3.wave Optics Final - PMD

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VOL--VI WAVE OPTICS
WAVE OPTICS
3. A screen is at a distance D=80cm from a
LEVEL - V diaphragm having two narrow slits S1 and
SINGLE CORRECT ANSWER TYPE S 2 which are d  2mm apart. Slit S1 is

covered by a transparent sheet of thickness
1. The YDSE apparatus is as shown in figure t1  2.5 m and slit S 2 is covered by another
below. The condition for point ‘P’ to be a
sheet of thickness t2  1.25 m as shown in
dark fringe is [  = wavelength of light used]
figure. Both sheets are made of same
material having refractive index   1.4 .
Screen Water is filled in the space between the
S1 diaphragm and screen. A monochromatic
light beam of wavelength   5000 is
l1 l3 Å
incident normally on the diaphragm.
Assuming intensity of beam to be uniform
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calcualte ratio of intensity at C to the

S P maximum intensity of interference patern
l2 S 2 l4
obtained on the screen?    4 3 
(A)  l1  l3    l2  l4   n
(B)  l1  l2    l3  l4   n S1 t1

(C)  l1  l3    l2  l4    2n  1
2 C
 S2
(D)  l1  l2    l3  l4    2n  1 t2
2
2. Intensity observed in an interference pattern
I  I 0 sin 2  . At  30o , intensity
I  5  0.002 . The percentage error in angle
3
if I 0  20 w / m 2 is (A)
4
(A) 4 3  10 2 % 2
(B)
4 3
(B)  102 %
 8
(C)
9
4 3
(C)  10 2 %
 5
(D)
7
(D) 3  10 2 %
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WAVE OPTICS JEE ADVANCED - VOL - V
4. In the given figure, S is a monochromatic 6. In a double slit experiment instead of taking
point source emitting light of wavelength slits of equal widths, one slit is made twice
  500nm . A thin lens of circular shape and as wide as the other, then in the interference
focal length 0.1m is cut into two identical pattern: (2000; 2M)
halves L1 and L2 by a plane passing through (A) the intensities of both the maxima and the
minima increases
daimeter. The two halves are placed
symmetrically placed about the central axis (B) the intensity of the maxima increases and the
SO with gap of 0.5mm. The distance along minima has zero intensity
the axis the from S to L1 and L2 is 0.15m, (C) the intensity of maxima decreases and that
of minima increases
while that from L1 and L2 to ‘O’ is 1.3m.
(D) the intensity of maxima decreases and the
The screen at O is normal to SO. If the third minima has zero intensity
maxima occurs at point A on the screen find
7. Two beams of light having intensities I and
the disntance OA?
4I interfere to produce a fringe pattern on a
screen. The phase difference between the

A beams is
2
at point A and  at point B.
L1
Then the difference between resultant
0.5mm intensities at A and B is: (2001; 2M)

S  O (A) 2 I (B) 4 I
(C) 5 I (D) 7 I
L2
8. In a Young’s double slit experiment, 12
fringes are observed to be formed in a
 0.15m 1.3m
certain segment of the screen when light of
wavelength 600 nm is used. If the
(A) 1 mm (B) 2 mm wavelength of light is changed to 400 nm,
(C) 1.5 mm (D) 2.5 mm number of fringes observed in the same
5. In a standard YDSE the region between segment of the screen is given by :
screen and slits is immersed in a liquid whose (2001; 2M)
refractvie index varies with time as (A) 12 (B) 18
5 T (C) 24 (D) 30
l   until it reaches a steady state 9. In the ideal double-slit experiment, when a
2 4
glass-plate (refractive index 1.5) of
5 thickness t is introduced in the path of one
value of . A glass plate of thickness 36  m
4 of the interfering beams (wavelength  ), the
3 intensity at the position where the central
and refractive index is introduced in front maximum occurred previously remains
2
of one of the slits. Find the velocity of central unchanged. The minimum thickness of the
maximuma when it is at ‘O’?(take d=2mm glass-plate is : (2002; 2M)
and D=1m) 2
(A) 2  (B)
(A) 2  103 ms 1 (B) 3  10 3 ms 1 3
(C) 4  103 ms 1 (D) 5  10 3 ms 1 

(C) (D) 
3
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10. In an interference arrangement similar to 13. In Young’s double slit experiment intensity
Young’s double-slit experiment, the slits S1
1
and S 2 are illuminted with coherent at a point is  4  of the maximum intensity..
 
microwave sources, each of frequency 106
Hz. The sources are synchronized to have Angular position of this point is:(2005; 2M)
   
zero phase difference. The slits are
(A) sin  d 
–1 
(B) sin  2d 
–1 
separated by a distance d = 150.0 m. The
intensity I (q) is measured as a function of      
q, where q is defined as shown. If I0 is the (C) sin–1  3d  (D) sin–1  4d 
maximum intensity, then I(q) for 0 < q < 90°
14. In the Young’s double slit experiment using
is given by : (1995; 42)
a monoclromatic light of wavelenght  , the
(A) I(q) = I0/2 for q = 450 path difference (in terms of an integer n)
(B) I(q) = I0/4 for q = 90° corresponding to any point having half the
(C) I(q) = I0 for q = 0° peak intensity is
(D) I(q) is constant for all values of q  
11. In a Young’s double slit experiment, the (A)  2n  1 (B)  2n  1
2 4
separation between the two slits is d and the
 
wavelength of the light is  . The intensity (C)  2n  1 (D)  2n  1
of light falling on slit-1 is four times the 8 16

intensity of light falling on slit 2. Choose the
correct choice(s), MULTIPLE ANSWER QUESTIONS
(A) if d =  , the screen will contain only one
15. White light is used to illuminate the two slits
maximum
in young’s double slit experiment. The
 seperation between the slits is b and the
(B) if  d   , at least one more maximum screen is at a distnace d(d>>>b) from the
2
(besides the central maximum) will be observed slits. At a point on the screen directly in front
on the screen of one of the slits, certain wavelength are
(C) if the intensity of light falling on slit 1 is reduced missing. Some of these missing wavelengths
so that it becomes equal to that of slit 2, the are
intensities of the observed dark and bright fringes b2 2b 2 b2 2b 2
will increase (A)   (B)   (C)   (D)  
d d 3d 3d
(D) if the intensity of light falling on slit 2 is reduced 16. Two point monochromatic and coherent
so that it becomes equal to that of slit 1, the sources of light of wavelength  are each
intensities of the observed dark and bright fringes
placed as shown in the figure below. The
will increase.
initial phase difference between the sources
12. In a YDSE be-chromatic light of
is zero.(Assume D>>>d). Select the correct
wavelengths 400 nm and 560 nm are used.
statement(s).
The distance between the slits is 0.1 mm and
the distance between the plane of the slits
and the screen is 1 m. The minimum distance D
between two successive regions of complete S1 S2
  O
darkness is : (2004; 2M) d
(A) 4 mm (B) 5.6 mm Screen
(C) 14 mm (D) 28 mm
7 D
(A) If d  , the point ‘O’ will be minima. 18. When Z  the intensity meausred at ‘O’
2 2d
(B) If d   , only one maxima canbe observed
2 D
on screen is I 0 . The intensity at ‘O’ when Z  is
(C) If d  4.8 ,then a total 10 minimas would d
be there on screen. (A) I 0 (B) 2I 0 (C) 3I 0 (D) 4I 0
5 19. The minimum value of Z for which the
(D) If d  , then Intensity at ‘O’ would be intensity at ‘O’ is zero is
2
minimum. 3 D D D D
17. The minimum value of ‘d’ so that there is a (A) (B) (C) (D)
2d 2d 3d d
dark fringe at ‘O’ is d min . For the value of
20. If a hole is made at O ' on AO ' O and the
d min the distance at which the next bright
slit S 4 is closed then the ratio of the
fringe is formed is x. Then
maximum to the minimum observed on
P
D
B screen at ‘O’ if O ' S3 is equal to is
4d
x (A) 1 (B) Infinity (C) 34 (D) 4
d
O' Comprehension-II:
O A monochromatic beam of light falls on Young’s
A
double slit experiment apparatus as shown in
D D figure. A thin sheet of glass is inserted in front of
D lower slit S 2 (   600nm is wavelength of
(A) d min   D (B) d min  source)
2
d min
(C) x  (D) x  d min
2
S1
COMPREHENSION TYPE
 d
O
Comprehension-I:
In the arrangement shown in figure, slits S3 and
S4 are having a variable separation Z. Point ‘O’ 
d  D
on the screen is at the common perpendicular S2 t
bisector of S1S2 and S3S4 . (Assume D>>>d).
21. The central bright fringe can be obtained

S1 S3
(A) At ‘O’ (B) At ‘O’ or below ‘O’
(C) at ‘O’ or above ‘O’
Z (D) Any where on the screen
d O 22. If central bright fringe is obtained onscreen
A O'
at ‘O’ then
S2 S4 (A)    1 t  d sin  (B)    1 t  d cos
D D t d
(C)  t   d (D)    1  sin 
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23. The phase difference between central
ASSERTION AND REASON TYPE
maxima and 5th minima is
 (A) A is true and R is true and R is the correct
(A) (B) 9
6 explanation of A.
(B) A and R are true but R is not the correct
3 
(C) (D) 8  explanation of A
2 6 (C) A is true, R is false
(D) A is false, R is true.
Comprehension-III: 27. Assertion(A): Thin films such as soap bubble
The figure shows a surface XY separating two or a thin layer of oil on water show beautiful
transparent media, medium-I and medium-II. colours when illuminated by white light.
The lines ab and cd represents wavefronts of a Reason(R): The colours are obtained by
light wave travelling in medium-I and incident on dispersion of light
XY. The lines ef and gh represent wavefronts of 28. Assertion(A): For best contrast between
light wave in medium-II after refraction the maxima and minima in the interference pattern of
b
Young’s double slit experiment the intensity of
d light emerging out of the two slits should be equal.
Reason(R): The intensity of interference pattern
medium  I is proportional to the square of the amplitude.
a c
X f h Y 29. Assertion(A): The film which appears bright in
reflected system will appear dark in the
e medium  II transmitted system and vice-versa.
g Reason(R): The condittions for film to appear
bright or dark in the reflected light are just reverse
24. Light travels as a
to those in the transmitted light
(A) Parallel beam in each medium
(B) Convergent beam in each medium
(D) divergent beam in each medium STATEMENT TYPE QUESTIONS
(D) divergent beam in one medium and
convergent beam in other medium. (A) Statement-I is true and Statement-II is
25. The phases of the light wave at c,d,e and f true and Statement-II is the correct
explanation of Statement-I.
are c , d , e ,  f respectively. It is given that
(B) Statement-I and Statement-II are true
c   f , then but Statement-II is not the correct
(A) c can not be equal to d explanation of Statement-I
(C) Statement-I is true, Statement-II is false
(B) d can not be equal to e
(D) Statement-I is false, Statement-II is true.
(C) d   f  is equal to  c  e 
(D) d  c  is not equla to  f  e  30. Statement-I: In Young’s double slit experiment
26. Speed of light is interference pattern disaapears when one of the
(A) same in medium-I and medium-II slits is closed
(B) Larger in medium-I than in medium-II Statement-II: Interference is observed due to
(C) Larger in medium-II than in medium-I superposition of light waves from two coherent
(D) Different at b and d sources

31. Statement-I : In Young’s double slit experiment, 19 2 D
the two slits are at distance ‘d’ apart. Interference C) Intensity is th of maximum R)
pattern is observed on a screen at a distance D 25 d
from the slits. At a point on the screen which is Intensity
directly opposite one of the slits, a dark fringe is 7 3 D
observed.Then the wavelength of wave is D) Intensity is th of maximum S)
25 d
proportional to the square of the distance
between the two slits. Intensity
Statement-II: For a dark fringe, intensity is zero. 34. In Young’s double slit experiment, the point
32. Statement-I : Fringe width depends on the source ‘S’ is placed slightly off the central
refractive index of the medium. axis as shown in figure. if   500nm , then
Statement-II: Refractive index changes optical match the following
path of ray of light forming fringe pattern.
MATRIX MATCHING TYPE P
33. Monochromatic light source having S1 x  10mm

S
wavelength  is used in YDSE. The
2mm
separation between the slits is ‘d’ and
O
separation between slit and screen is D 20mm
(>>>d) as shown in figure. The slits width

are different and individual intensities due
to upper and lower slits at the screen are
4I 0 and 9I 0 respectively. In column I,
1m 2m
intensity of the interference pattern on the
screen at certain points are given and in
Column-I
Column II the distance between two points
(A) Nature and order of interference at the point
on the screen which are having that intensity
P, (OP=10mm)
is given. Match the entries of Column I with
(B) Nature and order of interference at point
the enteries of Column II.
‘O’
(C) If a transparent paper (refractive index
  1.45 ) of thickness t  0.02mm is pasted
on S1 the nature and order of interference at P..
S d (D) After inserting transparent paper in front of
O
slit S1 ,the nature and order of interference at ‘O’
Column-II
D P) Bright fringe of order 80.
Screen
Q) Bright fringe of order 262.
Column-I Column-II R) Bright fringe of order 62.
S) Bright fringe of order 280.
1 D 35. In Column-I the effect on fringe pattern in
A) th of maximum intensity P)
25 d YDSE is mentioned when the changes
D mentioned in Column-II are made. Match
B) Intensity is 25I 0 Q) the entries of Column-I with entries of
3d
Column-II.
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VOL--VI WAVE OPTICS
Column-I
A) Angular Fringe width changes A
B) Fringe width [linear seperation between two
consecutive fringes] F

C) Angular Fringe width remains same
D) The fringe pattern disappear
B
Column-II
P) Screen is moved away from the plane of the 38. In Young’s double slit experiment
slits interference bands are produced on the
Q) Wavelength of light used is decreased. screen placed at 1.5m from two slits 0.15 mm
R) The seperation between the slits is increased apart and illuminated by light of wavelength
S) The width of the source slit is increased 6000Å . If the screen is now taken away from
T) The source slit is moved closer to the double the slits by 50cm then find the change in
slit plane. fringe width (in mm)?
39. An inteference is observed due to two
INTEGER TYPE QUESTIONS coherent sources A and B separated by a
distnce 4 along Y-axis, where  is the
36. In YDSE, the source is red light of wavelength of light . A detector ‘D’ is moved
wavelength 7  107 m . When a thin glass film along the positive X-axis. Find the total
of refractive index 1.5 at this wavelength is number of maxima observed on the X-axis
put in the path of one of the interfering excluding the points x  0 and x   ?
beams, the central bright fringe shifts by
10 3 m to the position previously occupies A
by the 5th bright fringe. Find the thickness
of glass plate in micrometers? 4
37. In a modified Young’s double slit experiment,
a monochromatic uniform and parallel beam 
B xn
of light of wavelength 6000 and intensity D
Å
40. Two coherent light sources each of
 10  2
  wm is incident normally on two wavelength  are separated by a distance

 
3 . Find the total number of minima formed
circular apertures A and B of radii 0.001m
on the line AB which runs from  to  ?
and 0.002m respectively. A perfectly
transparent film of thickness 2000 and B
Å P
refractive index 1.5 for the wavelength of 
6000Å is placed in front of aperture A as
shown in figure. Calculate the power (in
microwatt) received at the focal spot F of 3
the lens. The lens is symmetrically placed
with respect to the apertures. Assume that
10% of the power received by each aperture 
goes in the original direction and is brought Q
A
to the focus

LEVEL-V
LEVEL-V
KEY
HINTS
SINGLE OPTION TYPE QUESTIONS
1-C 2-C 3-A 4-A SINGLE CORRECT ANSWER QUESTIONS
5-B 6-A 7-B 8-B
1) For to obtain dark fringe
9-A 10-C 11-A 12-D
13-C 14-B 
path difference  x    2n  1
2
MORE THAN ONE OPTION TYPE 
 l1  l3    l2  l4    2n  1
15 - A,C; 16 - A,B,C,D; 17 - B,D; 2
2) I  I 0 sin 2 
COMPREHENSION-I:
dI
18 - B; 19 - D; 20 - C  I 0  2sin  cos 
d
dI I 0  2sin  cos  d
COMPREHENSION-II:   2 cot  d
I I 0 sin 2 
21 - D; 22 - A; 23 - B
% error in angle
COMPREHENSION-III: d  dI 1  1
 100       100
24 - A; 25 - C; 26 - B   I 2 cot   
d 0.002 6
 100  o
 100 
ASSERTION & REASON TYPE  2  5 cot 30 
27-C 28-B 29-A 6 1 1 6 1 1 1
    0.1    
 5 3  5 10 3
STATEMENT(S) TYPE
30-A 31-B 32-B 3 1 3
    0.04
 25 
MATRIX-MATCHING TYPE d 4 3
 100   102 %
33) A  P, R, S ; B  P, R, S ;  
C  P, Q , R, S ; D  P, Q , R , S 3) Path difference at ‘C”
34) x  t1   g 1  t2   g 1  g  t1  t2    t1  t2 

AS ; BP; C Q; D R
35) A  Q, R ; B  P , Q, R ; x   t1  t2     g  1
C  P , S , T ; D  Q, R , S , T
  1.4 
x   t1  t2   g  1   2.5  1.25   3  1
    4 
INTEGER TYPE QUESTIONS
36-7 37-7 38-2 39-3  42  2 2.5
= 1.25    1  1.25  
40-6  40  40 40

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VOL--VI WAVE OPTICS
25 1 d 0.3  0.15
x   m . 3

400 16 0.5  10 0.15
2 0.45
Phase difference      x d  0.5  103  1.5  10 3 m
 0.15
For third maximum
2 2  4 1
     x  10
  106 3 D 3  500 109  1
a a 1 5000  10  3 16 yn    109  106
3
d 1.5  10
 106  106 
   yn  103 m  y3  1mm
30 10 7 3  106 3
5)
intensity at ‘C’
I C  4 I 0 cos 2   2   4I cos  6   4I  34
0
2
0
P
I C  3I 0 S1 y

maximum intensity I max  4 I 0
d  2mm
IC 3I I 3 O
 0  C 
I max 4 I 0 I max 4 
S2
4) t
D  1m
A
S1 
yn Path difference
P1
d O  x    S2 P liq    g  l  t   S1P liquid 
S 
P2
S2
 x    S2 P  S1P liq   g  l  t
a  0.15m 0.3m D  1m

 l  S 2 P  S1 P  air   g  l t 
v
 yd 
 x  l      g  l  t
The images of S due to both parts of lenses are  D 
formed at S1 and S 2 whose distance ‘v’ from For central maxima x  0
the lens is give n by
 yd 
1 1 1 1 1 1
O  l      g  l  t
      D 
f v u 0.1 v  0.15 
   g  l  tD
1 1 1 10 100 150  100 50 10 y
       d l
v 0.1 0.15 1 15 15 15 3
3  3  5 T 
v  0.3m D      t
10 2  2 4 
y 
From similar les SS1S 2 and SPP 5 T 
1 2 d  
2 4 
2
 T Then, Imax= I 1    > 4 I and
D 1   t
4 D 4  T t
y  2
 Imin = I    1 > 0
 5 T  d 10  T 
d  
2 4  Intensity of both maxima and minima is increased.
The time when y becomes zero is 7. I(  ) = I1 + I2 + 2 I1 I 2 cos  …(1)

Here, I1 = I and I2 = 4I
D 4 T t
O 
d 10  T  At point A,  =
2
D  4  T  t  0  4  T  0  T  4sec IA = I + 4I = 5I
At point B,  = 
Speed of central maxima
IB = 1 + 4I – 4I = I
dy Dt d  4  T  IA – IB = 4I
V 
dt d dT 10  T  D
8. Fringe width, w =
tD  1
d
V   4  T 10  T  
d   n11  n2 2
9. Path difference due to slab should be integral
tD    4  T   10  T  
   multiple of l or  x = n 
2
d  10  T  
or    1 t  n , n  1, 2,3
tD  4  T  10  T  n
   or t =  1
d  10  T  2 
for minimum value of t, n = 1
6 Dt  
V
d 10  T 
2 t =   1  1.5  1 = 2 
Central maxima is at ‘O’ at T  4sec  

10. The intensity of light is I(q) = Io cos  2 
2
6  1 36  106 3  36  103 2  2 
V  where  =  (  x) =   (d sin q)
2   
2  10 3 10  4  36
(i) For q = 30°
3 1
V  3  10 ms c 3  108
 = = = 300 m and d = 150 m
2
v 106
6. In interference we know that Imax =  I1  I 2 
2
 2  1 
and Imin =  I1 ~ I 2   =   (150)   =
 300  2 2
Under normal conditions (when the widths of  
both the slits are equal) =
2 4
I1» I2 = I (say) \ Imax = 4I and Imin = 0
  I0
When the width of one of the slits is increased. I(q) = I0 cos2  4   2 (option a)
 
Intensity due to that slit would increase, while
(ii) For 0 = 90°
that of the other will remain same. So let :
I1 = I and I2 =  I (  > 1)  2 
d =  300  (150)(1) = p
 

JEEADVANCED
JEE MAINS - CW--VOL
VOL--VI WAVE OPTICS
 
Hence, the correct option is (D).
or =
2 2  
13. l = lmax cos2  2 
and I(q) = 0  
 lmax 
(iii) For q = 0°, d = 0 or =0 = lmax cos2
2 4 2
I(q) = I0 (option c)  1
11. Condition for maxima is d sin  = n  cos =
2 2
If d =  , sin  = n  
Possible value is  or =
2 3
only one maxima will be obtained  2 2 
(A) is correct. = 3     x …(1)
 
(B) is correct.
2 2
where  x= d sin 
Imax   I1  I 2  & I min   I1  I 2  Substituting in Eq. (1), we get
with I1 = 4I2    
I max  9I 2 , I min  I 2
sin  = or q = sin–1  3d 
3d
but when I1 = I2 correct answer is (C).
Imax = 4I2 and Imin = 0
I max  
(C) and (D) are wrong. 14.  Im cos 2  

12. Let nth minima of 400 nm coincides with mth 2  2
minima of 560 nm, then   1
 cos   
 400   560   2 2
(2n – 1)  2  = (2m – 1)  2 
   
 
2n  1 7 14  
or =   ... 2 4
2m  1 5 10
i.e., 4th minima of 400 nm coincides with 3rd 
   2n  1
minima of 560 nm. 2
Location of this minima is,   
6
 x    2n  1   2n  1
(2  4  1)(100)(400  10 ) 2 2 4
Y1 = 2  0.4
= 14 nm
Next 11th minima of 400 nm coincides with 8 th MULTIPLE ANSWER TYPE

minima of 560 nm, then
15) Solution:
 400   560 
(2n – 1)  2  = (2m – 1)  2  d
   
2n  1 7 14 S1 p
or 2m  1 =  = …
5 10
b
i.e., 11 minima of 400 nm coincides with 8th
th
S2
minima of 560 nm.
Location of this minima is, + Y2 =
(2  11  1)(100)(400  10 6 ) From the figure:
= 42 mm
2  0.1 S 2 p 2  S1P 2  S1S 22 ; S 2 p 2  S1 P 2  S1S22
Required distance = Y2 – Y1 = 28 mm
 S2 p  S1P  S2 p  S1 P   S1S22
S1 S22 b2 D
 S 2 p  S1P     S2 P  S1P  d   d min 
 S2 P  S1P  2d 2
Missing wavelengths means dark fringes,hence There bright fringe is formed at P if the path
difference
 b2
path difference  x    2n  1  x1  AO ' P  ABP  
2 2d
2
2
x1  D  D 2  x 2  D 2  d 2  D 2   x  d   
b

x 2 d 2  x  d   2 xd
2 2
 2n  1 d
   
2 D 2D 2D
b2
For n  1  1 
d d 2 xd
Given d  d min then    on solving
D D
b2
For n  2  2  x  d min
3d
b2 COMPREHENSION TYPE
For n  3  3  and so on.
5d
16) Path difference at ‘O’ is x  d , which is Comprehension-I:
maxima 18) intensity at ‘O’ is proportional to intensity at S3

7
(A) If d 
2
, the point ‘O’ will be minima. and at S 4 . I  k cos  
2 
2 where k is constnat,
(B) If d   , The point ‘O’ will be maxima
 is the phase difference.
(C) If d  4.8 ,then a total 10 minimas can be
observed on screen 5 above ‘O’ and 5 below 2 Z  z
   where  is the fringe width
‘O’, which corresponding to  2 
 3 5 7 9 D
x   , , , , . when Z 
2 2 2 2 2 2d
5 Z  D 
(D) If d  , then ‘O’ will be minimum and   d 
2  D 2d 2
hence intensity is minimum.
17) There is a dark fringe at ‘O’ is the path difference I  I 0  k cos 2  4   k  12 ; k  2I0
x  ABO  AO ' O   2
2 D
When Z ' 
d2 d
x  2 D 2  d 2  2 D  2 D 1   2D Phase difference
D2
2 Z ' 2 2 D
'   d  2
 d 2  1 2   d2   2 D 2d
x  2 D  1  2   1  2 D 1  2
 1
 D    2D  Required intensity at ‘O’
 
x  2 D 
d 2

d 2
   d2 
D  2   2I cos  22   2I  1
I '  k cos 2  ' 0
2
0
2
2D 2
D 2 2
I '  2I0
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JEE MAINS - CW--VOL
VOL--VI WAVE OPTICS
  22) Total path difference x     1 t  d sin  .
19) Intensity at ‘O’ os zero if 
2 2 For cent ral maxima x  0 hence
2 Z
     Z   ; Z    D    1 t  d sin 
 2 d 23) Phase difference
20) Intensity at S3 is    2n  1    2  5   1   9 ;
2 2 2
I S3  A  A  2 A cos 
0 0 0   9
2  D 2 D  Comprehension-III:
Where     d  
 4d  D 4d 2
I S3  2 A02  2 A02 cos  b
d
 2 A02 1  cos    2 A02  2cos2  
 2    X
a c
i medium  I
Y
f
IS3  4A cos 
2
0
2
 2   4A cos  4 2
0
2 r h
medium  II
1 e
I S3  4 A   2 A02
2 g
0
2
Amplitude of wave at S3 : AS3  2 A0
24) The wavefronts, in both the media are parallel.
2 The light will be a parallel beam.
Maximum intensity at O '  A0  2 A0   25) Points c&d are on the same wavefrnts hence
2 c  d .Similarly points e and f are on the same
Minimum intensity at O   2 A0  A0 
wavefront hence e   f .
2 2

I max  A0

 2 1  
 
  2 1  
2  1 Hence d   f  c  e
I min  A0
  2 1  
    2  1 2  1

26) From the diagram, it clearly indicates that on
entering medium-II, the ray bend towards normal
2 and hence medium-II is density medium.
2
 2 1 
I max

    4 Hence speed of light is larger in medium I than in
 2 1 
 2 1  medium-II.
I min
    ASSERTION & REASON TYPE
27) The beautiful colours are obtained due to
2 2
interference of light reflected from the upper and

 2  1 2 2   3  2 2  lower surfaces of thin film. As the conditions for
  3  2  1.414    3  2.828 
2 2 constructive and destructive interference are
dependent on wavelength of light. Hence
I max coloured interference fringes are observed. A is
  5.828   34 ; I  34
2
min
true and R is false
28) When intensity of light emerging out from two
Comprehension - II:
slits is equal then the intensity of minima.
21) Any where on the screen because there is no
2
relation b/n  &  . I min   I I   0 or absolutre dark. It

provides a better contrast. 101
 A & R are true but R isnot the correct 1 1

explanation of A I  I max   25I 0  I 0
25 25
29) For reflected system of the film, the consition for
1
maxima is 2 t cos r   2n  1  2 . While the th of maximum intenxity gives the intensity
25
maxima for transmitted system of film is of minima. Hence teh distance b/n such
2 t cos r  n and is reverse for minima. A & n D
R are true and R is the correct explanation of A points is equal to fringe width Yn 
d
STATEMENT TYPE D 2 D
n  1  y1  ; n  2  y2 ;
30) When one of the slits is closed, there appears d d
general illumination from a single source. 3 D 4 D
Interference does not take place. n  3  y3  ; n  4  y4 ;
d d
31) Path difference
The distnace b/n two successive minima points
S 2 P 2  S1 P 2  S1S 22
 S2 P  S1P  S2 P  S1P   S1S 22   D 2 D 3 D 4 D 
 d , d , d , d     
 
S1 S22 d2
 S 2 P  S1 P    (B) Intensity is 25I 0

 S2 P  S1 P  2 D
This represnts the maximum intensity. The
 distance b/n the points on the screen.
For dark fringe  x   S2 P  S1 P   2n  1
2 Which area having maximum intensity
d2    D 2 D 3 D 
  2n  1  d , d , d ,    
2D 2  
d 2   2n  1  D 19
(C) Intensity is th of maximum intensity
25
d2  
Intensity of dark fringe is zero. 19
I  25 I 0  19 I 0
D 1 25
32)  but  
d  I  I1  I 2  2 I1 I 2 cos 
19 I 0  4 I 0  9 I 0  2 36 I 0 cos 
MATRIX-MATCHING TYPE
19 I 0  13I 0  2  6 I 0 cos 
33) I1  4 I 0 ; I 2  9 I 0 6I0 1
6 I 0  12 I 0 cos   cos   
2 2 2
12 I 0 2
I max   I1  I 2   4 I 0  9I 0    I 0  2  3 
I max  I 0  25  25 I 0  5 7 11  
 , , ,       2n  
2 2 2
3 3 3 3  3
I min   I1  I 2    I 0  2  3    I 0    I0 path difference
I min  I 0
  5 7 11
1 x  , , , 
(A) th of maximum intensity: 2 6 6 6 6
25
JEEADVANCED
JEE MAINS - CW--VOL
VOL--VI WAVE OPTICS
Path difference
 1
x   2n   
 3 2 x =  SS 2  S 2 P    SS1  S1P 
 1  D   SS 2  SS1    S2 P  S1P 
yn   2n  
 3  2d  d sin  0  d sin 
D dy0 dy
For n  0  y0  x D  D
6d
1 2
 1   D 5 D
For n  1  y1   2    20  2 20  10
 3  2d 6d x    0.04  0.1  0.14mm
1000 2000
 1   D 11 D A) For bright fringe:
For n  2  y2   4   
 3  2d 6d x 0.14 14  102
x  n  n   
Required seperation  500 109  103 5  107  103
2 D 11 D 5 D n  2.8  102  n  280 A  S
  y 2  y0  ;   y2  y1  
3d 6d 6d B) At the centre (origin) ‘O’
6 D 3 D d  y0 20  2
  x   0.04mm
6d 3d D1 1000
7 For bright fringe x  n
(D) Intensity is th of maximum
25 x 0.04  10 3 4  10 5
n  
I
7
 25 I 0  7 I 0  500 10 9 5 107
25 n  0.8  102  80
I  I1  I 2  2 I1I 2 cos   7 I 0  13I 0  12 I 0 cos  n  80 ; BP
C) Due to transparent paper, the change in optical
1 2 4 8
cos       , , ,    path is
2 3 3 3
   1 t  1.45  1 0.02   0.45  0.02
 d 2 d 3 d
Required seraration , , ,     0.009mm
3d 3d 3d
path difference at P:
34)
 x   0.14  0.009  0.131mm
1
x1 131 106 13110 6 1310

P n     262
S1
 500 109 5  107 5
S  n  262 C Q
0  10mm
y0  2mm D) Due to transparent paper the path difference at
d
 O ‘O’ x   0.04  0.009   0.031mm
x1 x2 x 0.031 103 31 106

n  
S2  500 109 5  107
D1 D2
310
n  62 ; n  62 ; D  R
5

35) Angular fringe width For s: As the source slit wodth increases, the
d  distance b/n slits s 
condition  would be violated at some
  S d
 
d D instant and fringe pattern disappears but there is
D  distance b/n slits & screen no effect on  and  . s  C , D
The arrangement for YDSE is shown in figure. For t: As the distance b/n source slit and
  Wavelength of light used double slit ‘S’ is decreasing then fringe pattern
Let the size of source and separation between disappears but  and  has no effect.
  Fringe width t  C, D
source slit and double slit plane is S.
Then for interference fringes to be observed INTEGER TYPE QUESTIONS
s 
 .
S d 36) Given shift of central bright fringe  5th order
If this condition is not satisfied then interference D 5 D
pattern produced by different parts of the source bright fringe    1 t 
d d
overlap and fringe pattern disappears.
5 5  7 10 7 5  7  107
t  
Screen
s    1 1.5  1 0.5
t  7  106 m  t  7  m
37) The power transmitted through A
S1 S 2  10  2
PA  10% of     0.001
S D  
10 10
For p: As ‘D’ is increased, angular fringe width      10 6  106 watt
100 
 The power transmitted through ‘B’ is
 remains same as it is independnet of D.
d
 10  2
D PB  10% of     0.002 
As D increase,fringe width increases    
d
10 10
p  B, C      4 106  4 106 watt
100 
For q: When  is decreased, angular fringe
Let  be the phse difference t=introduced by film
width (  ) and fringe width  decreases and
2 2 2
fron the condition as  decreases and from the  x     1 t   0.5  2000 10 10
  6000 10 10
s  
condition  as  decreases this condition 
S d redian. The power received at F
3
would be failed and fringe pattern disappears.
q  A, B, D
P  PA  PB  2 PA PB cos 
For r: ‘d’ is increasing  and  both decreases 1

 106  4  106  2 106  4 106 
and fringe pattern disappears. r  A, B , D 2
 5  106  2  106  7  106  7 W
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38) D2  D1  50cm ; D1  1.5m  150cm B
  6000Å  6000 10 cm ; d  0.15mm
8 P
Change in fringe width    2  1
O
 50  6000  10 8 (mid point)
   D2  D1   R
d 0.15 10 1 Q
20 104 A
   2  101 cm  2mm
103
PQ  QR  PR [In any triangle sum of two sides
  2mm is more than the third side]
39) To obtain maxima the path difference
PR  QR  3
AD  BD  n
As  PR  QR  represrnt the path difference at
AD  xn  n
any point on AB  it can occurs at
AD  n  xn  1
 3 5
, , only..
From  le ABD, 2 2 2
SO 3 minimas below the mid point ‘O’ and 3
AD 2  AB 2  BD 2
2
minimas above ‘O’
 n  xn   16 2  xn2 Hence total no.of minimas formed is 6.
n2  2  xn2  2n xn  16 2  xn2
P
d
2n xn  16  n   2 2
S 
3d
xn 
16  n   2
D
2n
For n  0  x0  
LEVEL-VI
 16  1  15
n  1  x1      7.5
 2  2 SINGLE CORRECT ANSWER TYPE
 16  4  12
n  2  x2      3 1. If two coherent sources are placed at a
 4  4
distnace 3 from each other, symmetric to
 16  9  7 the centre of the circle as shown in figure,
n  3  x3     Find the number of fringes shown on the
 6  6
screen, placed along the circumference?
 16  16 
n  4  x4     0
 2 4 
So the total no.of maxima points observed are  
3. S1 S2
3
40) Let us take any general point R on the line AB,
For any position of R on line AB,
(A) 16 (B) 12
PQR is a  le . (C) 8 (D) 4

2. In standared YDSE slits were moved apart 3 

symmetrically with relative velocity v, (A) cos   (B) cos  
2d 4d
calculate the rate at which fringes pass a
point at a distance x from the centre of the 
fringe system formed on a screen at ' y ' (C) sec   cos  
d
distance away from the double slits if
wavelength of light is  , and distance 4
(D) sec  cos  
between the slits is ‘b’. Assuming d
y  b & b   5. A glass plate of refractive index 1.5 is coated
with a thin layer of thickness t and refractive
2xv xv xv
(A) (B) (C) (D) Zero index 1.8. light of wavelength  travelling
y 2 y y
in air is incident normally on the layer. It is
3. ABC is a spherical wavefront centred at ‘O’
partly reflected at the upper and lower
symmetric about BE is incident on slits S1 surfaces of the layer and the two reflected
and S 2 . BS1  3 ; S1S 2  4 ; BO  6 ; rays interfere. If   648nm , then find the
S1E  128 and  is the wavelength of least value of ‘t’ for which the rays interfere
incident light wave. A mica sheet of constructively?
refractive index 1.5 is pasted on S 2 . Find (A) 30 nm (B) 60 nm (C) 90 nm (D)120 nm.
6. In a Young’s experiment, the upper slit is
the minimum value of thickness of mica sheet

for which central fringe forms at E? covered by a thin glass plate of refractive
index. 1.4, while the lower slit is covered by
A
S2 another glass plate, having the same
thickness as the first one but having
B 
O
E refractive index 1.7. Interference pattern is
S1 observed using light of wavelength 5400 Å.
C It is found that the point P on the screen,
where the central maximum (n = 0) fall
31 15 5 7 before the glass plates were inserted, now
(A) (B) (C) (D)
8 8 8 8
3
4. In the adjacent diagram,CP represents a has the original intensity. It is further
4
wavefront and AO and BP, the corresponding
two rays. Find the condition on  for observed that what used to be the fifth
constructine interference at P between the maximum earlier lies below the point P while
ray BP and reflected rays OP the sixth minima lies above P. Calculate the
thickness of glass plate. (Absorption of light
by glass plate may be neglected). (1997; 5M)
(A) 9.3 m (B) 6.2 m (C) 8.5 m (D) 5.8 m
7. In a young’s double slit experiment, two
wavelengths of 500 nm and 700 nm were
used. What is the minimum distance from
the central maximum where their maximas
D
coincide again? Take = 103. Symbols
d
have their usual meanings. (2004; 4M)
(A) 6.5 mm (B) 3.5 mm
(C) 10.2 mm (D) 5.9 mm

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MULTIPLE ANSWER QUESTIONS COMPREHENSION TYPE
8. A transparents slab of thickness ‘t’ and Comprehension-I:

refractive index  is inserted in front of In YDSE arangement shown in figure, fringes are
upper slit of YDSE apparatus. The seen on screen using monochromatic source S
wavelength of light used is  . Assume that having wavelength 3000 (in air). S1 and S 2
Å
there is no absorption of light by the slab. are two slits separate by d  1mm and D  1m .
Mark the correct statments.
Left of slits S1 a nd S 2 medium if refractive index
(A) The intensity of dark fringes will be zero, if
slits are identical. 1  2 is present & to the right of slits S1 and
(B) The change in optical path due to insertion
3
of plate is t . S 2 medum of refractive index  2  is present.
2
(C) The change in optical path due to insertion
A thin slab of thickness ‘t’ is placed in front of
of plate is    1 t . S1 . The refractive index of slab 3 varies with
(D) For making intensity zero at centre of screen distance from its starting face as shown in figure.
5 y
the thickness can be 2    1 .
x 2
1 S1 
9. In a YDSE with two identical slits, when the 3 Screen
S 3
upper slit is covered with a thin, prefectly d
2

t
tranparent sheet of mica, the intensity at the O 3
d
2
centre of the screen reduces to 75% of the
S2 1
initial value second minima is observed to
be above this point and third maxima below D  1m D  1m 0 x
t
it which of the following can not be a possible
value of phase difference caused by the mica 11. In order to get central maxima at the centre
sheet of thescreen, the thickness of slab required
is
 13 5 11
(A) (B) (C) (D) (A) 1 m (B) 2 m
3 3 3 3
10. In Young’s double slit experiment, the two (C) 0.5 m (D) 1.5 m
slits are covered by slabs of same thickness 12. If the thickness of the slab is selected 1 m ,
but refractive index 1.4 and 1.7. The distance then the positionof the central maxima will
between slits and screen is 1m and distance
between slits is 1mm and wavelength of be  y  coordiante 
coherent source used is 4000 and the 1 1
Å (A) mm (B)  mm
central fringe shifts to the 3rd bright fringe 3 3
position, then 1 1
(A) Shift will be towards slab of R.I.-1.7 by 1.2 (C) mm (D)  mm
6 6
mm.
(B) Shift will be towards slab of R.I.-1.4 by 1.2 13. Fringe width on the screen is
mm (A) 0.4mm (B) 0.1mm
(C) Slabs are of thickness 4 m (C) 0.2mm (D) 0.3mm
(D) Slabs are of thickness 4
Å
Passage-II:
In the figure, light of wavelength   5000 is L
Å
incident on the slits (in a horozontally fixed place).
Here d  1mm and D  1m . Take origin ‘O’ S
and XY plane as shown in figure. The screen is 1mm
D M irro r C O
released from rest from the initial position as
shown. 5cm 5cm 190cm
  5000Å M
d
S1 S2
O X 17. Determine the width of the region where the
fringes will be visible
Y (A) 4cm (B) 6cm (C) 2cm (D) 3cm
D 18. Find the fringe width of the fringe pattern?
(A) 0.05cm (B) 0.25cm
Screen (C) 0.01cm (D) 0.1cm
19. Calculate the number of fringes
(A) 10 (B) 20
14. The velocity of central maxima at t  5sec

(C) 30 (D) 40
is
(A) 50ms 1 along Y-axis Comprehension-IV:
(B) 50ms along negative Y-axis
1
A double slit apparatus is immersed in a liquid of
(C) 25ms 1 along Y-axis refractive index 1.33. It has slit separation of 1
mm and distance between the plane of slits and
(D) 3  108 ms 1 along Y-axis screen is 1.33 m. The slits are illuminated by a
15. Velcoity of 2nd maxima w.r.t central maxima parallel beam of light whose wavelength in air is
at t  2sec is 6300 Å. (1996; 3M)
20. Calculate the fringe width.
(A) 8  cms 1  i  20  ms 1  j
(A) 0.95 mm (B) 0.36 mm
(B) 8  cms 1  i
(C) 0.63 mm (D) 0.56 mm
21. One of the slits of the apparatus is covered
(C) 2  cms 1  i (D) 86  ms 1  i by a thin glass sheet of refractive index 1.53.
Find the smallest thickness of the sheet to
16. Acceleration of 3rd maxima wrt 3rd maxima bring the adjacent minimum at the centre of
on the other side of central maxima at the screen the axis.
t  3sec is (A) 2.132  m (B) 2.512  m
(A) 0.02ms2 i (B) 0.03ms 2 i (C) 6.521 m (D) 1.579 m
Comprehension-V:
(C) 10 ms 2 i (D) 0.6 ms 2 i In Young’s experiment, the source is red light of
Comprehension-III: wavelength 7 × 10–7 m. When a thin glass plate
The arrangmeent for a mirror experiment is shown of refractive index 1.5 at this wavelength is put in
in figure. S is a point source of frequency the path of one of the interfering beams, the central
6  1014 Hz . D and C represnt the two ends of a bright fringe shifts by 10–3 m to the position
mirror placed horizontally and LOM represets previously occupied by the 5th bright fringe.
the screen. (1997C, 5M)
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22. Find the thickness of the plate.
(A) 7  106 m (B) 9  1010 m ASSERTION AND REASON TYPE
(C) 4  106 m (D) 6  1010 m
23. When the source is now changed to green (A) A is true and R is true and R is the correct
light of wavelength 5 × 10–7 m, the central explanation of A.
fringe shifts to a position initially occupied (B) A and R are true but R is not the correct
by the 6th bright fringe to red light. Find the explanation of A
refractive index of glass for green light. (C) A is true, R is false
(A) 1.4 (B) 1.2 (C) 1.6 (D) 1.3 (D) A is false, R is true.
24. Also estimate the change in fringe width due 28. Assertion(A): Interference pattern is made by
to the change in wavelength. using blue light instead of red light, the fringes
(A) 6.22  105 m (B) 6.22  105 m becomes narrower.
(C) 5.71 105 m (D) 5.71  105 m Reason(R):In Young’s double slit experiment,
Comprehension-VI: D
The Young’s double slit experiment is done in a fringe width is given by the relation  
d
medium of refractive index 4/3. A light of 600 29. Assertion(A): In YDSE, the fringes become
nm wavelength is falling on the slits having 0.45 indistinct if one of the slits is covered with
mm separation. The lower slit S2 is covered by a
cellophane paper.
thin glass sheet of thickness 10.4  m and
Reason(R): The cellophane paper decreases

refractive index 1.5. The interference pattern is
observed on a screen placed 1.5 m from the slits the wavelength of light.
as shown in the figure. (1999; 10M) 30. Assertion(A): If the whole apparatus of YDSE
25. Find the location of central maximum (bright is immersed in a liquid, then the fringe width will
fringe with zero path difference) on the y- decrease
axis. Reason(R): The wavelength of light in water is
more than that of air.
y
S1
STATEMENT TYPE QUESTIONS
S O2
(A) Statement-I is true and Statement-II is
S2 true and Statement-II is the correct
explanation of Statement-I.
(A) 4.33 mm (B) 2.56 mm (B) Statement-I and Statement-II are true
(C) 3.26 mm (D) 5.16 mm but Statem ent-II is not the correct
26. Find the light intensity of point O relative to explanation of Statement-I
the maximum fringe intensity. (C) Statement-I is true, Statement-II is false
3 5 3 7 (D) Statement-I is false, Statement-II is true.
(A) I max (B) I max (C) I max (D) I max
2 4 4 5
27. Now, if 600 nm light is replaced by white light 31. Statement-I: In YDSE, if separation between
of range 400 to 700 nm, find the wavelengths the slits is less than wavelength of light used, then
of the light that form maxima exactly at no interference pattern is observed.
point O. [All wavelengths in the problem are
4 Statement-II: For interference pattern to be
for the given medium of refractive index .
3 observed, light sources have to be coherent.
Ignore dispersion]
(A) 650 nm, 6.32 nm (B) 350 nm, 4.33 nm
(C) 452 nm, 3.53 nm (D) 650 nm, 433 nm
32. Statement-I: In YDSE, if intensity of each D) 1   2 S) Film 1 appears dark
source is I 0 then minimum and maximum intensity from the transmitted
is zero and 4I 0 respectively.. system
35. Column-I shows four situations of standard
Statement-II: In YDSE, energy conservation young’s double slit arrangement with the
is not followed.
33. Statement-I: For the situtation shown in figure, screen placed far away from the slits S1 and
two identical coherent sources of light produce S 2 . In each of these cases S1 P0  S2 P0 ;
interference pattern on the screen. The intensity
of minima nearest to S1 is not exactly zero. S1 P1  S 2 P1   4 S1 P2  S2 P2   3
and
Statement-II: Minimum intensity is zero when where  is the wavelength of the light used.
interfering waves have same intensity at the In the cases, B,C,D a transparent sheet of
location of superposition refractive index  and thickness ‘t’ is pasted
on slit S 2 . The thickness of the sheets are
S1  P dfferent in different cases. The phase
difference between the light waves reaching
a point P on the screen from the two slits is
S2  denoted by   p  and the intensity by I  p  .
Screen Match the Column-I with Column-II, valid
for that situtation. (2009-JEE)

Column-I Column-II
MATRIX MATCHING TYPE
 p2
34. For the situtation shown in figure below, S2 p1
match the entries of Column-I with that of  p0
Column-II A) S P)   P0   0
1
Assume thickness of films to be very small
compared to wavelength of incident light.

Reflected System    1 t   p2
4
Air S2 p1
 p0
Film-1
B) Q)   P1   0
1 S1
2 Film-2

Air    1 t   p2
2
S2 p1
Transmitted System
 p0
C) R) I  P1   0
S1
Column-I Column-II
A) 1  2 P) Film 1 appears shiny
from the reflected    1 t 
3
 p2
system 4
S2 p1
B) 1  2 Q) Film1 appears dark  p0
from the reflected D) S) I  P0   I  P1 
S1
system.
C) 1  2 R) Film 1 appears shiny
T) I  P2   I  P1 
from the transmitted
110 system NISHITH Multimedia India (Pvt.) Ltd.,
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36. A double slit interference pattern is produced
on a screen as shown in figure. Using INTEGER TYPE QUESTIONS
monochromatic light of wavelength 500nm.
Point P is the location of the central bright 37. In YDSE, the intensity of light at a point on
fringe that is produced when light waves the screen is I for a path difference  . The
arrive in phase without any path differene. intensity of light at a point where the path
A choice of three strips A,B,C of tranparent
 I
materials with different thickness and difference becomes is . Find the value
refractive indices is available as shown in 3 P
table.These are placed over one or both of of P?
38. In YDSE, the slits separation is 0.6mm and
the slits singularly or in conjunction,causing
the separation between slit and screen is
the interference pattern to be shifted across 1.2m. The wavelength of light used is
the screen from the original pattern. In
4800 . The angular fringe width is
Column-I, how the strips have been placed Å
is mentioned whereas in Column-II order of P  10-4 radian. FInd the value of P?
the fringe at point P on the screen that will 39. In YDSE, the wavelength of red light is
be produced due to the placement of the 7.5  10-5 cm and that of blue light is
strips as shown. Correctly match both the 5  10-5 cm . Find the value if ‘n’ for which
Columns. th
Film A B  n  1 blue bright band coincides with nth

C red bright band?
Thickness(in  m ) 5 1.5 0.25 40. In YDSE, the slits have different widths. As
Refractive index 1.5 2.5 2 a result, amplitude of waves from two slits
are A and 2A respectively. If I 0 be the
maximum intensity of the interference
pattern then the intensity of the pattern at a
Slit  I
point where the phase difference between
P
I
Slit  II  Screen waves is  is given by 0  5  4 cos   .
P
Where P is in in integer. Find the value of
P?
41. Consider the optical system shown in figure.
Column-I The point source of light S is having
(A) Only strip B is placed over slit-I wavelength equal to  . The light is reaching
(B) Strip A is placed over slit-I and strip C is the screen only after reflection. For point
placed over slit-II
 Kd 2
(C) Strip A is placed over slit-I and strips B and ‘P’ to be 2nd maxima, the value of  .
C are placed over slit-II in conjnction 2 D
(D) Strips A and C are placed over slit I in Where K is an integer. Find value of K?
conjunction and strips B is placed on slit-II {Assume D>>>d and d>>>  ]
Column-II P
(P) First bright d
S
(Q) Fourth dark 3d
(R) Fifth dark
(S) Central bright
D
LEVEL-VI LEVEL-VI
KEY
HINTS
SINGLE OPTION TYPE
1-B 2-C 3-A 4-B Single Option Type Questions
5-C 6-A 7-B 1) Path dirfference at any point ‘P’ on the
circumference
MULTIPLE ANSWER TYPE
8 - A,C,D; 9 - A,C; 10 - A,C; B P
P
r2
COMPREHENSION TYPE 
 A r1
Comprehension-I: C R
11 - B; 12 - B; 13 - C S1 O S2  
S1 O S2
Comprehension-II: 3 3
2
2
14 - A; 15 - C; 16- B
Comprehension-III:
17 - C; 18 - A; 19 - D x  S1P  S2 P
Comprehension-IV: 3
S1 P  R  cos 
20 - C; 21 - D; 2
Comprehension-V:
22 - A; 23 - C; 24 - C 3
S2 P  R  cos 
Comprehension-VI: 2
25 - A; 26 - C; 27 - D x  S1P  S2 P
x  3 cos 
ASSERTION & REASONING TYPE
For maxima at ‘P’
28-A 29-C 30-C
x  n  3 cos 
STATEMENT(S) TYPE n
31-B 32-C 33-A cos    n  0,1, 2,3,    
3
MATRIX-MATCHING TYPE 1 2
cos   0, , ,1 [ maximum value of cos   1 ]
34) A  Q, R ; B  Q, S ; 3 3
 In each quadrant there are 3 maximas. Hence in
C  P, S ; D  P, Q , R , S 4 quadrants there willbe 12 maximas. Hence
35) A  P, S ; B  Q ; 12 minimas in between so there are 12 fringes
C  T ; D  R, S ,T 2) Conceptual
36) 3) Let S11 and S 21 are the points on the wavefront
A R; B R;
C  S; D  P where perpendiculars can be drawn from S1 and
S2 .
INTEGER TYPE For central fringe formed ar E the path difference
37-4 38-8 39-2 40-9 should be zero.
41-6
S11  S 2     1 t  S 2 E  S11S1  S1 E
2 2 
 O S 1  O S     1 t  
D 2  d 2  O S1  O S  D
1 1 
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    1 t   OS2  OS1    D2  d 2  D  cos  


4d
  d 2  12 
    1 t  2   D 1  2   D 
  D  
 
  d2  
 2   D  1  2 
 D
  2D  
 d2 
 2   D   D
 2D 
d2
   1 t  2 
2D 5) Conducting for costructive reference
3  16 2 
  1  t  2    2 t   2n  1  n  1, 2,3,    
2  2 128 2

1  t   2n  1
t  2  4
2 16
For least value of ‘t’, n = 1
1 31 31
t  
2 16 8 tmin   2 1  1
4
31
Thickness of mica sheet t 
8
4) PR=d
PO  d sec , and   1.8 t
CO  PO cot 2  d sec  cos 2   1.5
Optical path difference between the two ras is
 x  C O  P O  d sec   d sec  cos 2 
  648 109
2 tmin  
4 4 1.8
Path difference between the two rays is
tmin  90  109 m
  2n , n  0,1, 2,3... tmin  90 nm
Condition for constructive interference should be 6.  1 = 1.4 and  2 = 1.7 and let t be the thickness
..... of each glass plates.
x  n , n  0,1, 2,3.... Path difference at O, due to insertion of glass
 plates will be
d sec  1  cos 2  
2
S1 1
 d   th
6 Minima
  2 cos   
2
 O
 cos   2 th
5 Maxima
S2 2

x = (  2 –  1)t
= (1.7 – 1.4)t = 0.3 t …(1) MULTIPLE ANSWER QUESTIONS
Now, since 5th maxima (earlier) lies below O and
6th minima lies above O. 8) Solution: Path difference at ‘P’
This path difference should lie between 5l and 5l x   S1P  t  air  tmed  S1Pair

+  S1P  S 2 P  t  t
2
So, let x = 5l = + D …(2) x   S1P  S2 P      1 t

where x <
2
t
Due to the path difference Dx, the phase S1
difference at O will be
2 2 S P
  x = (5  + x ) = (10  +
 
S2
2
. )…(3)
 x
3 When there is no absorption of light by slab then
Intensity at O is given I and since

4 max path difference at P earlier is S1P  S 2 P  0 . so
 
change in optical path due to inserion of slab is
I(f) = Imax cos2  2 
 
   1 t .
3
Imax = Imax cos2  2  For Intensity to be zero at ‘P’ we have
4
3   x   2n  1

    1 t  t 
 2n  1 
or = cos  2 
-2 
…(4) 2 2    1
4
  3 5
From Eqs. (3) and (4), we find that D = Then t  2    1 , 2    1 , 2    1 ,     
6
 31
i.e., x = 5 +  l = 0.3 t(Dx = 0.3 t) 
6 6 9) I  4 I 0 cos 2
2
31 (31)(5400  1010 )
t = 6(0.3) = m
1.8  75  2 
4I0    4 I 0 cos
or t = 9.3 × 10–6 m = 9.3 mm  100  2
7. Let n1 bright fringe corresponding to wavelength
l1 = 500 nm coincides with n2 bright fringe  3  3 3
cos 2  ; cos  
corresponding to wavelength l2 = 700 nm. 2 4 2 4 2
n1 2 7   5 7 11
n1 1 D = n2 2 D or n =  =  , , , ......
d d 2 1 5 2 6 6 6 6
th
This implies that 7 maxima of l1 coincides with
 5 7 11
5th maxima of l2 . Similarly 14th maxima of l1 will   , , , .......
coincide with 10th maxima of l2 and so on. 3 3 3 3
the posibel value of
n11 D
\ Minimum distance = 11 13 17
d   , ,
–7 3
= 7 × 5 × 10 × 10 = 3.5 × 10 m = 3.5 mm –3 3 3 3
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10) Shift due to the introduction of slab is 2

2  1 103  3
D O  0  2t  t
   1 t . On the side of the slit where the 2 1 2
d
slab is placed. On placing the two slabs of smae 3t t
1 106  2t  
thickness, the shift should be 2 2
D t  2  106 m  t  2  m
 2  1  t . 12) From the above equation
d
Since three bright fringes shift is seen, 3 yd
O  1 106   0.5t
D 3 D 2 D
 2  1  t 
d d 3 yd
  0.5t  1 106
3 D 3  4000  1010 2 D
t 
 2  1  1  7  1.4   0.5  10 6  1 106  0.5  106
10 6 2 D 106  1 1
3  4 107 y       10 3
  4  106  4 m 2 3 d 3 110 3
3
0.3
t  4 m 1
y   mm
Shift 3
D 1  D 3000 1010 1 2
  2  1  t  3  0.3  4 10 6  1.2 103 13)   3
 2  104 m
d 10 2 d 3  110
Shift = 1.2 mm   0.2mm
Comprehension-II:
COMPREHENSION TYPE
Comprehension-I: S1 O S2
X
3
3
1 Y
D1  D gt2
1 2
0 x
t
1  3 t  2t
Area  A  3  dx   P0 P
2
v  gt x
11. Path difference
t
 
x  1SS2  2 S2 P   1SS1  2 S1 P    3   2  dx  14) At any instant of time ‘t’, the situtation is shown
 0  is figure.
t t The central maxima always lies along Y-axis at
 1  SS 2  SS1   2  S2 P  S1 P    3 dx   2 dx P0 . Its velcoity at any time ‘t’ is v  gt along
0 0
positive Y-axis.
In order to get central maxima at the centre ‘O’
of screen is x  0 v  gt  10  5  50ms 1
15) Path difference corresponding to P. Fringes will be observed in the region between
dx P1 and P2 because the reflected rays lie only in
x  d sin   d tan  ; x  D1
this region.
For 2nd maxima From similar les : BDS1 and S1P2 A .
xd 2 D1 AP2 AS1
x  2   2  ; x  
D1 d 1
BS1 BD ; BS1  BS  1mm  10 cm
 1 2
Location of central maxima is O, D  gt  AS1  BS1 200  0.1
 2  AP2    4cm
BD 5
 2 D1 1  AP2  4cm
Location of 2 maxima is  d , D 
nd
  Similarly, BCS1 and S1P1 A a re similar les
nd
Velocity of 2 maxima wrt central maxima, AP1 AS1
 
2 2
V rel   0  gt  i   gt i BS1 BD
d d
AS1  BS1 200  0.1
AP1    2cm
2  5000  10 10
 10  2i
Vrel   2  102 ms 1.i BD 10
1 10 3 Width of teh region wher fringes oberved is


V rel  2  cms 1  .i ; 1 2  2cm
1 2  AP2  AP1  4  2  2cm PP
PP
18) Wavelength of light used is
1
 3 D 1 C 3 108
16) Location of 3rd maxima  d , D     5  10 7 m
  D 6 1014
Location of 3rd maxima on the other side is D

Fringe width    
 3 D1 1  d
 d , D  D  S1 A  200cm  2m ;
 
 d  SS1  2mm  2  103 m
3  3    3 3 g  
a rel  g  g i   g  i
d  d  d d  5 10 7  2
  5  104 m
 2 10 3
6  6  5000  1010  10 
a rel  gi  i   5  102 cm    0.05cm
d 110 3
  19) No of fringes formed (N)
arel  3102 ms2 i ; arel  0.03  ms  i
2
P1P2 2 200
Comprehension - III:     40
 0.05 5
Comprehension - IV:
P2
20. Given —  = 1.33, d = 1 mm, D = 1.33 m
l = 6300 Å
S
P1 Wavelength of light in the given liquid —
1mm
O  6300
B ' =  Å » 4737 Å = 4737 × 10–10 m
17) D C  1.33
S1 A  D
Fringe width, w =
d
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(4737  10 10 m)(1.33 m) 24. In part (i), shifting of 5 bright fringes was equal
w= (1  10 3 m)
= 6.3 × 10–4 m to 10–3 m. Which implies that
5wred = 10–3 m
21. Let t be the thickness of the glass slab.
[Here w = Fringe width]
Path difference due to glass slab at center O.
103
 µ glass  1.53 wred = m = 0.2 × 10–3 m
  1 t =  
 1 t 5
x =  µ liquid

  1.33 
D
or x = 0.15 t Now since w = or w µ l
d
ωgreen green
ω red
= 
red
green  5  107 
O wgreen = wred  –3
= (0.2 × 10 )  7  107 
red  
wgreen = 0.143 × 10–3 m
Dw = wgreen – wred = (0.143 – 0.2) × 10–3 m
Dw = –5.71 × 10–5 m
Now, for the intensity to be minimum at O, this Comprehension - VI:
 25. Given l = 600 nm = 6 × 10–7 m
path difference should be equal to

2 d = 0.45 mm = 0.45 × 10–3 m ; D = 1.5 m
 Thickness of glass sheet, t = 10.4 mm
x = 2 = 10.4 × 10–6 m
4737 4
or 0.15 t = Å Refractive index of medium, mm =
2 3
t = 15790 Å or t = 1.579 mm And refractive index of glass sheet, mg = 1.5
Comprehension - V: Let central maximum is obtained at a distance y
22. Path difference due to the glass slab, yd
Dx = (m – 1)t = (1.5 – 1)t = 0.5t below point O. Then Dx1 = S1P – S2P =
D
Due to this slab, 5 red fringes have been shifted
upwards, Therefore S1
t O

y
P
S2
g  
–7 Path difference due to glass sheet Dx2=    1 t
x = 5  red or 0.5t = (5)(7 × 10 m)  m 
t = thickness of glass slab = 7 × 10–6 m Net path difference will be zero when Dx1= Dx2
23. Let m¢ be the refractive index for green light then
yd g  g   D
x = 6  Red or =   1 t  y=    1 t d
D  m   m 
(   1 )t = 6  Red
Substituting the values, we have
(6)(7  107 )
(  1) = = 0.6  1.5  10.4  10 6 (1.5)
7  10 6 y=   1 3 ; y = 4.33 × 10–3 m
 4 3  0.45  10
  1.6 or we can say y = 4.33 mm

g   D  1
26. At O, Dx1 = 0 and Dx2 =    1 t 30)        
 m  d  
Net path difference, Dx = Dx2    air hence   air thus    air
Corresponding phase difference, Df or simple f
A is true and R is false.
2
= . Dx Substituting the values, we have
 STATEMENT(S) TYPE
2  1.5   13 
f = 6  107  4 3  1 =  3  
    31) path difference x  d sin   xmax  d , if
   13 
Now, I(f) = Imax cos2  2  ; I = Imax cos2  6  d   then xmax   so maxima can be
present and interference pattern cannot be
3 observed. Statement-II is true,but not explaining
I = Imax
4 Statement-I
 g  32) In YDSE, law of converstion of energy is
27. At O : path difference is Dx = Dx2 =    1 t obeyed.
 m 
33) At the location of minima, the two waves have
For maximum intensity at O different intensities hence minimum intensity is not
Dx = nl (Here n = 1, 2, 3, …) equal to zero.
x x x  1.5 
l= , , … and so on Dx =  4 3  1 MATRIX-MATCHING TYPE
1 2 3  
 1.5 
(10.4 × 10–6 m) =  4 3  1 (10.4 × 103 nm) 34) A) For 1  2 , the two waves in the reflected
 
Maximum intensity will be corresponding to system differ by optical phase difference of 
and hence due to destructive interference, the
1300 1300 1300
l = 13000 nm, nm, m, nm …. film appears to be dark. Intransmitted system,
2 3 4
the two waves are in phase and hence due to
= 1300 nm, 650nm, 433.33 nm, 325 nm nm …. constructive interference the film appears to be
The wavelengths in the range 400 to 700 nm are
shiny. A  Q, R
650 nm and 433.33 nm.
Similarly we can go for the other conditions.
ASSERTION & REASON TYPE 35) A)
  
I  P1   I1  I 2  2 I1 I 2 cos   
28) b  R hence  b   R hence fringes becomes 2  2
D I  P1   I 0  I 0  0  2 I 0
narrower. Fringe width   .
d
 2 
A & R are true and R is the correct explanation I  P2   I1  I 2  2 I1 I 2 cos  
29) When one of the slits is covered with cellophane  3 
paper, the intensity of light imerging out from the  1
slit decreases. Now the two interfering beam have  I0  I0  2I0   
 2
different intensities or amplitudes. Hence intensity
at minima will not be zero and fringe will become I  P2   2 I 0  I 0  I 0
indistinct. A is true, R is flase.
 I  P1   I  P2    P0   0 A  P, S

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2  2     2  3   2
B)   P0      1 t       P1      1 t      
 4  2  4   4 4  
   2     2 2 2
  P1      1 t      0  
 4   4 4   4 
   2     2   2     2  3   2
  P2      1 t            P2      1 t     
 3   4 3   12  6
 3   4 3  
I  P1   I1  I 2  2 I1 I 2 cos 00  I 0  I 0  2 I 0  4 I 0 5 2 5
  
   12  6
I  P2   I1  I 2  2 I1 I 2 cos  
 6  I  P1   I1  I 2  2 I1 I 2 cos 
 3  3 
 I 0  I 0  2 I 0   I  P0   I1  I 2  2 I1 I 2 cos    2I0
 2   2 
 2I0  2I0  0

I  P2   2  3 I 0  BQ
 5 
2  2 I  P2   I1  I 2  2 I1 I 2 cos  
  P0      1 t   6 
C)   
 2 
   3
   2     2  I 0  I 0  2 I 0    cos   2 I 0  2 I 0    
  P1      1 t     
 4   2 4    6  2 
 2 
  
I  P2   2  3 I 0 
4  2
I  P0   I  P1  ; I  P2   I  P1  ;
   2     2
  P2      1 t      
 3   2 3   36) By using    1 t  n we can find the value
of n.i.e., order of fringe produced at P, is that
 2 
   particular strip has been placed over any of the
6  3 slit. If two strips are used in conjunction, path
I  P0   I1  I 2  2 I1 I 2 cos  difference due to each is added to get net path
difference created. If two strips are used over
 I0  I0  2I0  1  2I0  2I0  0 different slits, their path differences are subtracted
to get net path difference. For A : When strip
 
I  P1   I 1  I 2  2 I 1 I 2 cos   B is placed on slit-I alone then
 2
 I0  I0  0  2 I0   B  1 t B  2.5  11.5  106
nB  
B 500  10 9
 
I  P2   I1  I 2  2 I1 I 2 cos  
3 1.5 1.5  10 6
  4.5
5 10 7
1
 I0  I0  2I0   3 I 0 ( I  P2   I  P1  ) Order of fringe is 4.5 i.e., 5th dark Film A:
2
nA 
  A  1 t A  1.5  1 5 106
2 3 2 3
D)   P0      1 t      500  10 9
 4  2
5 1.5 10 6   8  104  P  104  P  8

nA  5
5 10 7 1 D D
39) n1 1  n2  2  n1  n2 2
Film C: Order of fringe d d
 C  1 tC  2  1 0.25  106 n11  n22 ; nb b  nR R
nC  
 500  109  n  1  5 105  n  7.5 105
1 2.5 108 1
  0.5 n  1  1.5n  0.5n  1  n 
5 10 7 0.5
nC  0.5 10
For B: When strip A is palced over slit-I and n 2n2
5
strip C is placed on slit-II then 40) As the amplitude are A and 2A then the ratio of
Net order of fringe: 5  0.5  4.5 i.e., 5th dark intensities is 1:4
For C: When strip A i splaced over slit-I and
strips B and C are placed over slit-II then I max  I 0  I1  I 2  2 I1  I 2  I  4 I  2  2 I
Net order of fringe 5   4.5  0.5   5  5  0 I0

I0  9I  I 
i.e., central bright fringe 9
Intensity at any point:
For D: When slits A&C are placed over slit-I

and stipr B is placed over slit-II then Net order I 1  I1  I 2  2 I1  I 2 cos 
fringe  5  0.5  4.5  5.5  4.5  1
I 1  I  4 I  2 I  4 I cos 
i.e.,1st bright
I 1  5I  4 I cos  ; I  I 5  4cos  
1
INTEGER TYPE QUESTIONS I0 I

I1  5  4 cos    0 5  4 cos  
9 P
37) For  path difference phase difference  2
P 9
 2
 
I  I 0 cos 2 
2
 I 0 cos 2 
 2

  I 0  1  I 0
 41) At P: Path difference x 
8d  3d
D
 24d 2 12d 2
For path differece phase difference For 2nd maxima x  2  ;
3
D D
2  2
1     6d 2 kd 2
 3 3   k 6
2 D D
 1   2  2 
I 1  I 0 cos 2     I 0 cos 2    I 0 cos  
 2   3 2  3 * * *
I0 I I
I1 
  P4
4 4 P
38) Angular fringe width is
 
 
D d
4800  1010 48  108
 
0.6  103 6  104

3.wave Optics Final - PMD

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JEEADVANCED

JEE MAINS - CW--VOL

SINGLE CORRECT ANSWER TYPE S 2 which are d  2mm apart. Slit S1 is

calcualte ratio of intensity at C to the

NISHITH Multimedia India (Pvt.) Ltd., 89

0.5mm intensities at A and B is: (2001; 2M)

(C) 4  103 ms 1 (D) 5  10 3 ms 1 

of light falling on slit-1 is four times the 8 16

21. The central bright fringe can be obtained

NISHITH Multimedia India (Pvt.) Ltd., 93

MATRIX MATCHING TYPE P

33. Monochromatic light source having S1 x  10mm

(>>>d) as shown in figure. The slits width

NISHITH Multimedia India (Pvt.) Ltd., 95

34) x  t1   g 1  t2   g 1  g  t1  t2    t1  t2 

96 NISHITH Multimedia India (Pvt.) Ltd.,

The time when y becomes zero is 7. I(  ) = I1 + I2 + 2 I1 I 2 cos  …(1)

Central maxima is at ‘O’ at T  4sec  

98 NISHITH Multimedia India (Pvt.) Ltd.,

(C) and (D) are wrong. 14.  Im cos 2  

Next 11th minima of 400 nm coincides with 8 th MULTIPLE ANSWER TYPE

maxima 18) intensity at ‘O’ is proportional to intensity at S3

NISHITH Multimedia India (Pvt.) Ltd.,

 A & R are true but R isnot the correct 1 1

 S 2 P  S1 P    (B) Intensity is 25I 0

x1 131 106 13110 6 1310

x1 x2 x 0.031 103 31 106

NISHITH Multimedia India (Pvt.) Ltd., 103

For r: ‘d’ is increasing  and  both decreases 1

NISHITH Multimedia India (Pvt.) Ltd., 105

2. In standared YDSE slits were moved apart 3 

the minimum value of thickness of mica sheet

106 NISHITH Multimedia India (Pvt.) Ltd.,

MULTIPLE ANSWER QUESTIONS COMPREHENSION TYPE

8. A transparents slab of thickness ‘t’ and Comprehension-I:

14. The velocity of central maxima at t  5sec

Reason(R): The cellophane paper decreases

for that situtation. (2009-JEE)

Film A B  n  1 blue bright band coincides with nth

    1 t   OS2  OS1    D2  d 2  D  cos  

NISHITH Multimedia India (Pvt.) Ltd., 113

Intensity at O is given I and since

10) Shift due to the introduction of slab is 2

1 10 3 Width of teh region wher fringes oberved is

Location of 3rd maxima on the other side is D

path difference should be equal to

NISHITH Multimedia India (Pvt.) Ltd., 117

118 NISHITH Multimedia India (Pvt.) Ltd.,

5 1.5 10 6   8  104  P  104  P  8

Net order of fringe 5   4.5  0.5   5  5  0 I0

For D: When slits A&C are placed over slit-I

INTEGER TYPE QUESTIONS I0 I

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