Page 1 of 5 CPP - SANKALP_WO–7-PH-II

CPP
WAVE OPTICS - SHEET: 7 (PRACTICE SHEET-1)

1. Four identical coherent sound sources S1, S2, S3 & S4 are placed
collinearly at a separation d from each other as shown in the figure.
Each one can produce sound of intensity I and wavelength . When d d d
the separation d is equal to minimum possible value d1, no sound S1 S2 S3 S4 P
was detected at a point P on the line joining the sources. But when
the separation d is equal to minimum possible value d2, the intensity
of sound at P is 3I. Find the value of (4d1 – 6d2)

2. Consider a plane inclined at an angle 45 with the horizontal

has two slits (S1 and S2) separated by a distance d = 2 mm. S2
The screen is placed at a distance of D = 10 m. A parallel
monochromatic light beam of wavelength 5000 Å is incident 10 meter
S1
on the slits as shown. If the fringe width of interference
pattern on the screen is k  103 meter, then find the value of
k.
45

screen

3. Figure shows wave fronts coming form a

10
source at equal interval of milli second.
3
Frequency of wave emitted by source is 300
Hz. XX’ is line along common diameter of
wave fronts passing through ‘A’. Distance
between two consecutive wave fronts along
line XX’ is 0.9 m and 1.3 m in the right and left
of point A respectively. At some instant
source is at point A. Detector is placed at point
‘Q’. AQ makes an angle 60° with line XX’ If
frequency of wave received by detector is f
f
Hz, then (in Hz) is equal to
110

4. Two mirrors M1 and M2 make an angle  with line AB. A point M1

b Surface
source S is kept at a distance r from the point of intersection of  Sc
mirrors. A small hemispherical annular mask is kept close to S so r b
that no ray emanating from S directly reaches the screen kept at  S
a distance b from the source. Interference pattern of rays hemispherical
reflected from M1 and M2 is observed on the screen. Find the annular mask

fringe width of this interference pattern.  = wavelength of light

being emitted by source.
5. A parallel beam of wavelength  = 4000 A falls on a Young’s Screen

double slit apparatus. The slits and the screen are kept fixed and
a converging lens is moved between the prism and the screen. S1
For two positions of the lens (between the slits and the screen),
we get two sharp images of slits on the screen in each case. The
images are separated from each other by distance 4.5 mm in one S2
case and 2 mm in the other. If the lens is removed and D = 150 cm
interference pattern is observed on the screen. Then
(a) Find the fringe width of the pattern on the screen.
(b) The focal length of the lens.

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Page 2 of 5 CPP - SANKALP_WO–7-PH-II
6. A point light source is situated at a distance of 5 mm away from plane
mirror and at a distance 1 m from the screen as shown in the figure. If the S P
1m
wavelength of the source is  = 5000 A, then find the answer of following 5 mm
questions 1 cm
(a) Find the number of bright fringes formed on the screen (approximately).
(b) If intensity of the source is I0 at a distance of ½ m from the source, then
the intensity of fringe formed at point P is (approximately)
(c) Find the fringe width (approximately).

7. ABC is a spherical wavefront centred at O symmetric about BE is A S

incident on slits S1 & S2. BS1 = 3, S1S2 = 4, BO = 6, S1E = 128 S2
and  is the wavelength of incident wave. A mica sheet of refractive
index 1.5 is pasted on S2. Find the minimum value of thickness of
B E
mica sheet for which central fringe forms at E. S1 O

C
8. Two light rays going through different media interfere at a point. One of the rays is reflected by a
mirror before the interference. The phase difference between the rays will depend on
(A) path lengths of the rays (B) the media
(C) the reflection (D) None of these

PARAGRAPH -1
In a young double slit experiment. The two slits are illuminated by a
monochromatic light source S of wavelength  = 500 nm. Distance 1 P
S S1
between slits and screen is D = 2m and distance between two slits
S1S2  10 m. 1 = 10 m and 2 = 30 m, as shown in the figure. Answer d O C
2
the following questions on the basis above data.
S2

9. The value of  relative to the central line OC, where maxima appear on the screen
  n    n 
(A)   sin1  2   1  (B)   sin1  2   1 
  40     20  
 n   n 
(C)   sin1   1  (D)   sin1   1 
  40    20 

10. How many maxima will appear on the screen

(A) 80 (B) 60 (C) 40 (D) 20

11. What should be minimum thickness of a slab of refractive index  = 1.5 will be placed on the path of
one ray so that minima occurs at C?
(A) 1000 nm (B) 250 nm (C) 1500 nm (D) 500 nm

PARAGRAPH -2
Interference from light reflected by step structures or partially overlapping scale produces the iridescent
colours seen in many butterflies, moths, birds and fish. A stunning example is the shimmering blue of the
MORPHO butterfly. Figure shows the MORPHO wing as viewed under an electron microscope.
1
2
 = 1.5 t1 air = 1
t2 Air
fig(a)  = 1.5
fig(b)
fig(c)

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Page 3 of 5 CPP - SANKALP_WO–7-PH-II
The tree like structure that projected up from the top surface of the wing [in the figure (a)] are made of a
transparent material. Light is thus reflected from the tops of successive steps thickness t1, with spacing t2
between the step where the index of refraction is  = 1.5 [in figure (c)]. Light is assumed to incident
normally on the wings and reflections and refraction are shown in the figure (c)

12. What is the optical path difference (in air) between the ray 1 and ray 2 as shown in figure (c).
(A) 2t1 + 2t2 (B) 3t1 + 2t2 (C) 2t1 + 3t2 (D) 3t1 + 3t2

13. Choose the correct option for a relation between the constructively interfering wavelength 0 (in air)
and the optical path difference in air
0 0
(A) 2t1 + 2t2 = (2n  1) , where n = 0, 1, 2, … (B) 3t1  2t 2  (2n) , where n = 1, 2, …….
2 2
 
(C) 3t1  2t 2  (2n  1) 0 , where n = 0, 1, 2, … (D) 2t1  2t 2  (2n) 0 , where n = 1, 2, …….
2 2

14. For a typical MORPHO butterfly t1 and t2 are nearly equal to 70 nm and 130 nm respectively. Then
choose the dominant colour of the wing, when we see when looking at the butterfly wing at normal
incidence.
Colour Wavelength(nm)
Red 620-780
Orange 590-620
Yellow 570-590
Green 500-570
Blue 450-500
Violet 350-450

(A) Red (B) Yellow (C) Blue (D) Violet

PARAGRAPH-3
In an YDSE setup, a point source of light moves in a circle in the y- y
z plane having centre at the origin and radius d/2. The angular
S1
velocity  of the source is constant and at time t = 0, the S
P
coordinates of the source are (0, 0, d/2). The card board having
x
slits S1 and S2 is located in the plane x = D. The coordinates of the S2
slits are (D, d/2, z) and (D, d/2, z). Assume D >> d and
D 2D
wavelength of the light is . Neglecting the Doppler effect of light, z
answer the following question.

15. The central fringe will execute

(A) SHM (B) Oscillatory but not SHM
(C) circular motion (D) elliptical motion.

16. The velocity of the central fringe at time t = / is

d
(A) d ĵ (B)  ĵ
2
d
(C)  ĵ (D) impossible to calculate from the given data
2
 
17. The number of fringes that will cross (3D, 0, 0) from the instant to is equal to
6 2
d2 d2 d2
(A) (B) (C) (D) None of these
D 2D 4D

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Page 4 of 5 CPP - SANKALP_WO–7-PH-II
WAVE OPTICS - SHEET: 7 (PRACTICE SHEET-1)
ANSWERS
1. 0
For d1 phase difference between two sound sources should be 90o and for d2 phase difference
should be 60o
 2    2  
   d1  and   d2 
   2    3
(4d1 – 6d2) = 0

2. 5
Path difference = S1N  MS2 P
d M S2
y
= dsin(45  ) 
2 N

 cos  sin   d 
 d  S1
 10 meter
 2 2  2
 1   d
n  d    (as ‘’ is small)
 2 2 2 45
2n screen

d
2nD
y ,
d
Fringe width =
2D
yn1  yn   5  10 3 m
d

3. 3
 0 – Vs t = x 1
0 + V s t = x 2
x 2 – x1 x x
Vs = = 60 m/s & 0 = 1 2 = 1.1 m
2T 2
Speed of sound V = n × 0 = 330 m/s
Frequency detected at point Q
V
n=  300Hz  330Hz
V – Vs cos 60
4. ?
5. (a) 0.2 mm.
(b) 36 cm
u + v = 150 cm
d u 
In case I 2     4.5 mm
2 v 
d v 
In case II 2     2 mm
2u
On solving u = 90 cm, v = 60 cm and d = 3mm
So, f = 36 cm and  = 0.2 mm
6. (a) 4  104
(b) zero

(c) 5  105 m

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Page 5 of 5 CPP - SANKALP_WO–7-PH-II
31
7.
8
Let S1 and S2 are the points on the wavefront where perpendiculars S2
can be drawn from S1 and S2 S2
S 2 S2 + (  1)t + S2E = S1 S1 + S1E d

 (O S2  OS2) + (  1)t + D  d =  OS1  OS1  + D

2 2
S1 S1 O E

 (  1)t = (OS2  OS1)   D  d  D  2 2 D

C
d2
= 2  [Bionomial approximation)
2D
31
 t=
8
8. A, B, C

9. A
Optical path difference between the beans arriving at P
x  (  2   1 )  dsin 
for maxima
( 2   1 )  dsin   n
1
sin   n   (  2   1 ) 
d
  n 
  sin1  2   1 
  40 
10. C
sin   1
n 
1  2   1  1
 40 
or 20  n  60
Hence number of maxima = 60 – 20 = 40

11. D
At C, phase difference
2
 (  2   1 ) = 80 

Hence for maximum intensity will appear at C.
Now for minimum intensity at C

(  1)t 
2
t
t  500 nm
2(  1)
12. (B)
13. (B)
14. (C)
15. A
y-coordinate of the source at any time t is y = (d/2) sin t
16. A
The amplitude A of the fringe is given by
A d2
 Ad
2D D

17. C

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