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 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
Read More
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FAQs on HC Verma Solutions: Chapter 17 - Light Waves - Physics Class 11 - NEET

1. What is the concept of light waves in physics?
Ans. Light waves in physics refer to the electromagnetic waves that can be seen by the human eye. These waves have both electric and magnetic fields oscillating perpendicular to each other, and they travel at the speed of light.
2. How are light waves produced?
Ans. Light waves are produced when an object or substance emits or reflects light. This can occur through various processes such as incandescence, fluorescence, phosphorescence, or by the interaction of light with a medium or material.
3. What are the properties of light waves?
Ans. Light waves possess several properties, including wavelength, frequency, amplitude, and speed. The wavelength determines the color of light, the frequency determines the number of waves per unit time, the amplitude determines the intensity or brightness of light, and the speed is always constant in a vacuum.
4. How do light waves interact with matter?
Ans. When light waves interact with matter, they can be absorbed, reflected, or transmitted. The interaction depends on the properties of the material, such as its transparency, reflectivity, and absorptive abilities. Different materials interact differently with light waves, leading to various phenomena such as refraction, diffraction, and polarization.
5. What are the applications of light waves?
Ans. Light waves have numerous applications in various fields. They are used in telecommunications for transmitting information through optical fibers. Light waves are also utilized in photography, microscopy, astronomy, and laser technology. Additionally, they play a crucial role in the functioning of the human eye, allowing us to see the world around us.
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