Class 11-science H C VERMA Solutions Physics Chapter 17: Light Waves
Light Waves Exercise 380
Solution 1
Hz
Hz
So, Range of frequency
Solution 2
(a)
Hz
(b) 
nm
(c) 
[∵frequency
depends only upon source, not an medium]
Hz
(d)
m/s
Solution 3
For 
m/s
For 
m/s
Solution 4
1.25
Solution 5
(a) 
D=1m
0.05mm
(b) 
0.5mm
Light Waves Exercise 381
Solution 6
m
λ = 400 nm
Solution 7
(a)
distance=
=0.25 mm
(b) Fringe
width=
mm
No. of
fringes obtained in 1 cm=
=20
fringes.
Solution 8
Distance between consecutive bright
=
m
=1.47 mm
Solution 9
Angular
fringe width, 
Solution 10
Distance of first maxima from
center=
Linear separation between first
maxima, 
mm
Solution 11
Central maxima for both wavelength forms at same position at centre of screen.
Let mth maxima
of violet coincides with 
maxima of red light.
Least value of m is 7.
Solution 12
Solution 13
(a) Change
in optical path=
(b) 
For minimum thickness, D-1 will be formed.
Solution 14
Hence, 14.5 fringes will shift.
Solution 15
Distance moved by fringes after introduction of slab
Now, distance of screen is doubled, new fringe width
Given,
589.2
nm
Solution 16
(a)
m
(b) No. of fringes shifted
Path difference due to sheets,
m
25 Fringes will be shifted completely
Position of first maxima on one side= 0.43β
=0.021 cm
and on other side=0.57β
0.028 cm
Solution 17
Path difference at centre of screen
Minima is formed at the centre of screen
For minimum thickness, n=1
Solution 18
(a) 
(b) Initially path difference at
center of screen=
So, B-15 is formed.
Now, when identical paper piece is pasted, path difference at center will be zero. So, B-0 will be formed.
Hence, 15 fringes will shift.
Solution 19
Fringe width
m
=0.9
mm
Solution 20
Path difference between waves meeting at center of screen will be due to pre-slit path difference.
So, dark fringe is obtained at centre of screen.
Solution 21
(a) The phase of a light wave reflecting from plane mirror will change by π.
At point just
above O, the 
and
So, minima is formed i.e., zero intensity.
(b) Let y distance above O, maxima is obtained at point P.
For maxima, 
Put n=1 for 
Light Waves Exercise 382
Solution 22
Fringe width,
0.35
mm
Solution 23
Intensity of
light from source 
Intensity of
light after reflection 
Solution 24
Distance between slits=d
Distance between apparent sources and screen
=
=
Fringe
width
Solution 25
(a) The wavelength which forms dark (minima) at hole will be absent.
nm
For n=1, λ=2000 nm
For n=2, λ=667 nm
For n=3, λ=400 nm
Thus, 667 nm and 400 nm are absent in the range of (400 nm-700 nm).
(b) The wavelength which forms bright (maxima) will have strong intensity.
(in nm)
For n=1; λ=1000 nm
For n=2; λ=500 nm
For n=3; λ=334 nm
So, in Range of (400-700 nm) , light of wavelength 500 nm will have maxima.
Solution 26
(a) For minima at O
By binomial expansion
For minima d, n=1
(b) At point O, D-1 is formed
Bright fringe next to D-1 will be B-0.
So, point at which path difference is zero B-0 will be formed.
By symmetry the point is in front of slit B.
So, minimum distance x=d.
(c) distance between B-0 and D-1 is d
Solution 27
Let point P at distance x will have minimum intensity i.e., D-2
Path difference=
Solution 28
(a) Path
difference 
By binomial expansion
(b) Path difference between wave from slit A and C is
By binomial expansion
By superposition principle
Thus, the resultant intensity is three times the intensity due to the individual slits.
Solution 29
For point at position, y=0.5cm
Path
difference=
m
Phase
difference, 
Solution 30
(a) 
(b)
Solution 31
m
Solution 32
Distance of the point from central maxima where intensity falls by half.
Line-width= 2y
Light Waves Exercise 383
Solution 33
(a) 
At β/2 distance from central maxima, minima will be obtained.
So, intensity from S4=0
(b)
At 
distance
from central maxima, first order bright will be obtained.
So,
intensity from 
(c) 
So, intensity at 
Now,
Solution 34
(a)
Distance of
slit 
and 
from perpendicular bisector is y=
.
Path
difference from 
and
wave at =
.
Similarly,
at path difference=
i.e., dark
fringes are formed at and .
So,
intensity of light at and are zero.
Hence, intensity at P is also zero.
(b) 
Distance of
slit and from perpendicular bisector is y=
.
Path
difference from and wave at =
.
Similarly,
at path difference=
Phase
difference at and will be=
When ,
i.e., distance from perpendicular bisector is 
intensity was I at
point P.
At phase difference 
,
intensity at point P is I.
(c) 
Distance of
slit and from perpendicular bisector is y=.
So, at and ,
bright fringe will be formed.
Intensity at P is 2I.
Solution 35
Minima is obtained in reflected light
n
For refractive index between 1.2 and 1.5, put n=5
1.32
Solution 36
Maxima is obtained in reflected light.
for 
, put π=0
100
nm
Solution 37
Maxima in reflected light
For n=4, λ=666 nm
For n=5, λ=532 nm
For n=6, λ=443 nm
Solution 38
Condition for maxima in reflected light
nm
For n=3, λ=714 nm
For n=4, λ=556 nm
For n=5, λ=455 nm
Solution 39
For minima, in diffraction
λ=2.5 cm
Solution 40
m
R=0.683 cm
Diameter of central bright spot=2R
=2(0.683)
=1.37 cm
Solution 41
m
∴ Diameter of the central bright spot, 
m.