Quizrr Chapter-Wise Test for JEE Advanced - 2025

By: C I P H Ξ R

Q.1

In YDSE, one of the slits is covered by a thin sheet of variable refractive index, whose variation with time is
represented by graph as shown. The variation of intensity (I) at a point on the screen with time, where first minima
is obtained at t = 0 will be best represented by

(A)

(B)

(C)

(D)

Q.2

In Young's double slit experiment, double slit of separation 0.1 cm is illuminated by white light. A coloured
interference pattern is formed on a screen 100 cm away. If a pin hole is located on this screen at a distance of
2 mm from the central fringe, the wavelength in the visible spectrum which will be absent in the light transmitted
through the pin-hole are

(A) 5714Å and 4444Å

(B) 6000Å and 5000Å
(C) 5500Å and 4500Å
(D) 5200Å and 4200Å

Q.3

Young's double slit experiment is first performed in air and then by immersing the whole setup in a liquid. The 10th
bright fringe when in liquid is formed at the point where 8th dark fringe is located when in air. The refractive index
of the liquid is :

(A) 1.25
(B) 1.33
(C) 1.40
(D) 1.20

Q.4

In a Young's experiment, two coherent sources are placed 0.90 mm apart and the fringes are observed one metre
away. It produces the second dark fringe at a distance of 1 mm from the central fringe, the wavelength of
monochromatic light used would be.

(A) 60 × 10−4 cm
(B) 10 × 10−4 cm
(C) 10 × 10−5 cm
(D) 6 × 10−5 cm

Q.5

The young's double slits experiment is performed with blue and with green light of wavelength 4360Å and 5460Å
respectively. It x is the distance of the 4th maxima from the central one, then

(A) xblue = xgreen

​ ​

(B) xblue > xgreen

​ ​

(C) xblue < xgreen

​ ​

(D) None of the above

Q.6

A beam of light consisting of two wavelength 6500Å & 5200Å is used to obtain interference fringes in a young's
double slit experiment. The distance between the slits is 2.0 mm and the distance between the plane of the slits and
the screen is 120 cm. What is the least distance from the central maximum where the bright fringes due to both the
wave length coincide ?

(A) 0.156 cm
(B) 0.152 cm
(C) 0.17
(D) 0.16 cm.

Q.7

In an experiment similar to young's experiment, interference is observed using waves associated with electrons. The
electrons are being produced in an electron gun. In order to increase the fringe width.

(A) electron gun voltage be increased.

(B) electron gun voltage be decreased.

(C) the slit be moved away.

(D) the screen be moved closer to interfering slits.

Q.8

Interference fringes were produced in young's double slit experiment using light of wave length 5000Å. When a
film of material 2.5 × 10−3 cm thick was placed over one of the slits, the fringe pattern shifted by a distance equal
to 20 fringe width. The refractive index of the material of the film is-

(A) 1.25
(B) 1.33
(C) 1.4
(D) 1.5

Q.9

If the first minima in a Young's slit experiment occurs directly in front of one of the slits, (distance between slit and
screen D = 12 cm and distance between slits d = 5 cm.) then the wavelength of the radiation used can be -

(A) 2 cm.
(B) 4 cm.
(C) 2/3 cm.
(D) 4/3 cm.

Q.10

Choose the correct options -

(A) A point object is kept in front of a plane mirror. The plane mirror is performing SHM of amplitude 2 cm. The
plane mirror moves along the x-axis and y -axis is normal to the mirror. The amplitude of the mirror is such that
the object is always infront of the mirror. The amplitude of SHM of the image is 4 cm.
(B) In Young's double slit experiment 7 maxima can be obtained on screen including central maxima if d =
7000Å and λ = 2000Å.
(C) A convex lens of focal length 30 cm and a concave lens of focal length 10 cm are placed at some distance. If a
parallel beam of light incident on convex lens emerges as parallel beam from convex lens than the distance
between two lenses is 20 cm.
(D) In Young's double slit experiment 5 maxima can be obtained on screen including central maxima if d =
7000Å and λ = 2000Å.

Q.11

Two coherent point light sources are placed at x = −d/2 and x = d/2. These sources are surrounded by a
spherical screen having equation x3 + y 2 + z 2 = R2 (R ≫ d). The sources emit monochromatic light of
wavelength λ = d/3

(A) The number of maxima produced on the screen are 7

(B) The shape and length of second order maxima is circle, 2πR 5 ​

3
​

(C) If radius of screen is doubled the intensity of maxima will become one fourth
(D) All maxima are equally spaced along x-axis

Q.12

Passage
The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength
600 nm.1, 2, 3, 4 and 5 are marked on five fringes.

Question
The third order bright fringe is -

(A) 2

(B) 3
(C) 4
(D) 5

Q.13

Passage
The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength
600 nm.1, 2, 3, 4 and 5 are marked on five fringes.

Question
Which fringe results from a phase difference of 4π between the light waves incidenting from two slits -

(A) 2
(B) 3
(C) 4
(D) 5

Q.14

Passage
The figure shows the interference pattern obtained in a double-slit experiment using light of wavelength
600 nm.1, 2, 3, 4 and 5 are marked on five fringes.

Question
Let ΔXA and ΔXC represent path differences between waves interfering at 1 and 3 respectively then
​ ​

(∣ΔXC ∣ − ∣ΔXA ∣) is equal to -

​ ​

(A) 0
(B) 300 nm
(C) 600 nm
(D) 900 nm

Q.15

Passage
= 5000Å is incident on the slits (in a horizontal fixed plane) S1 and S2
In the figure shown light of wavelength λ ​ ​

separated by distance d = 1 mm. A horizontal screen S is released from rest from initial distance D0 = 1 m
​
from the plane of the slits. Taking origin at O and positive x and y axis as shown, ( Use g = 10 m/s2 )

Question
Velocity of central maxima in vector form at t = 2sec. is −
(A) 20^jm/s ​

(B) 5^jm/s
​

(C) 10^jm/s ​

(D) 25^jm/s ​

Q.16

Passage
In the figure shown light of wavelength λ = 5000Å is incident on the slits (in a horizontal fixed plane) S1 and S2
​ ​

separated by distance d = 1 mm. A horizontal screen S is released from rest from initial distance D0 = 1 m
​

from the plane of the slits. Taking origin at O and positive x and y axis as shown, ( Use g = 10 m/s2 )

Question
Acceleration of central maxima in vector form at t = 2sec. is-
(A) 5^j m/s2
​

(B) 15^jm/s2 ​

(C) 10^jm/s2 ​

(D) 20^jm/s2 ​

Q.17
Passage
In the figure shown light of wavelength λ = 5000Å is incident on the slits (in a horizontal fixed plane) S1 and S2
​ ​

separated by distance d = 1 mm. A horizontal screen S is released from rest from initial distance D0 = 1 m ​

from the plane of the slits. Taking origin at O and positive x and y axis as shown, ( Use g = 10 m/s2 )

Question
Relative acceleration of second maxima with respect to first minima, on the same side of central maxima is -

(A) 2.5 × 10−3 m/s2^i

(B) 7.5 × 10−3 m/s2^j ​

(C) 5.5 × 10−3 m/s2^i

(D) 7.5 × 10−3 m/s2^i

Q.18

A monochromatic parallel beam of light of wavelength λ is incident normally on the plane containing slits S1 and ​

S2 . The slits are of unequal width such that intensity only due to one slit on screen is four times that only due to other
​

slit. The screen is placed along y axis is as shown. The distance between slits is d and that between screen and slit is
D. Match the statements in column I with results in column-II.
(A) (A) − q, r, s(B) − p, q, r, s(C) − q, r, s(D)p, q, r, s
(B) (A) − s, p(B) − p, (C) − s, (D) − q
(C) (A) − q, r(B) − q, r(C) − p, q, r, s(D) − p, s
(D) (A) − p, r, (B) − q, r, (C) − q, r, (D) − q, r

Q.19

A narrow monochromatic beam of light of intensity I is incident on a glass plate A as shown in figure. An other
identical glass plate B is kept close to A and parallel to it. Each glass plate reflects 25% of the light intensity incident
on it and transmits the remaining. Find the ratio of the maximum and minimum intensities in the interference pattern
formed by the two beams obtained after one reflection at each plate.

Q.20

Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an
index of refraction of n = 1.50 and a glass plate with an index of n = 1.80, the diameter of the third bright ring is
0.640 mm. If water (n = 1.33) now fills the space between the lens and the glass plate, what is the new diameter
of this ring? Assume the radius of curvature of the lens is much greater than the wavelength of the light.

Answers & Solutions

Q.1 Answer:
—
Solution:

The path difference due to the slab is, (Δx)0 ​ = (μ − 1)t (t = thickness )
And its corresponding phase difference is,
2π
Δϕ = (μ − 1)t ​

λ
​

Δϕ ​

Ires = 4I0 cos2

2
​ ​ ​

For Imax ; cos2 ( 2 ) = 1

Δϕ
​ ​

⇒ Δϕ = 0
2π
⇒ Δϕ = (μ − 1)t = 0
​ ​

λ
⇒ μ = 1 (and function will be cosine)

Q.2 Answer:
5714Å and 4444Å
Solution:

Absent wavelengths correspond to interference minima

2yd
∴ d sin θ = (2n − 1) λ2 ⇒ d Dy = (2n − 1) λ2 ⇒ λ = ​ ​ ​

D(2n−1)
​ = 40000
2(n−1)
​

= 13333Å, 80000Å, 5714Å, 4444Å, 3636Å ​

∴ (a)

Q.3 Answer:
1.33
Solution:
λD
Fringe width w = d
where symbols havo their usual meanings. Note that wavelchgth λ in air changes to
​

λ
μ
in a liquid of refractive index μ.
​

Now, we can write 10 × wliq = 7.5 × wair ⇒ 10 λliqd D = 7.5 λaird D ⇒ 10 λμair = 7.5λair .
​
​ ​

​ ​ ​ ​ ​ ​

10
This then gives μ = 7.5 ​ = 1.33.
Q.4 Answer:
6 × 10−5 cm
Solution:

D = 1 m., d = .90 mm = .9 × 10−3 m

The distance of the second dark ring from centre = 10−3 m

∴ Xn = (2n − 1) λ2 Dd
​ ​ ​

2×10−3 ×.9×10−3
For n = 2, Xn = 3λ D
2 d ⇒λ= ​ ​ ​
2Xnd
3D ​ = 3 ​

λ = 6 × 10−7 m.
​

λ = 6 × 10−5 cm.

Q.5 Answer:
xblue < xgreen
​ ​

Solution:

Distance of nth maximum, x = nλD

d
​

xblue 4360
= ∴ xblue < xgreen
​

xgreen 5460
​ ​ ​ ​

Q.6 Answer:
0.156 cm
Solution:

Suppose the mth bright fringe of 6500Å coincides with the nth bright fringe of 5200Å
mλ1 D nλ2 D
Xn = =
​ ​

​ ​ ​

d d
m × 6500 × D n × 5200 × D m 5200 4
⇒ = ⇒ = =
6500 5
​ ​ ​ ​ ​

d d n
​ ​

10
mλ1 D 4 × 6500 × 10 × 1.2
∴ Distance y is y = = = 0.156 cm.
​

2 × 10−3
​ ​

Q.7 Answer:
electron gun voltage be decreased.
Solution:
λD
Fringe width β = d ​

h h
According to de Broglie, wavelength λ = p ​ = 2mv
​

As V decreases, λ increases, β increases.

Q.8 Answer:
1.4
Solution:
 (μ−1)tD λD D β
n= d but β ​
= d
​
⇒ d
​
= λ
​

n = (μ − 1)tβ/λ
20β = (μ − 1)2.5 × 10−3 {β/5000 × 10−8 } ​

−8
μ − 1 = 20×5000×10
2.5×10 −3 = 0.4; μ = 1.4 ​

Q.9 Answer:
2 cm.
Solution:

Path difference = D2 + d2 − D = 1 cm. ​

2(1)
Also, [ D2 + d2 − D] = (2n − 1) λ2 ⇒ λ = ​ ​

2n−1
​

For n = 1, 2, 3.
λ = 2 cm, 23 cm, 25 cm, ​ ​

Q.10 Answer:
In Young's double slit experiment 7 maxima can be obtained on screen including central maxima if d = 7000Å and
λ = 2000Å.
Solution:

(A) Since mirror is moving perpendicular to y-axis to image will remain stationary and hence its
amplitude is zero.
(B) Δx = d sin θ = nλ where n = 0, ±1, ±2 ……
sin θ = nλd
= n × 2000 7000
= 2n
​

7
​ ​

sin θ = 0, ± 27 , ± 47 , ± 67
​

​ ​ ​

sin θ = nλ
d
= n × 20
​

70
sin θ = 0, ± 27 , ± 47 Total = 7 maxima ​ ​ ​

(C) 1F = f1 + f1 − f df ; ∞
​ ​
1
= 301 1
+ (−10) ​
d
− 30×−10 ; d = 20 cm. ​ ​ ​ ​ ​

1 2
​

1 2 ​ ​ ​

Q.11 Answer:
The number of maxima produced on the screen are 7
Solution:

Order of maxima are 3, 2, 1, 0, 1, 2, 3

Along x-axis : ( d2 ​
+ x) − ( d2 − x) = nλ ​

2π = nλ
λ 2λ 3λ
x1 = x2 = x3 =
​

2 2 3
​ ​ ​ ​ ​ ​

hence x2 ​ − x1 = x3 − x2 also ​ ​ ​
 d cos θ = nλ
3λ cos θ = 2λ
2
cos θ =
3
​ ​
​

5
sin θ =
​

3
​

r = R sin θ
r = R3 5 ​

Q.12 Answer:
5
Solution:

Order of the fringe can be counted on either side of the central maximum. For example no. 3 is first order
bright fringe.
ΔXC = λ, ΔXA = λ2
​ ​ ​

ΔXC − ΔXA = λ2 = 300 nm

​ ​ ​

Q.13 Answer:
4
Solution:

Order of the fringe can be counted on either side of the central maximum. For example no. 3 is first order
bright fringe.
ΔXC = λ, ΔXA = λ2
​ ​ ​

ΔXC − ΔXA = λ2 = 300 nm

​ ​ ​

Q.14 Answer:
300 nm
Solution:

Order of the fringe can be counted on either side of the central maximum. For example no. 3 is first order
bright fringe.
ΔXC = λ, ΔXA = λ2
​ ​ ​

ΔXC − ΔXA = λ2 = 300 nm

​ ​ ​
Q.15 Answer:
20^jm/s
​

Solution:

Velocity and acceleration of central maximum = velocity and acceleration of screen

[∵ it does not move to the left or right on the screen]
vscreen = 0 + gt = 20 m/s
​

Q.16 Answer:
10^jm/s2
​

Solution:

Velocity and acceleration of central maximum = velocity and acceleration of screen

[∵ it does not move to the left or right on the screen]
vscreen = 0 + gt = 20 m/s
​

v s = 20^jm/s ( in vector form ) and a s = 10^jm/s2

​ ​ ​ ​
​

D = D0 + 12 gt2 [ where D in distance of screen at a time t along y-axis ]

​ ​

Q.17 Answer:
7.5 × 10−3 m/s2^i
Solution:

D = D0 + 12 gt2 [ where D in distance of screen at a time t along y-axis ]

​ ​

Position vector of 2nd maximum is given by r x = ±2 λD ^i + D^j

d
​ ​ ​ ​

Position vector of 1st minimum is given by r n = ± 12 λD ^i ^j

d +D
​ ​ ​ ​

Relative position vector of 2nd maximum with respect to 1st minimum is given by r = ± 2 d ^i
3 λD
​ ​

2
Differentiating with respect to time, a = ddt2r = ± 32 d ^i = ±7.5 × 10−3 m/s2^i
λg
​ ​ ​

Q.18 Answer:
(A) − q, r, s(B) − p, q, r, s(C) − q, r, s(D)p, q, r, s
Solution:

(A) - qrs, (B) - pqrs, (C) - qrs, (D) pqrs

Initially at a distance x from central maxima on screen is I = I0 + 4I0 + 2 I0 4I0 cos 2πx
​

β
, where
​ ​ ​ ​ ​

Dλ
β= I
d max
​ ​ = 9I0 and Imin = I0
​ ​ ​

(A) At points where intensity is 19 th of maximum intensity, minima is formed.​

∴ Distance between such points is β, 2β, 3β, 4β, …… ..

3 2πx
(B) At points where intensity is 9 th of maximum intensity, cos β ​ ​
= − 12 or x =​
β
3
​

2πx β
(C) cos β = 0 or x = 4 ​ ​

∴ Distance between such points is 2 , β, β

β
​ + β2 , 2β, … .
​

(D) cos 2πx

β
= 12 or x = β6 ​ ​ ​

2β 2β
∴ Distance between such points is 3 , 3
β
​ ​
, β, β + β3 , β + ​

3 , 2β, …
​

Q.19 Answer:
—
Solution:

25 I 3I 25 3I
I1 = I × = , × =
100 4 4 100 16
​ ​ ​ ​ ​ ​

3I 75 9I
I2 = × = , I1 = ka21 and I2 = ka22
16 100 64
​ ​ ​ ​ ​ ​ ​ ​

a1 = 4 units and a2 = 3 units

​ ​

​ ​

2 2
a1 + a2 4+3
=( ) =( ) = 49
Imax ​ ​ ​

a1 − a2 4−3
​ ​ ​

Imin ​ ​ ​

Q.20 Answer:
—
Solution:

IDENTIFY: Consider the interference between light reflected from the top and bottom surfaces of the air
film between the lens and the glass plate. Introducing a liquid between the lens and the plate just changes
the wavelength from λ0 to λn0 , where n is the refractive index of the liquid.
​
​

SET UP: For maximum intensity, with a net half-cycle phase shift due to reflections, 2t = (m + 12 ) λ,
​

where λ is the wavelength in the film. t = R − R2 − r 2 . ​

(2m+1)λ (2m+1)λ
EXECUTE: 4
​
=R− R2 − r 2 ⇒ R2 − r 2 = R − ​ ​

4
​

2 2
(2m + 1)λ (2m + 1)λR (2m + 1)λR (2m + 1)λ
⇒R −r =R +[ 2
] − 2
⇒r= 2
−[ ]
4 2 2 4
​ ​ ​ ​ ​

​ ​

(2m + 1)λR
⇒r= , for R ≫ λ.
2
​ ​

λ = λ0 /n, where λ0 is the wavelength in air. Therefore, if r0 is the radius of the third bright ring when
​ ​ ​

air is between the lens and plate, the radius with water between the lens and plate is

r0 0.640 mm
r= = = 0.555 mm.
​

1.33
​ ​

n ​
​
Evaluate: The refractive index of the water is less than that of the glass plate, so the phase changes on
reflection are the same as when air is in the space.