
?
?
= 5.09 ? 10
14 1
c
sec f
?
? ?
?
? ?
? ? ?
?
b)
a w w
w
9
w a
1
1.33 589 10
?
? ? ?
? ? ? ? ?
? ? ?
= 443 nm
c) f
w
= f
a
= 5.09 ? 10
14
sec
–1
[Frequency does not change]
d)
8
a a a w
w
w a w
v v 3 10
v
v 1.33
? ? ?
? ? ? ?
? ?
= 2.25 ? 10
8
m/sec.
3. We know that, 
2 1
1 2
v
v
?
?
?
So, 
8
8
400
400
1472 3 10
v 2.04 10 m/ sec.
1 v
?
? ? ? ?
[because, for air, ? = 1 and v = 3 ? 10
8
m/s]
Again, 
8
8
760
760
1452 3 10
v 2.07 10 m/ sec.
1 v
?
? ? ? ? ?
4.
8
t
8
1 3 10
1.25
(2.4) 10
? ?
? ? ?
?
velocity of light in vaccum
since,  = 
velocity of light in the given medium
? ?
?
? ?
? ?
5. Given that, d = 1 cm = 10
–2
m, ? = 5 ? 10
–7
m and D = 1 m
a) Separation between two consecutive maxima is equal to fringe width.
So, ? = 
7
2
D 5 10 1
d 10
?
?
? ? ?
? m = 5 ? 10
–5
m = 0.05 mm.
b) When, ? = 1 mm = 10
–3
m
10
–3
m = 
7
5 10 1
D
?
? ?
? D = 5 ? 10
–4
m = 0.50 mm. ?
6. Given that, ? = 1 mm = 10
–3
m, D = 2.t m and d = 1 mm = 10
–3
m
So, 10
–3
m = 
3
25
10
?
? ?
? ? = 4 ? 10
–7
m = 400 nm. ?
7. Given that, d = 1 mm = 10
–3
m, D = 1 m.
So, fringe with = 
D
d
?
= 0.5 mm.
a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm
b) No. of fringes = 10 / 0.5 = 20.
8. Given that, d = 0.8 mm = 0.8 ? 10
–3
m, ? = 589 nm = 589 ? 10
–9
m and D = 2 m.
So, ? = 
D
d
?
= 
9
3
589 10 2
0.8 10
?
?
? ?
?
= 1.47 ? 10
–3
m = 147 mm. ?
Chapter 17
17.2
9. Given that, ? = 500 nm = 500 ? 10
–9
m and d = 2 ? 10
–3
m
As shown in the figure, angular separation ? = 
D
D dD d
? ? ?
? ?
So, ? = 
9
3
500 10
D d 2 10
?
?
? ? ?
? ?
?
= 250 ? 10
–6
= 25 ? 10
–5
radian = 0.014 degree. ?
10. We know that, the first maximum (next to central maximum) occurs at y = 
D
d
?
Given that, ?
1
= 480 nm, ?
2
= 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10
–3
m
So, y
1
= 
9
1
3
D 1.5 480 10
d 0.25 10
?
?
? ? ?
?
?
= 2.88 mm
y
2 
= 
9
3
1.5 600 10
0.25 10
?
?
? ?
?
= 3.6 mm.
So, the separation between these two bright fringes is given by,
? separation = y
2
– y
1
= 3.60 – 2.88 = 0.72 mm.
11. Let m
th
bright fringe of violet light overlaps with n
th
bright fringe of red light.
?
m 400nm D n 700nm D m 7
d d n 4
? ? ? ?
? ? ?
? 7
th
bright fringe of violet light overlaps with 4
th
bright fringe of red light (minimum). Also, it can be 
seen that 14
th
violet fringe will overlap 8
th
red fringe.
Because, m/n = 7/4 = 14/8.
12. Let, t = thickness of the plate
Given, optical path difference = ( ? – 1)t = ?/2
? t = 
2( 1)
?
? ?
?
13. a) Change in the optical path = ?t – t = ( ? – 1)t
b) To have a dark fringe at the centre the pattern should shift by one half of a fringe.
? ( ? – 1)t = t
2 2( 1)
? ?
? ?
? ?
. ?
14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10
–3
m and ? = 620 nm = 620 ? 10
–9
m
We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t.
Again, for shift of one fringe, the optical path should be changed by ?.
So, no. of fringes crossing through the centre is given by,
n = 
3
9
( 1)t 0.45 0.02 10
620 10
?
?
? ? ? ?
?
? ?
= 14.5 ?
15. In the given Young’s double slit experiment, 
? = 1.6, t = 1.964 micron = 1.964 ? 10
–6
m
We know, number of fringes shifted = 
( 1)t ? ?
?
So, the corresponding shift = No.of fringes shifted ? fringe width
= 
( 1)t D ( 1)tD
d d
? ? ? ? ?
? ?
?
… (1)
Again, when the distance between the screen and the slits is doubled,
Fringe width = 
(2D)
d
?
…(2)
From (1) and (2), 
( 1)tD
d
? ?
= 
(2D)
d
?
? ? = 
( 1)t ? ?
?
= 
6
(1.6 1) (1.964) 10
2
?
? ? ?
= 589.2 ? 10
–9
= 589.2 nm. ?
B
S 1
S 2
? ?
D
Chapter 17
17.3
16. Given that, t
1
= t
2
= 0.5 mm = 0.5 ? 10
–3
m, ?
m
= 1.58 and ?
p
= 1.55, 
? = 590 nm = 590 ? 10
–9
m, d = 0.12 cm = 12 ? 10
–4
m, D = 1 m
a) Fringe width = 
9
4
D 1 590 10
d 12 10
?
?
? ? ?
?
?
= 4.91 ? 10
–4
m.
b) When both the strips are fitted, the optical path changes by 
?x = ( ?
m
– 1)t
1
– ( ?
p
– 1)t
2
= ( ?
m
– ?
p
)t
= (1.58 – 1.55) ? (0.5)(10
–3
) = 0.015 ? 10
–13
m.
So, No. of fringes shifted = 
3
3
0.015 10
590 10
?
?
?
?
= 25.43.
? There are 25 fringes and 0.43 th of a fringe.
? There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.
So, position of first maximum on both sides will be given by
? x = 0.43 ? 4.91 ? 10
–4
= 0.021 cm
x ? = (1 – 0.43) ? 4.91 ? 10
–4
= 0.028 cm (since, fringe width = 4.91 ? 10
–4
m)
17. The change in path difference due to the two slabs is ( ?
1
– ?
2
)t (as in problem no. 16).
For having a minimum at P
0
, the path difference should change by ?/2.
So, ? ?/2 = ( ?
1
– ??
2
)t ? t = 
1 2
2( )
?
? ? ?
. ?
18. Given that, t = 0.02 mm = 0.02 ? 10
–3
m, ?
1
= 1.45, ? = 600 nm = 600 ? 10
–9
m
a) Let, I
1
= Intensity of source without paper = I
b) Then I
2
= Intensity of source with paper = (4/9)I
?
1 1
2 2
I r 9 3
I 4 r 2
? ? ? [because I ? r
2
]
where, r
1
and r
2
are corresponding amplitudes.
So, 
2
max 1 2
2
min 1 2
I (r r )
I (r r )
?
?
?
= 25 : 1
b) No. of fringes that will cross the origin is given by,
n = 
( 1)t ? ?
?
= 
3
9
(1.45 1) 0.02 10
600 10
?
?
? ? ?
?
= 15.
19. Given that, d = 0.28 mm = 0.28 ? 10
–3
m, D = 48 cm = 0.48 m, ?
a
= 700 nm in vacuum 
Let, ?
w
= wavelength of red light in water
Since, the fringe width of the pattern is given by,
??= 
9
w
3
D 525 10 0.48
d 0.28 10
?
?
? ? ?
?
?
= 9 ? 10
–4
m = 0.90 mm. ?
20. It can be seen from the figure that the wavefronts reaching O from S
1
and S
2
will
have a path difference of S
2
X.
In the ? S
1
S
2
X,
sin ???= 
2
1 2
S X
S S
So, path difference = S
2
X = S
1
S
2
sin ? = d sin ? = d ? ?/2d = ?/2
As the path difference is an odd multiple of ?/2, there will be a dark fringe at point P
0
.
21. a) Since, there is a phase difference of ? between direct light and 
reflecting light, the intensity just above the mirror will be zero.
b) Here, 2d = equivalent slit separation 
D = Distance between slit and screen.
We know for bright fringe, ?x = 
y 2d
D
?
= n ?
But as there is a phase reversal of ?/2.
?
y 2d
D
?
+ 
2
?
= n ? ?
y 2d
D
?
= n ? –
2
?
? y = 
D
4d
?
?
mica
Screen
polysterene
S 2
S 1
? ?
P 0 ? ?
x
S 2
S 1
Dark
fringe
(1 – 0.43) ??
0.43 ??
S 2
S 1
2d
D
Screen 
Page 4


17.1
SOLUTIONS TO CONCEPTS 
CHAPTER 17
1. Given that, 400 m < ? < 700 nm.
1 1 1
700nm 400nm
? ?
?
?
8 8
7 7 7 7
1 1 1 3 10 c 3 10
7 10 4 10 7 10 4 10
? ? ? ?
? ?
? ? ? ? ?
? ? ? ? ? ?
(Where, c = speed of light = 3 ? 10
8
m/s)
? 4.3 ? 10
14
< c/ ? < 7.5 ? 10
14
? 4.3 ? 10
14
Hz < f < 7.5 ? 10
14
Hz. ?
2. Given that, for sodium light, ? = 589 nm = 589 ? 10
–9
m
a) f
a
= 
8
9
3 10
589 10
?
?
?
= 5.09 ? 10
14 1
c
sec f
?
? ?
?
? ?
? ? ?
?
b)
a w w
w
9
w a
1
1.33 589 10
?
? ? ?
? ? ? ? ?
? ? ?
= 443 nm
c) f
w
= f
a
= 5.09 ? 10
14
sec
–1
[Frequency does not change]
d)
8
a a a w
w
w a w
v v 3 10
v
v 1.33
? ? ?
? ? ? ?
? ?
= 2.25 ? 10
8
m/sec.
3. We know that, 
2 1
1 2
v
v
?
?
?
So, 
8
8
400
400
1472 3 10
v 2.04 10 m/ sec.
1 v
?
? ? ? ?
[because, for air, ? = 1 and v = 3 ? 10
8
m/s]
Again, 
8
8
760
760
1452 3 10
v 2.07 10 m/ sec.
1 v
?
? ? ? ? ?
4.
8
t
8
1 3 10
1.25
(2.4) 10
? ?
? ? ?
?
velocity of light in vaccum
since,  = 
velocity of light in the given medium
? ?
?
? ?
? ?
5. Given that, d = 1 cm = 10
–2
m, ? = 5 ? 10
–7
m and D = 1 m
a) Separation between two consecutive maxima is equal to fringe width.
So, ? = 
7
2
D 5 10 1
d 10
?
?
? ? ?
? m = 5 ? 10
–5
m = 0.05 mm.
b) When, ? = 1 mm = 10
–3
m
10
–3
m = 
7
5 10 1
D
?
? ?
? D = 5 ? 10
–4
m = 0.50 mm. ?
6. Given that, ? = 1 mm = 10
–3
m, D = 2.t m and d = 1 mm = 10
–3
m
So, 10
–3
m = 
3
25
10
?
? ?
? ? = 4 ? 10
–7
m = 400 nm. ?
7. Given that, d = 1 mm = 10
–3
m, D = 1 m.
So, fringe with = 
D
d
?
= 0.5 mm.
a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm
b) No. of fringes = 10 / 0.5 = 20.
8. Given that, d = 0.8 mm = 0.8 ? 10
–3
m, ? = 589 nm = 589 ? 10
–9
m and D = 2 m.
So, ? = 
D
d
?
= 
9
3
589 10 2
0.8 10
?
?
? ?
?
= 1.47 ? 10
–3
m = 147 mm. ?
Chapter 17
17.2
9. Given that, ? = 500 nm = 500 ? 10
–9
m and d = 2 ? 10
–3
m
As shown in the figure, angular separation ? = 
D
D dD d
? ? ?
? ?
So, ? = 
9
3
500 10
D d 2 10
?
?
? ? ?
? ?
?
= 250 ? 10
–6
= 25 ? 10
–5
radian = 0.014 degree. ?
10. We know that, the first maximum (next to central maximum) occurs at y = 
D
d
?
Given that, ?
1
= 480 nm, ?
2
= 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10
–3
m
So, y
1
= 
9
1
3
D 1.5 480 10
d 0.25 10
?
?
? ? ?
?
?
= 2.88 mm
y
2 
= 
9
3
1.5 600 10
0.25 10
?
?
? ?
?
= 3.6 mm.
So, the separation between these two bright fringes is given by,
? separation = y
2
– y
1
= 3.60 – 2.88 = 0.72 mm.
11. Let m
th
bright fringe of violet light overlaps with n
th
bright fringe of red light.
?
m 400nm D n 700nm D m 7
d d n 4
? ? ? ?
? ? ?
? 7
th
bright fringe of violet light overlaps with 4
th
bright fringe of red light (minimum). Also, it can be 
seen that 14
th
violet fringe will overlap 8
th
red fringe.
Because, m/n = 7/4 = 14/8.
12. Let, t = thickness of the plate
Given, optical path difference = ( ? – 1)t = ?/2
? t = 
2( 1)
?
? ?
?
13. a) Change in the optical path = ?t – t = ( ? – 1)t
b) To have a dark fringe at the centre the pattern should shift by one half of a fringe.
? ( ? – 1)t = t
2 2( 1)
? ?
? ?
? ?
. ?
14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10
–3
m and ? = 620 nm = 620 ? 10
–9
m
We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t.
Again, for shift of one fringe, the optical path should be changed by ?.
So, no. of fringes crossing through the centre is given by,
n = 
3
9
( 1)t 0.45 0.02 10
620 10
?
?
? ? ? ?
?
? ?
= 14.5 ?
15. In the given Young’s double slit experiment, 
? = 1.6, t = 1.964 micron = 1.964 ? 10
–6
m
We know, number of fringes shifted = 
( 1)t ? ?
?
So, the corresponding shift = No.of fringes shifted ? fringe width
= 
( 1)t D ( 1)tD
d d
? ? ? ? ?
? ?
?
… (1)
Again, when the distance between the screen and the slits is doubled,
Fringe width = 
(2D)
d
?
…(2)
From (1) and (2), 
( 1)tD
d
? ?
= 
(2D)
d
?
? ? = 
( 1)t ? ?
?
= 
6
(1.6 1) (1.964) 10
2
?
? ? ?
= 589.2 ? 10
–9
= 589.2 nm. ?
B
S 1
S 2
? ?
D
Chapter 17
17.3
16. Given that, t
1
= t
2
= 0.5 mm = 0.5 ? 10
–3
m, ?
m
= 1.58 and ?
p
= 1.55, 
? = 590 nm = 590 ? 10
–9
m, d = 0.12 cm = 12 ? 10
–4
m, D = 1 m
a) Fringe width = 
9
4
D 1 590 10
d 12 10
?
?
? ? ?
?
?
= 4.91 ? 10
–4
m.
b) When both the strips are fitted, the optical path changes by 
?x = ( ?
m
– 1)t
1
– ( ?
p
– 1)t
2
= ( ?
m
– ?
p
)t
= (1.58 – 1.55) ? (0.5)(10
–3
) = 0.015 ? 10
–13
m.
So, No. of fringes shifted = 
3
3
0.015 10
590 10
?
?
?
?
= 25.43.
? There are 25 fringes and 0.43 th of a fringe.
? There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.
So, position of first maximum on both sides will be given by
? x = 0.43 ? 4.91 ? 10
–4
= 0.021 cm
x ? = (1 – 0.43) ? 4.91 ? 10
–4
= 0.028 cm (since, fringe width = 4.91 ? 10
–4
m)
17. The change in path difference due to the two slabs is ( ?
1
– ?
2
)t (as in problem no. 16).
For having a minimum at P
0
, the path difference should change by ?/2.
So, ? ?/2 = ( ?
1
– ??
2
)t ? t = 
1 2
2( )
?
? ? ?
. ?
18. Given that, t = 0.02 mm = 0.02 ? 10
–3
m, ?
1
= 1.45, ? = 600 nm = 600 ? 10
–9
m
a) Let, I
1
= Intensity of source without paper = I
b) Then I
2
= Intensity of source with paper = (4/9)I
?
1 1
2 2
I r 9 3
I 4 r 2
? ? ? [because I ? r
2
]
where, r
1
and r
2
are corresponding amplitudes.
So, 
2
max 1 2
2
min 1 2
I (r r )
I (r r )
?
?
?
= 25 : 1
b) No. of fringes that will cross the origin is given by,
n = 
( 1)t ? ?
?
= 
3
9
(1.45 1) 0.02 10
600 10
?
?
? ? ?
?
= 15.
19. Given that, d = 0.28 mm = 0.28 ? 10
–3
m, D = 48 cm = 0.48 m, ?
a
= 700 nm in vacuum 
Let, ?
w
= wavelength of red light in water
Since, the fringe width of the pattern is given by,
??= 
9
w
3
D 525 10 0.48
d 0.28 10
?
?
? ? ?
?
?
= 9 ? 10
–4
m = 0.90 mm. ?
20. It can be seen from the figure that the wavefronts reaching O from S
1
and S
2
will
have a path difference of S
2
X.
In the ? S
1
S
2
X,
sin ???= 
2
1 2
S X
S S
So, path difference = S
2
X = S
1
S
2
sin ? = d sin ? = d ? ?/2d = ?/2
As the path difference is an odd multiple of ?/2, there will be a dark fringe at point P
0
.
21. a) Since, there is a phase difference of ? between direct light and 
reflecting light, the intensity just above the mirror will be zero.
b) Here, 2d = equivalent slit separation 
D = Distance between slit and screen.
We know for bright fringe, ?x = 
y 2d
D
?
= n ?
But as there is a phase reversal of ?/2.
?
y 2d
D
?
+ 
2
?
= n ? ?
y 2d
D
?
= n ? –
2
?
? y = 
D
4d
?
?
mica
Screen
polysterene
S 2
S 1
? ?
P 0 ? ?
x
S 2
S 1
Dark
fringe
(1 – 0.43) ??
0.43 ??
S 2
S 1
2d
D
Screen 
Chapter 17
17.4
22. Given that, D = 1 m, ? = 700 nm = 700 ? 10
–9
m
Since, a = 2 mm, d = 2a = 2mm = 2 ? 10
–3
m (L loyd’s mirror experiment)
Fringe width = 
9
3
D 700 10 m 1 m
d 2 10 m
?
?
? ? ?
?
?
= 0.35 mm.
23. Given that, the mirror reflects 64% of energy (intensity) of the light.
So, 
1 1
2 2
I r 16 4
0.64
I 25 r 5
? ? ? ?
So, 
2
max 1 2
2
min 1 2
I (r r )
I (r r )
?
?
?
= 81 : 1.
24. It can be seen from the figure that, the apparent distance of the screen from the slits is, 
D = 2D
1
+ D
2
So, Fringe width = 
1 2
(2D D ) D
d d
? ? ?
?
25. Given that, ? = (400 nm to 700 nm), d = 0.5 mm = 0.5 ? 10
–3
m, 
D = 50 cm = 0.5 m and on the screen y
n
= 1 mm = 1 ? 10
–3
m
a) We know that for zero intensity (dark fringe)
y
n
= 
n
D 2n 1
2 d
? ? ? ?
? ?
? ?
where n = 0, 1, 2, …….
? ?
n
= 
3 3
6 3 n
d 2 2 10 0.5 10 2 2
10 m 10 nm
(2n 1) D 2n 1 0.5 (2n 1) (2n 1)
? ?
?
? ? ?
? ? ? ? ? ?
? ? ? ?
If n = 1, ?
1
= (2/3) ? 1000 = 667 nm
If n = 1, ?
2
= (2/5) ? 1000 = 400 nm
So, the light waves of wavelengths 400 nm and 667 nm will be absent from the out coming light.
b) For strong intensity (bright fringes) at the hole
? y
n
= 
n n
n
n D y d
d nD
?
? ? ?
When, n = 1, ?
1
= 
n
y d
D
= 
3 3
6
10 0.5 10
10 m 1000nm
0.5
? ?
?
? ?
? ? .
1000 nm is not present in the range 400 nm – 700 nm
Again, where n = 2, ?
2
= 
n
y d
2D
= 500 nm
So, the only wavelength which will have strong intensity is 500 nm. ?
26. From the diagram, it can be seen that at point O.
Path difference = (AB + BO) – (AC + CO) 
= 2(AB – AC) [Since, AB = BO and AC = CO] = 
2 2
2( d D D) ? ?
For dark fringe, path difference should be odd multiple of ?/2.
So, 
2 2
2( d D D) ? ? = (2n + 1)( ?/2)
?
2 2
d D ? = D + (2n + 1) ?/4 
? D
2
+ d
2
= D
2
+ (2n+1)
2
?
2
/16 + (2n + 1) ?D/2
Neglecting, (2n+1)
2
?
2
/16, as it is very small
We get, d = 
D
(2n 1)
2
?
?
For minimum ‘d’, putting n = 0 ? d
min
= 
D
2
?
. ?
C
O
P
B
d
A
x
D
D
D
y
n
d=0.5mm
50cm
1 mm
Page 5


17.1
SOLUTIONS TO CONCEPTS 
CHAPTER 17
1. Given that, 400 m < ? < 700 nm.
1 1 1
700nm 400nm
? ?
?
?
8 8
7 7 7 7
1 1 1 3 10 c 3 10
7 10 4 10 7 10 4 10
? ? ? ?
? ?
? ? ? ? ?
? ? ? ? ? ?
(Where, c = speed of light = 3 ? 10
8
m/s)
? 4.3 ? 10
14
< c/ ? < 7.5 ? 10
14
? 4.3 ? 10
14
Hz < f < 7.5 ? 10
14
Hz. ?
2. Given that, for sodium light, ? = 589 nm = 589 ? 10
–9
m
a) f
a
= 
8
9
3 10
589 10
?
?
?
= 5.09 ? 10
14 1
c
sec f
?
? ?
?
? ?
? ? ?
?
b)
a w w
w
9
w a
1
1.33 589 10
?
? ? ?
? ? ? ? ?
? ? ?
= 443 nm
c) f
w
= f
a
= 5.09 ? 10
14
sec
–1
[Frequency does not change]
d)
8
a a a w
w
w a w
v v 3 10
v
v 1.33
? ? ?
? ? ? ?
? ?
= 2.25 ? 10
8
m/sec.
3. We know that, 
2 1
1 2
v
v
?
?
?
So, 
8
8
400
400
1472 3 10
v 2.04 10 m/ sec.
1 v
?
? ? ? ?
[because, for air, ? = 1 and v = 3 ? 10
8
m/s]
Again, 
8
8
760
760
1452 3 10
v 2.07 10 m/ sec.
1 v
?
? ? ? ? ?
4.
8
t
8
1 3 10
1.25
(2.4) 10
? ?
? ? ?
?
velocity of light in vaccum
since,  = 
velocity of light in the given medium
? ?
?
? ?
? ?
5. Given that, d = 1 cm = 10
–2
m, ? = 5 ? 10
–7
m and D = 1 m
a) Separation between two consecutive maxima is equal to fringe width.
So, ? = 
7
2
D 5 10 1
d 10
?
?
? ? ?
? m = 5 ? 10
–5
m = 0.05 mm.
b) When, ? = 1 mm = 10
–3
m
10
–3
m = 
7
5 10 1
D
?
? ?
? D = 5 ? 10
–4
m = 0.50 mm. ?
6. Given that, ? = 1 mm = 10
–3
m, D = 2.t m and d = 1 mm = 10
–3
m
So, 10
–3
m = 
3
25
10
?
? ?
? ? = 4 ? 10
–7
m = 400 nm. ?
7. Given that, d = 1 mm = 10
–3
m, D = 1 m.
So, fringe with = 
D
d
?
= 0.5 mm.
a) So, distance of centre of first minimum from centre of central maximum = 0.5/2 mm = 0.25 mm
b) No. of fringes = 10 / 0.5 = 20.
8. Given that, d = 0.8 mm = 0.8 ? 10
–3
m, ? = 589 nm = 589 ? 10
–9
m and D = 2 m.
So, ? = 
D
d
?
= 
9
3
589 10 2
0.8 10
?
?
? ?
?
= 1.47 ? 10
–3
m = 147 mm. ?
Chapter 17
17.2
9. Given that, ? = 500 nm = 500 ? 10
–9
m and d = 2 ? 10
–3
m
As shown in the figure, angular separation ? = 
D
D dD d
? ? ?
? ?
So, ? = 
9
3
500 10
D d 2 10
?
?
? ? ?
? ?
?
= 250 ? 10
–6
= 25 ? 10
–5
radian = 0.014 degree. ?
10. We know that, the first maximum (next to central maximum) occurs at y = 
D
d
?
Given that, ?
1
= 480 nm, ?
2
= 600 nm, D = 150 cm = 1.5 m and d = 0.25 mm = 0.25 ? 10
–3
m
So, y
1
= 
9
1
3
D 1.5 480 10
d 0.25 10
?
?
? ? ?
?
?
= 2.88 mm
y
2 
= 
9
3
1.5 600 10
0.25 10
?
?
? ?
?
= 3.6 mm.
So, the separation between these two bright fringes is given by,
? separation = y
2
– y
1
= 3.60 – 2.88 = 0.72 mm.
11. Let m
th
bright fringe of violet light overlaps with n
th
bright fringe of red light.
?
m 400nm D n 700nm D m 7
d d n 4
? ? ? ?
? ? ?
? 7
th
bright fringe of violet light overlaps with 4
th
bright fringe of red light (minimum). Also, it can be 
seen that 14
th
violet fringe will overlap 8
th
red fringe.
Because, m/n = 7/4 = 14/8.
12. Let, t = thickness of the plate
Given, optical path difference = ( ? – 1)t = ?/2
? t = 
2( 1)
?
? ?
?
13. a) Change in the optical path = ?t – t = ( ? – 1)t
b) To have a dark fringe at the centre the pattern should shift by one half of a fringe.
? ( ? – 1)t = t
2 2( 1)
? ?
? ?
? ?
. ?
14. Given that, ? = 1.45, t = 0.02 mm = 0.02 ? 10
–3
m and ? = 620 nm = 620 ? 10
–9
m
We know, when the transparent paper is pasted in one of the slits, the optical path changes by ( ? – 1)t.
Again, for shift of one fringe, the optical path should be changed by ?.
So, no. of fringes crossing through the centre is given by,
n = 
3
9
( 1)t 0.45 0.02 10
620 10
?
?
? ? ? ?
?
? ?
= 14.5 ?
15. In the given Young’s double slit experiment, 
? = 1.6, t = 1.964 micron = 1.964 ? 10
–6
m
We know, number of fringes shifted = 
( 1)t ? ?
?
So, the corresponding shift = No.of fringes shifted ? fringe width
= 
( 1)t D ( 1)tD
d d
? ? ? ? ?
? ?
?
… (1)
Again, when the distance between the screen and the slits is doubled,
Fringe width = 
(2D)
d
?
…(2)
From (1) and (2), 
( 1)tD
d
? ?
= 
(2D)
d
?
? ? = 
( 1)t ? ?
?
= 
6
(1.6 1) (1.964) 10
2
?
? ? ?
= 589.2 ? 10
–9
= 589.2 nm. ?
B
S 1
S 2
? ?
D
Chapter 17
17.3
16. Given that, t
1
= t
2
= 0.5 mm = 0.5 ? 10
–3
m, ?
m
= 1.58 and ?
p
= 1.55, 
? = 590 nm = 590 ? 10
–9
m, d = 0.12 cm = 12 ? 10
–4
m, D = 1 m
a) Fringe width = 
9
4
D 1 590 10
d 12 10
?
?
? ? ?
?
?
= 4.91 ? 10
–4
m.
b) When both the strips are fitted, the optical path changes by 
?x = ( ?
m
– 1)t
1
– ( ?
p
– 1)t
2
= ( ?
m
– ?
p
)t
= (1.58 – 1.55) ? (0.5)(10
–3
) = 0.015 ? 10
–13
m.
So, No. of fringes shifted = 
3
3
0.015 10
590 10
?
?
?
?
= 25.43.
? There are 25 fringes and 0.43 th of a fringe.
? There are 13 bright fringes and 12 dark fringes and 0.43 th of a dark fringe.
So, position of first maximum on both sides will be given by
? x = 0.43 ? 4.91 ? 10
–4
= 0.021 cm
x ? = (1 – 0.43) ? 4.91 ? 10
–4
= 0.028 cm (since, fringe width = 4.91 ? 10
–4
m)
17. The change in path difference due to the two slabs is ( ?
1
– ?
2
)t (as in problem no. 16).
For having a minimum at P
0
, the path difference should change by ?/2.
So, ? ?/2 = ( ?
1
– ??
2
)t ? t = 
1 2
2( )
?
? ? ?
. ?
18. Given that, t = 0.02 mm = 0.02 ? 10
–3
m, ?
1
= 1.45, ? = 600 nm = 600 ? 10
–9
m
a) Let, I
1
= Intensity of source without paper = I
b) Then I
2
= Intensity of source with paper = (4/9)I
?
1 1
2 2
I r 9 3
I 4 r 2
? ? ? [because I ? r
2
]
where, r
1
and r
2
are corresponding amplitudes.
So, 
2
max 1 2
2
min 1 2
I (r r )
I (r r )
?
?
?
= 25 : 1
b) No. of fringes that will cross the origin is given by,
n = 
( 1)t ? ?
?
= 
3
9
(1.45 1) 0.02 10
600 10
?
?
? ? ?
?
= 15.
19. Given that, d = 0.28 mm = 0.28 ? 10
–3
m, D = 48 cm = 0.48 m, ?
a
= 700 nm in vacuum 
Let, ?
w
= wavelength of red light in water
Since, the fringe width of the pattern is given by,
??= 
9
w
3
D 525 10 0.48
d 0.28 10
?
?
? ? ?
?
?
= 9 ? 10
–4
m = 0.90 mm. ?
20. It can be seen from the figure that the wavefronts reaching O from S
1
and S
2
will
have a path difference of S
2
X.
In the ? S
1
S
2
X,
sin ???= 
2
1 2
S X
S S
So, path difference = S
2
X = S
1
S
2
sin ? = d sin ? = d ? ?/2d = ?/2
As the path difference is an odd multiple of ?/2, there will be a dark fringe at point P
0
.
21. a) Since, there is a phase difference of ? between direct light and 
reflecting light, the intensity just above the mirror will be zero.
b) Here, 2d = equivalent slit separation 
D = Distance between slit and screen.
We know for bright fringe, ?x = 
y 2d
D
?
= n ?
But as there is a phase reversal of ?/2.
?
y 2d
D
?
+ 
2
?
= n ? ?
y 2d
D
?
= n ? –
2
?
? y = 
D
4d
?
?
mica
Screen
polysterene
S 2
S 1
? ?
P 0 ? ?
x
S 2
S 1
Dark
fringe
(1 – 0.43) ??
0.43 ??
S 2
S 1
2d
D
Screen 
Chapter 17
17.4
22. Given that, D = 1 m, ? = 700 nm = 700 ? 10
–9
m
Since, a = 2 mm, d = 2a = 2mm = 2 ? 10
–3
m (L loyd’s mirror experiment)
Fringe width = 
9
3
D 700 10 m 1 m
d 2 10 m
?
?
? ? ?
?
?
= 0.35 mm.
23. Given that, the mirror reflects 64% of energy (intensity) of the light.
So, 
1 1
2 2
I r 16 4
0.64
I 25 r 5
? ? ? ?
So, 
2
max 1 2
2
min 1 2
I (r r )
I (r r )
?
?
?
= 81 : 1.
24. It can be seen from the figure that, the apparent distance of the screen from the slits is, 
D = 2D
1
+ D
2
So, Fringe width = 
1 2
(2D D ) D
d d
? ? ?
?
25. Given that, ? = (400 nm to 700 nm), d = 0.5 mm = 0.5 ? 10
–3
m, 
D = 50 cm = 0.5 m and on the screen y
n
= 1 mm = 1 ? 10
–3
m
a) We know that for zero intensity (dark fringe)
y
n
= 
n
D 2n 1
2 d
? ? ? ?
? ?
? ?
where n = 0, 1, 2, …….
? ?
n
= 
3 3
6 3 n
d 2 2 10 0.5 10 2 2
10 m 10 nm
(2n 1) D 2n 1 0.5 (2n 1) (2n 1)
? ?
?
? ? ?
? ? ? ? ? ?
? ? ? ?
If n = 1, ?
1
= (2/3) ? 1000 = 667 nm
If n = 1, ?
2
= (2/5) ? 1000 = 400 nm
So, the light waves of wavelengths 400 nm and 667 nm will be absent from the out coming light.
b) For strong intensity (bright fringes) at the hole
? y
n
= 
n n
n
n D y d
d nD
?
? ? ?
When, n = 1, ?
1
= 
n
y d
D
= 
3 3
6
10 0.5 10
10 m 1000nm
0.5
? ?
?
? ?
? ? .
1000 nm is not present in the range 400 nm – 700 nm
Again, where n = 2, ?
2
= 
n
y d
2D
= 500 nm
So, the only wavelength which will have strong intensity is 500 nm. ?
26. From the diagram, it can be seen that at point O.
Path difference = (AB + BO) – (AC + CO) 
= 2(AB – AC) [Since, AB = BO and AC = CO] = 
2 2
2( d D D) ? ?
For dark fringe, path difference should be odd multiple of ?/2.
So, 
2 2
2( d D D) ? ? = (2n + 1)( ?/2)
?
2 2
d D ? = D + (2n + 1) ?/4 
? D
2
+ d
2
= D
2
+ (2n+1)
2
?
2
/16 + (2n + 1) ?D/2
Neglecting, (2n+1)
2
?
2
/16, as it is very small
We get, d = 
D
(2n 1)
2
?
?
For minimum ‘d’, putting n = 0 ? d
min
= 
D
2
?
. ?
C
O
P
B
d
A
x
D
D
D
y
n
d=0.5mm
50cm
1 mm
Chapter 17
17.5
27. For minimum intensity 
? S
1
P – S
2
P = x = (2n +1) ?/2
From the figure, we get
?
2 2
Z (2 ) Z (2n 1)
2
?
? ? ? ? ?
?
2
2 2 2 2
Z 4 Z (2n 1) Z(2n 1)
4
?
? ? ? ? ? ? ? ?
? Z = 
2 2 2 2 2 2
4 (2n 1) ( / 4) 16 (2n 1)
(2n 1) 4(2n 1)
? ? ? ? ? ? ? ?
?
? ? ? ?
…(1)
Putting, n = 0 ? Z = 15 ?/4 n = –1 ? Z = –15 ?/4
n = 1 ? Z = 7 ?/12 n = 2 ? Z = –9 ?/20
? Z = 7 ?/12 is the smallest distance for which there will be minimum intensity. ?
28. Since S
1
, S
2
are in same phase, at O there will be maximum intensity.
Given that, there will be a maximum intensity at P.
? path difference = ?x = n ?
From the figure,
(S
1
P)
2
– (S
2
P)
2
= 
2 2 2 2 2 2
( D X ) ( (D 2 ) X ) ? ? ? ? ?
= 4 ?D – 4 ?
2
= 4 ?D ( ?
2
is so small and can be neglected)
? S
1
P – S
2
P = 
2 2
4 D
2 x D
?
?
= n ?
??
2 2
2D
x D ?
?????
? n
2
(X
2
+ D
2
) = 4D
2
= ?X = 
2
D
4 n
n
?
when n = 1, x = 3 D (1
st
order)
n = 2, x = 0 (2
nd
order)
? When X = 3 D, at P there will be maximum intensity. ?
29. As shown in the figure, 
(S
1
P)
2
= (PX)
2
+ (S
1
X)
2
…(1)
(S
2
P)
2
= (PX)
2
+ (S
2
X)
2
…(2)
From (1) and (2), 
(S
1
P)
2
– (S
2
P)
2
= (S
1
X)
2
– (S
2
X)
2
= (1.5 ? + R cos ?)
2
– (R cos ? – 15 ?)
2
= 6 ? R cos ?
? (S
1
P – S
2
P) = 
6 Rcos
2R
? ?
= 3 ? cos ?.
For constructive interference, 
(S
1
P – S
2
P)
2
= x = 3 ? cos ? = n ?
? cos ? = n/3 ? ? = cos
–1
(n/3), where n = 0, 1, 2, ….
? ? = 0°, 48.2°, 70.5°, 90° and similar points in other quadrants. ?
30. a) As shown in the figure, BP
0
– AP
0
= ?/3
?
2 2
(D d ) D / 3 ? ? ? ?
? D
2
+ d
2
= D
2
+ ( ?
2
/ 9) + (2 ?D)/3
? d = (2 D)/ 3 ? (neglecting the term ?
2
/9 as it is very small)
b) To find the intensity at P
0
, we have to consider the interference of light 
waves coming from all the three slits.
Here, CP
0
– AP
0
= 
2 2
D 4d D ? ?
S 2
Z
S 1
P
2 ??
Screen
P 0
x
d 
d 
D 
C
B 
A 
Screen
x
2 ? ?
O
S 2
S 1
P
D
? ?
x
S 1
1.5 ?? O
P
S 2
R