Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Q.

P
Physics
31. In Young’s double slit arrangement, water is filled in the space between screen and slits.
Then
1) fringe pattern shifts upwards but fringe width remains unchanged
2)fringe width decreases and central bright fringe shifts upwards.
3) fringe width increases and central bright fringe does not shift.
4) fringe width decreases and central bright fringe does not shift
32. In a Young’s double-slit experiment, let A and B be the two slits. A thin film of thickness
t and refractive index µ is placed in front of A. Let β = fringe width. The central
maximum will shift by:
t β µ β β β
1) ( µ − 1) 2) t 3) t ( µ − 1) 4) µ t
2 λ 2 λ λ λ
33. In the previous question, films of thicknesses t and t B and refractive indices µ A and
A
µ B , are placed in front of A and B respectively. If µ A t A = µ B tB , the central maximum will:
1) not shift
2) Shift towards A
3) Shift towards B
4) Shift towards A if t B > t A ; and Shift towards B if t B < t A
34. In YDSE how many maxima can be obtained on the screen if wavelength of light used is
200 nm and d=700 nm:
1) 12 2) 7 3) 18 4) None of these
35. Two coherent monochromatic light beams of intensities I and 4I are superposed. The
maximum and minimum possible intensities in the resulting beam are :
1) 5I and I 2) 5I and 3I 3) 9I and I 4) 9I and 3I
36. A parallel beam of light 500 nm is incident at an angle 300 with the normal to the slit
plane in a young’s double slit experiment. The intensity due to each slit is I0. Point O is
equidistant from S1 and S2 . The distance between slits is 1 mm
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A
300 d D=3 m o
B

Screen

1) the intensity at O is 4I0

2) the intensity at O is Zero
3) the intensity at a point on the screen 4mm from O is 4I0
4) the intensity at a point on the screen 4 mm from O is zero.
37. Light of wavelength λ in air enters a medium of refractive index µ . Two points in this
medium, lying along the path of this light, are at a distance x part. The phase difference
between these points is :
2πµ x 2π x 2π ( µ − 1) x 2π x
1) 2) 3) 4)
λ µλ λ ( µ − 1) λ
0
38. In Young’s double slit experiment, the wavelength of red light is 7800 A and that of blue
0
light is 5200 A . The value of n for which nth bright band due to red light coincides with
th
( n + 1) bright band due to blue light, is:
1) 1 2) 2 3) 3 4) 4
39. Two monochromatic and coherent point sources of light are placed at a certain distance
from each other in the horizontal plane. The locus of all those points in the horizontal
plane which have construct interference will be
1) a hyperbola 2) family of hyperbolas
3) Family of straight lines 4) Family of parabolas
40. Two points monochromatic and coherent sources of light of wavelength λ are placed on
the dotted line in front of an large screen. The source emit waves in phase with each
other. The distance between S1 and S2 is ‘d’ while their distance from the screen is much
larger. Then,

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S1 S2 o

1. → If d = 7λ / 2, O will be a minima
2. → If d = 4.3λ , there will be a total of 8 minima on y axis
3. → If d = 7λ , O will be a maxima
4. → If d = λ , there will be only one maxima on they y aixs
Which is the set of correct statement.
1) 1, 2 & 3 2) 2, 3 & 4 3) 1, 2 , 3& 4 4) 1, 3 & 4
41. Two identical narrow slits S1 and S2 are illuminated by light of wavelength λ from a
point source P. If, as shown in the diagram above the light is then allowed to fall on a
screen, and if n is a positive integer, the condition for destructive interference at Q is that

S1
l1 l3
P
l2 Q
l4
S2

1) ( l1 − l2 ) = ( 2n + 1) λ / 2 2) ( l3 − l4 ) = ( 2n + 1) λ / 2
3) ( l1 + l2 ) − ( l2 + l4 ) = nλ 4) ( l1 + l3 ) − ( l2 + l4 ) = ( 2n + 1) λ / 2
42. Figure shown plane waves refracted for air to water using Huygen’s principal a,b,c,d,e
are lengths on the diagram. The refractive index of water wrt air is the ratio.

a
air b
c
d e
water

1) a/e 2) b/e 3) b/d 4) d/b

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 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Q.P
43. To make the central fringe at the centre O, a mica sheet of refractive index 1.5 is
introduced. Choose the correct statements (s)
d
S S1

d
O

S2
D>>d

1) The thickness of sheet is 2 ( 2 − 1) d S1

2) The thickness of sheet is ( )

2 − 1 d in front of S 2

3) The thickness of sheet is 2 2 d in front of S1

4) The thickness of sheet is 2 ( 2 − 1) d in front of S1

44. In a young double slit experiment D equals the distance of screen and d is the separation
between the slit. The distance of the nearest point to the central maximum where the
intensity is same as that due to a single slit, is equal to
Dλ Dλ Dλ 2Dλ
1) 2) 3) 4)
d 2d 3d d
45. Two point source separated by d=5 µ m emit light of wavelength λ = 2 µ m in phase. A
circular wire of radius 20 µ m is placed around the sources as shown in figure.
A

20µm
D B
5µm

1) Point A and B are dark and point C and D are bright

2) Point A and B are bright and point C and D are dark
3) Point A and B are dark and point B and D are bright
4) Point A and C are bright and point B and D are dark
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 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Q.P
46. In a YDSE experiment if a slab whose refractive index can be varied is placed in front of
one of the slits then the variation of resultant intensity at mid-point of screen with ' µ '
will be best represented by ( µ ≥ 1) . [Assume slits of equal width and there is no
absorption by slab]
I0 I0

µ µ

1) µ =1 2) µ =1

I0
I0

µ
3) µ =1 µ 4) µ =1
0 0
47. A beam of light consisting of two wave lengths 6500 A and 5200 A is used to obtain
interference fringes in Young’s double slit experiment. The distance between slits is 2
mm and the distance of screen from slits is 120 cm. What is the least distance from
central maximum where the bright due to both wavelength coincide?
1)0.156 cm 2) 0.312 cm 3) 0.078 cm 4) 0.468 cm
48. Young’s double slit experiment is carried with two thin sheets of thickness 10.4 µ m each
and refractive index µ1 = 1.52 and µ 2 = 1.40 covering the slits S1 and S2 , respectively. If
white light of range 400 nm to 780 nm is used then which wavelength will form maxima
exactly at point O, the centre of the screen?

S1
O
S2

1) 416 nm only 2) 624 nm only

3) 416 nm and 624 nm only 4) None of these

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0
49. In a biprism experiment using sodium light λ = 6000 A an interference pattern is obtained
in which 20 fringes occupy 2cm. On replacing sodium light by another source of
wavelength λ2 without making any other change 30fringes occupy 2.7 cm on the screen.
What is the value of λ2 ?
0 0 0 0
1) 4500 A 2) 5400 A 3) 5600 A 4) 4200 A
50. Two coherent narrow slits emitting light of wavelength λ in the same phase are placed
parallel to each other at a small separation of 3λ . The light is collected on a screen S
which is placed at a distance D ( >> λ ) from the slits. The smallest distance x such that the
P is a maxima.
P

S1 S2 O

D
D
1) 3D 2) 8D 3) 5D 4) 5
2
51. In a Young’s Double slit experiment, first maxima is observed at a fixed point P on the
screen. Now the screen is continuously moved away from the plane of slits. The ratio of
intensity at point P to the intensity at point O (centre of the screen)

P
O

1) remains constant 2) Keeps on decreasing

3) first decreases and then increases 4) First decreases and then becomes constant
52. In a double slit experiment, the separation between the slits is d=0.25 cm and the distance
0
of the screen D=100 cm from the slits. If the wavelength of light used is λ = 6000 A and I 0
is the intensity of the central bright fringe, the intensity at a distance x = 4 × 10−5 m from
the central maximum is
1) I 0 2) I 0 / 2 3) 3I 0 / 4 4) I 0 / 3
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 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Q.P
53. A parallel coherent beam of light falls on Fresnel biprism of refractive index µ and angle
α .The fringe width on a screen at a distance D from biprism will be (wavelength = λ )
λ λD D
1) 2) 3) 4) none
2 ( µ − 1) α 2 ( µ − 1) α 2 ( µ − 1) α
54. A monochromatic light source of wavelength λ is placed at S. Three slits S1 , S2 and S3 are
equidistant from the source S and the point P on the screen.
S1 P − S 2 P = λ / 6 and S1 P − S3 P = 2λ / 3 . If I be the intensity at P when only one slit is open,
the intensity at P when all the three slits are open is

S1

S2
P
S3
S
Screen
D D
( D>>λ)
1) 3 I 2) 5 I 3) 8 I 4) zero
55. In young’s double slit experiment, the value of λ = 500 nm . The value of d=1mm, D=1 m.
Then the minimum distance from central maximum for which the intensity is half the
maximum intensity will be
1) 2.5 × 10−4 m 2) 2 × 10−4 m 3) 1.2 × 10−4 m 4) 10−4 m
56. In a YDSE with two identical slits, when the upper slits is covered with a thin, perfectly
transparent sheet of mica, the intensity at the centre of screen reduces to 75% of the
initial value. Second minima is observed to be above this point and third maxima below
it. Which of the following can not be a possible value of phase difference caused by the
mica sheet
π 13π 17π 11π
1) 2) 3) 4)
3 3 3 3
57. Radio waves coming at ∠α to vertical are received by a radar after reflection from a
nearby water surface & directly. What should be minimum height of antenna from water
surface so that it records a maximum intensity. (wavelength = λ )

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 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Q.P

h α

λ λ λ λ
1) 2) 3) 4)
2 cos α 2sin α 4sin α 4 cos α
58. A thin film of thickness t and index of refraction 1.33 coats a glass with index of
refraction 1.50. What is the least thickness t that will strongly reflect light with
wavelength 600 nm incident normally?
1) 225 nm 2) 300 nm 3) 400 nm 4) 450 nm
59. In the figure shown if a parallel beam of white light is incident on the plane of the slits
then the distance of the white spot on the screen form O is [Assume d << D, λ << d ]

d 2d/3
O

1) 0 2) d/2 3) d/3 4)d/6

60. In the above question if the light incident is monochromatic and point O is a maxima,
then the wavelength of the light incident cannot be
1) d 2 / 3D 2) d 2 / 6D 3) d 2 / 12D 4) d 2 / 18D

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 Sri Chaitanya IIT Academy., India.
A.P, TELANGANA, KARNATAKA, TAMILNADU, MAHARASHTRA, DELHI, RANCHI
A right Choice for the Real Aspirant
ICON Central Office – Madhapur – Hyderabad
Sec: Sri Chaitanya-Sr.Chaina(Coming) Jee-Main Date: 04-05-19
Time: 3 Hr’s WTM-31 Max.Marks:360
Key Sheet
MATHEMATICS:

1) 1 2) 3 3) 1 4) 4 5) 1 6) 2 7) 4 8) 3 9) 3 10) 3
11) 2 12) 4 13) 4 14) 3 15) 2 16) 1 17) 3 18) 2 19) 2 20) 3
21) 3 22) 4 23) 2 24) 1 25) 1 26) 3 27) 2 28) 3 29) 2 30) 3

PHYSICS:

31) 4 32) 3 33) 4 34) 2 35) 3 36) 1 37) 1 38) 2 39) 2 40) 3
41) 4 42) 3 43) 1 44) 3 45) 4 46) 3 47) 1 48) 3 49) 2 50) 4
51) 3 52) 3 53) 1 54) 1 55) 3 56) 1 57) 4 58) 1 59) 4 60) 1

CHEMISTRY:

61) 2 62) 2 63) 2 64) 4 65) 2 66) 3 67) 3 68) 4 69) 4 70) 3
71) 4 72) 3 73) 3 74) 2 75) 2 76) 2 77) 3 78) 2 79) 2 80) 2
81) 4 82) 2 83) 3 84) 2 85) 2 86) 3 87) 4 88) 1 89) 4 90) 4
 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Key & Sol’s
PHYSICS

31. Central bright finge does not shift, however the pattern is compressed.
32.
yd ( µ − 1) tD
( µ − 1) t = ⇒y=
D d
 ( µ − 1) t 
=  ×β
 λ 
33. Central maximum will not shift it
( µA − 1) t A = (µ B − 1) t B
⇒ µA t A − t A = µ Bt B − t B
∵t A < t B , ( µ A t A − t A ) > ( µ B t B − t B )
so shifts towards A

d
34. = 3.5
λ
So, number of maxima will be 7.

( ) ( )
2 2
35. Imax = I1 + I 2 and Imax = I1 − I 2
d
36. ∆x = d sin θ = = 0.5mm = 107 × λ
2
So ‘O’ is a maximum.
2π
37. φ= × ∆x here ∆x = µx
λ
n × λ R × D ( n + 1) λ R × D
38. = ⇒n=2
d d
39. S1P − S2 P = nλ is a hyperbola for a given n.
40. Here we get circular finger. As these circular intersect y-axis at two place, so 8 points on y-axis are
minima.s
λ
41. Conceptual, ∆x = ( odd no ) ×
2
42.

i0
r

sin i b / c b
µ= = =
sin r d / c d
43. ( µ − 1) t = ( ) (
2 −1 d ⇒ t = 2 2 −1 d )
It should be infront of S1
44.

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 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Key & Sol’s
φ φ 1 2π
I0 = 4I0 cos 2   ⇒ cos   = ⇒ φ =
2 2 2 3
d λ λD
y = ⇒y=
D 3 3d
45. Conceptual.
46. Phase difference
φ = ( µ − 1) t
 ( µ − 1) t 
so I = 4I0 cos 2  
 2 
47. n × 65 = ( n + 1) × 52 ⇒ n = 4
4 × 6500A 0 × 120 cm
So y = = 0.156cm
2 mm
48.
( µ1 − µ2 ) t = n
λ
0.12 × 10.4 µm
⇒λ=
n
n is an integral number.
49. even is a bi-prism experiment
λD
Fringe width =
d
50.
d cos θ = 2λ
D D2 4
d× = 2λ ⇒ 2 2
=
D2 + x 2 D +x 9
⇒ 5D 2 = 4x 2
D
⇒x = 5×
2
51. Conceptual.
52.
d
∆x = y ×
D
2π
φ= × ∆x
λ
φ
so I = I0 cos 2  
2
λ
53. 8 = ( µ − 1) α , β =
2 ( µ − 1) α
54.

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 Sri Chaitanya IIT Academy 04-05-19_Sri Chaitanya-Sr.Chaina(Coming)_Jee-Main_WTM-31_Key & Sol’s
λ 2π
SS1P − SS2 P = ⇒ φ1 =
3 3
4λ 2π
also, SS1P − SS3P = ⇒φ2 = 4 ×
3 3
2π
= 2π +
3
A
A

So A

A res = 3A so I res = 3I
55. Conceptual
56.
1
cos φ =
2
But 3π < φ< 6 π
λ
57. 2h cos α =
2
As due to reflection, additional phase charge of π takes place.
58. Conceptual
59. Conceptual
60. Conceptual

CHEMISTRY

61. NO2− and SCN − are ambidentate ligands.

62. Mohr’salt ( FeSO4 . ( NH 4 )2 SO4 .6 H 2 O ) is a double salt.
+
63.  Fe ( NH 3 )4 Cl2  is a heteroleptic complex because it has more than one type of ligands.

+2
 +1

64. Sodium nitroprusside is Na2  Fe ( CN )5 NO 
 
EAN of Fe =26-2+12=36
 +4 
 P t  N H 3   Br    Cl   Cl
a b n c
65.  N O 2  
  3     
 
Triamminebromochloronitroplatinum chloride
66.
CrCl3.6H2O+ AgNO3 →AgCl
1 mol excess 2 mol
The formula of the complex is Cr ( H 2 O )5 Cl  Cl2 .H 2 O
+2
67. According to VBT, geometry of Cu ( NH 3 ) 4  is tetrahedral but its actual geometry is square planar.
68. Cause of colour of a compound: Polarisation, d-d transition, charge transfer spectra.
69. Ma4 b2 complex does not show optical isomerism.

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 Final Key
GIVEN FINAL
S.NO SUB Q.NO REMARKS
KEY KEY
We have circular fringes so please above & below x-axis are the same minima and not to be
3 PHY 40 3 Delete canted. Points at ∞, − ∞ twice should also be considered because screen is large.
4 PHY 43 1 1 or 4 1 and 4 are same