Ray Optics and Optical

Instruments

Single Correct Type Questions A spherical mirror is obtained as shown in the figure from
a hollow glass sphere. if an object is positioned infront of
1. Two plane mirrors M1 and M2 are at right angle to each the mirror, what will be the nature and magnification of
shown. A point source ‘P’ is placed at ‘a’ and ‘2a’ meter the image of the object? (Figure drawn as schematic and
away from M1 and M2 respectively. The shortest distance not to scale) [2 Sep, 2020 (Shift-I)]
between the images thus formed is : (Take 5 = 2.3 ) (a) Inverted, real and magnified
 [31 Aug, 2021 (Shift-I)] (b) Erect, virtual and unmagnified
(c) Inverted, real and unmagnified
a (d) Erect, virtual and magnified
P
4. An object is gradually moving away from the focal
point of a concave mirror along the axis of the mirror.
M1 2a The graphical representation of the magnitude of linear
magnification (m) versus distance of the object from
the mirror (x) is correctly given by (Graphs are drawn
M2 schematically and are not to scale)
[8 Jan, 2020 (Shift-II)]
(a) 2 10 a (b) 2.3 a
m m
(c) 4.6 a (d) 3 a
2. A point source of light, S is placed at a distance L in front
of the centre of plane mirror of width d which is hanging 1 1
vertically on a wall. A man walks in front of the mirror (a) x (b) x
along a line parallel to the mirror, at a distance 2 L as f 2f f 2f
shown below. The distance over which the man can see
the image of the light source in the mirror:
[12 Jan, 2019 (Shift-I)]
m
m
1 1
d S
(c) x (d) x
f 2f f 2f
L
2L
(a) d (b) 2d
(c) 3d (d) d 5. A vessel of depth ‘d’ is half filled with oil of refractive
3. index n1 and the other half is filled with water of refractive
index n2. The apparent depth of this vessel when viewed
Object
from above will be [13 Apr 2023 (Shift-I)]
dn1n2 d (n1 + n2 )
(a) (b)
20 16 12 8 4 (n1 + n2 ) 2n1n2
(cm) dn1n2 2 d (n1 + n2 )
(c) (d)
2(n1 + n2 ) n1n2
 6. Time taken by light to travel in two different materials A 11. A ray of light entering from air into a denser medium of
and B refractive indices µA and µB of same thickness is 4
t1 and t2 respectively. If t2 – t1 = 5 × 10–10 s and the ratio refractive index , as shown in figure. The light ray
3
of µA and µB is 1 : 2. Then, the thickness of material, in
suffers total internal reflection at the adjacent surface as
meter is: (Given υA and υB are velocities of light in A and
shown. The maximum value of angle θ should be equal
B materials respectively.) [25 July, 2022 (Shift-I)]
to : [25 July, 2021 (Shift-II)]
(a) 5 × 10 υA m
–10
(b) 5 × 10–10 m
7
(c) 1.5 × 10–10 m (d) 5 × 10–10 υB m (a) sin −1 5 (b) sin −1
7. Consider a light ray travelling in air is incident into a 4 3
medium of refractive index 2n . The incident angle is 5
(c) sin −1 7 (d) sin −1
twice that of refracting angle. Then, the angle of incidence 4 3
will be [27 June, 2022 (Shift-I)]
 n θ
(a) sin −1 ( n) −1
(b) cos  
 2 θ′
 n θ″ 4
(c) sin −1 ( 2n ) (d) 2 cos −1   µ=
3
 2
8. A ray of laser of a wavelength 630 nm is incident at an 12. A ray of light passes from a denser medium to a rarer medium
angle of 30º at the diamond- air interface. It is going from at an angle of incidence i. The reflected and refracted rays
diamond to air. The refractive index diamond is 2.42 and make an angle of 90º with each other. The angle of reflection
that of air is 1. Choose the correct option. and refraction are respectively r and r′. The critical angle is
 [25 July, 2021 (Shift-I)] given by: [22 July, 2021 (Shift-II)]
(a) Angle of refraction is 53.4º
(b) Angle of refraction is 30º
(c) Angle of refraction is 24.41º i r
(d) Refraction is not possible
r'
9. A vessel of depth 2 h is half filled with a liquid of refractive
index 2 2 and the upper half with another liquid of
refractive index 2 . The liquids are immiscible. The
apparent depth of the inner surface of the bottom of vessel (a) sin–1 (cot r) (b) tan–1 (sin i)
will be [9 Jan, 2020 (Shift-I)] (c) sin–1 (tan r ') (d) sin–1 (tan r)

h 13. In figure, the optical fiber is l = 2 m long and has a diameter

h
(a)
2
(b) 2 ( 2 +1) of d = 20 µm. If a ray of light is incident on one end of the
fiber at angle θ1 = 40°, the number of reflection it makes
before emerging from the other end is close to:
3 h  [8 April, 2019 (Shift-I)]
(c) h 2 (d)
4 3 2 (refractive index of fibre is 1.31 and sin 40° = 0.64)

10. The speed of light in media ‘A’ and ‘B’ are 2.0 ×1010 cm/s
and 1.5 ×1010 cm/s respectively. A ray of light enters from
the medium B to A an incident angle ‘θ’. If the ray suffers
total internal reflection, then [29 June, 2022 (Shift-II)] d
2
−1  3  −1  2 
(a) θ =sin   (b) θ > sin   40º
4 3
3 3 (a) 55000 (b) 57000
(c) θ < sin −1   (d) θ > sin −1   (c) 66000 (d) 45000
4 4

2 JEE PYQs Physics

 14. A transparent cube of side d, made of a material of refractive 17. One plano-convex and one plano-concave lens of same
index µ2, is immersed in a liquid of refractive index radius of curvature ‘R’ but of different materials are joined
µ1(µ1 < µ2). A ray is incident on the face AB at an angle θ side by side as shown in the figure. If the refractive index
of the material of 1 is µ1 and that of 2 is µ2, then the focal
(shown in the figure). Total internal reflection takes place length of the combination is: [10 April, 2019 (Shift-I)]
at point E on the face BC. [12 April, 2019 (Shift-II)]
E 1 2
B C

 2
1
1 2
A D
The θ must satisfy: R 2R
(a) (b)
2 − (µ1 − µ 2 ) µ1 − µ 2
µ1 2
(b) θ < sin −1 µ 2 − 1
−1
(a) θ < sin R
µ2 µ12
R
(c) (d)
2(µ1 − µ 2 ) µ1 − µ 2
µ1 µ 22
(c) θ > sin −1 (d) θ > sin −1 −1 18. Curved surfaces of a plano-convex lens of refractive
µ2 µ12 index m1 and a planoconcave lens of refractive index m2
have equal radius of curvature as shown in figure. Find
15. Region I and II are separated by a spherical surface of radius the ratio of radius of curvature to the focal length of the
25 cm. An object is kept in region I at a distance of 40 cm combined lenses. [27 Aug, 2021 (Shift-II)]
from the surface. The distance of the image from the surface
is: [20 July, 2021 (Shift-I)]
I II
I=1.25 II=1.4 1 2

25cm
O C

1
(a) (b) m2 – m1
µ 2 − µ1
(a) 9.52 cm (b) 18.23 cm
(c) 37.58 cm (d) 55.44 cm 1
(c) (d) m1 – m2
16. A concave mirror has radius of curvature of 40 cm. It is µ1 − µ 2
at the bottom of a glass that has water filled up to 5 cm 19. Find the distance of the image from object O, formed by
(see figure). If a small particle is floating on the surface the combination of lenses in the figure:
of water, its image as seen, from directly above the glass, [27 Aug, 2021 (Shift-I)]
is at a distance d from the surface of water. The value of
(a) 75 cm (b) 20 cm
d is close to: (Refractive index of water = 1.33)
(c) 10 cm (d) infinity
 [12 April, 2019 (Shift-I)]

particle

5cm
O

(a) 8.8 cm (b) 11.7 cm

(c) 6.7 mm (d) 13.4 cm 30cm 3cm 10cm

3 JEE PYQs Physics

 20. A microscope is focused on an object at the bottom of a 24. A point source of light S, placed at a distance 60 cm infront
5 of the centre of a plane mirror of width 50 cm, hangs
bucket. If liquid with refractive index is poured inside
3 vertically on a wall. A man walks infront of the mirror
the bucket, then microscope have to be raised by 30 cm
along a line parallel to the mirror at a distance 1.2 m from
to focus the object again. The height of the liquid in the
it (see in the figure). The distance between the extreme
bucket is: [31 Jan, 2023 (Shift-II)]
points where he can see the image of the light source in
(a) 75 cm
the mirror is _____ cm. [26 Feb, 2021 (Shift-II)]
(b) 50 cm
(c) 18 cm
S
(d) 12 cm 50 cm
60 cm
21. In normal adjustment, for a refracting telescope, the
distance between objective and eye piece is 30 cm. 1.2 m
The focal length of the objective, when the angular
magnification of the telescope is 2 , will be:
[28 July, 2022 (Shift-I)] 25. A small bulb is placed at the bottom of a tank containing
(a) 20 cm (b) 30 cm water to a depth of 7 m. The refractive index of water
(c) 10 cm (d) 15 cm 4
is . The area of the surface of water through which light
Integer Type Questions 3
22. Two vertical parallel mirrors A and B are separated by 10 from the bulb can emerge out is xπ m2. The value of x is
cm. A point object O is placed at a distance of 2 cm from ________. [26 June, 2022 (Shift-II)]
mirror A. The distance of the second nearest image behind
mirror A from the mirror A is ____ cm 
26. As shown in the figure, a combination of a thin plano
[8 Apr, 2023 (Shift-I)]
A B
concave lens and a thin plano convex lens is used to
image an object placed at infinity. The radius of curvature
of both the lenses is 30 cm and refractive index of the
2cm material for both the lenses is 1.75. Both the lenses are
O placed at distance of 40 cm from each other. Due to the
combination, the image of the object is formed at distance
x = _________ cm, from concave lens. [24 Jan,
10cm 2023 (Shift-I)]
23. As shown in the figure, a plane mirror is fixed at a height 40 Cm,
of 50 cm from the bottom of tank containing water
 4
 µ =  . The height of water in the tank is 8 cm. A small
 3
bulb is placed at the bottom of the water tank. The distance
of image of the bulb formed by mirror from the bottom
of the tank is _________ cm. [11 Apr, 2023 (Shift-II)]
O r
27. Two transparent media having refractive indices 1.0 and
1.5 are separated by a spherical refracting surface of radius
of curvature 30 cm. The centre of curvature of surface is
50 cm towards denser medium and a point object is placed on
the principle axis in rarer medium at a distance of 15 cm
from the pole of the surface. The distance of image from
8 cm the pole of the surface is _______ cm.
[8 Apr, 2023 (Shift-II)]

4 JEE PYQs Physics

28. An object is placed on the principal axis of convex lens 29. A point object, ‘O’ is placed in front of two thin
of focal length 10 cm as shown. A plane mirror is placed symmetrical coaxial convex lenses L1 and L2 with focal
on the other side of the lens at a distance of 20 cm. The length 24 cm and 9 cm respectively. The distance between
image produced by the plane mirror is 5 cm inside the two lenses is 10 cm and the object is placed 6 cm away
mirror. The distance of the object from the lens is ______ from lens L1 as shown in the figure. The distance between
cm.  [25 Jan, 2023 (Shift-II)] the object and the image formed by the system of two
lenses is _____cm. [10 Apr, 2023 (Shift-II)]

L1 L2

6m
10m

5 JEE PYQs Physics

 ANSWER KEY
1. (c) 2. (c) 3. (c) 4. (d) 5. (b) 6. (a) 7. (d) 8. (d) 9. (c) 10. (d)
11. (b) 12. (d) 13. (a, b) 14. (b) 15. (c) 16. (a) 17. (d) 18. (d) 19. (a) 20. (a)
21. (a) 22. [18] 23. [98] 24. [150] 25. [9] 26. [120] 27. [30] 28. [30] 29. [34]

EXPLANATIONS
1. (c) M1 Therefore, the image formed is real, inverted and
diminished or unmagnified.
I1 2a a P
Hence, option (c) is correct answer
2a 4. (d) At focus, magnification is ∞.
5a
real depth
M2 5. (b) From the formula n =
2a apparent depth
5a
d1 d 2

d app
= +
I2 n1 n2
The shortest distance between the image thus formed, d n + n 
I1 I 2 = 2 5a d app =  1 2 
2  n1n2 

= 2 × 2.3 a = 4.6 a
c 1
2. (c) So, the final distance is 6. (a) Since, µ = , therefore µ ∝
V v
d d
h =d + + + d = 3d .
2 2 mA vB 1
= =
h = d + d + d = 3d. mB v A 2

d Let the thickness is d

Given: t1 – t2 = 5 × 10–10
d/2
S
d L L h d d
− =5 × 10−10
d/2 vB v A

d 5 × 10−10 × v A vB
d=
v A − vB

3. (c)
5 × 10−10 × 2v A .vB
As vA = 2vB ⇒ d = = 5×10–10 ×2VB
ho 2v A − vB
O C hi F P or d = 5×10–10 ×vA
Image 7. (d) From Snell’s law
i

hi < ho sin i ×=
1 sin × 2n
2

6 JEE PYQs Physics

 2
i i i 3
2sin ⋅ cos = sin × 2n 1 −  sin θ  > sin 2 θc (By equation (i))
2 2 2 4 
i n  n 9  1 3
cos = −1 1− sin 2 θ > sin 2 θc ∴sin θc= = 
⇒ i = 2 cos   16  µ 4
2 2  2
9 9 9 9
8. (d) From snell's law 1− sin 2 θ > ⇒ 1 − > sin 2 θ
16 16 16 16

m1 sin30º = m2 sinq 7 9 7
> sin 2 θ ⇒ sin θ <
1 16 16 3

⇒ 2.42 ×= (1) sin θ
2 12. (d) r + r' = 90º
sinq = 1.21 > 1

⇒ r' = 90º – r
 Law of refraction i = r

air q
\ r' = 90º – i

diamond
30º i r
 90º
Refraction is not possible
= 1 r′
9. (c) Apparent depth of the inner surface of the bottom of
h h
the vessel,=
d +
2 2 2
h 3 3 2h
⇒d= × =
2 2 4 Snell's law:

10. (d) We have given, m sin i = 1 sin r' ⇒ m sin i = sin(90º – i)
vA = 2 × 10 cm/s, vB = 1.5 × 10 cm/s
10 10 1
⇒ tan i = = tan r
µ
We know that, for total internal reflection,
Now critical angle:
sin θC sin 90 v 1.5 × 1010 3
m sin C = 1 sin 90º = 1
= ⇒ sin θC = B = =
v vA v A 2.0 × 1010 4
B 1
⇒ sin C == tan r
3 µ
⇒ θC =sin −1  
4 C = sin–1 (tan r)
13.(a, b)
According to question, we can write l

3
θ > θc =sin −1  
4 d

11. (b) Applying snell’s law on first surface 2

4 3 40º x
1 × sinθ = × sin θ′ ⇒ sin θ′ = sin θ ... (i)
3 4
Exact solution
For TIR on second surface by Snells’ law 1.sin 40° = (1.31) sin q2
sinθ" > sinθc (θ′ + θ″ = 90) 0.64 64
sin q2 = = ≈ 0.49
sin(90 – θ′) > sinθc 1.31 131
cosθ′ > sinθc 64
Now tan q2 =
cos2θ' > sin2θc ⇒ 1 – sin2θ' > sin2θc (131) 2 − (64) 2

7 JEE PYQs Physics

 64 64 d µ 2 µ1 µ 2 − µ1
≈ = 15. (c) − =
13065 114.3 x v u R
2
Now no. of reflections = 1.4 1.25 1.4 − 1.25
x
⇒ − =
v −40 −25
2 × 64 64 × 105
= =
114.3 × 20 × 10 −6
114.3 1.4 0.15 1.25 7.45

⇒ −= + =
≈ 55991 ≈ 55000 v 25 40 200
Approximate solution 200 × 1.4

⇒ v =− =−37.58cm
By Snells’ law 1. sin 40° = (1.31) sin q2 7.45
II II=1.4
I I=1.25
0.64 64
sin q2 = = ≈ 0.49 DIR
1.31 131
If assume ⇒ q2 ≈ 30° 25cm

d I O C P
tan 30° = ⇒ x = 3d
x
Now number of reflections
l 2 105 16. (a) Light incident from particle P will be reflected at mirror
= = =
3d 3 × 20 × 10−6 3
P
≈ 57735 ≈ 57000
Note : If we approximate the angle q2 as 30° initially
5cm
then answer will be closer to 57000. But if we solve water
thoroughly, answer will be close to 55000. So both (=4/3)
the answers must be awarded. Detailed solution is as
following. R= 40 cm
R
2 u = –5cm, f =− = −20 cm
1 C 2
14. (b) 90–C
1 1 1
 + =
v u f

µ1 20
sin c = v1 = + cm
µ2 3
This image will act as object for light getting refracted
m1 sin q = m2 sin (90º – C) = m2 cos C at water surface
20 35
µ12 So, object distance d = 5 + =cm
µ2 1 − 3 3
µ 22
sin θ = below water surface.
µ1
After refraction, final image is at

µ 22 − µ12 µ 
θ =sin −1 d′ = d  2 
µ12  µ1 

For TIR  35   1 
=   
 3  4 / 3 
µ 22 35
θ < sin −1 −1 = = 8.75 cm ≈ 8.8 cm
µ12 4

8 JEE PYQs Physics

 1  µ − 1  1 1  µ1 − 1 22. [18] I1 is the first nearest image to A.
17. (d) For 1st lens
=  1   −= 
f1  1   ∞ − R  R
d1(A) = 2 cm

1  µ2 − 1   1  µ −1
for 2nd lens =  − 0 =
− 2 I2 is the second nearest image to A.

f 2  1   − R  R
d2(A) = 18 cm (Desired)

1 1 1
= +
f eq f1 f 2
A B
1 µ − 1 µ 2 − 1 µ1 − µ 2
= 1 − =
f eq R R R
2cm 2cm
R O
⇒ feq = I3 I1 I2
µ1 − µ 2 18cm 8cm
1 µ1 − 1
18. (d) =
f1 R 10cm
1  µ − 1 d1(A) d2(A)
= −  2 
f2  R 

1 1 1 µ1 − µ 2 R I2
= + = ⇒ = µ1 − µ 2 23. [98]
f eq f1 f 2 R f eq
48 cm
1 1 1
19. (a) + =
v1 30 10
⇒ v1 = 15cm
1 1 1
⇒ − = − 50 cm
v2 10 10

⇒ v2 = ∞ ⇒ v3 = 30cm ⇒ OV3 = 75cm 6 cm

8 cm I1
 1
d h 1 − 
20. (a) Using formula,= O
 µ
real depth d
Apparent depth of O= = = 6
refractire index µ
 3  5

⇒=30 h 1 −    Given,
= d 30cm,
= µ
 5 3  Distance between O and I2 = 48 + 50 = 98 cm

2 x

⇒ 30= h × 24. [150] I
5 q S

∴ h = 75 cm
60 cm 120 cm
21. (a) |f0| + |fe| = 30 cm (Given)

f0 25 x
= 2 (given) tan =
θ =
As, 60 180
fe
x = 75 cm
⇒ |f0| = 2 |fe|
⇒ Maximum Distance = 2x = 2 × 75 = 150 cm
f0 2
∴ f0 + =30 ⇒ f 0 = × 30 =20 cm
2 3

9 JEE PYQs Physics

 25. [9] Given, h = 7 m, µ =4 / 3 µ 2 µ1 µ 2 − µ1
− =
v u R
r 1.5 1 1.5 − 1
− =
v (−15) 30
C
1.5 1 1
h + = ⇒ v = – 30 cm
v 15 60
C

28. [30]

r
tan C =
h
r = h tan C
1 3
sin C= = Image distance,
µ 4
v = 20 – 5
3
tan C =   = 15 cm
7
3 1 1 1
r = 7× =3 − =
7 v u f
Area of surface = πr2 = 9π m2
1 1 1
1 1  − =
26. [120] For the plano concave lens, = (1.75 − 1)  −  15 −u 10
f1  30 
−40 cm
⇒ f1 =
1 1 1

⇒ = −
For the plano convex lens, u 10 15

1 1 
⇒ u = 30cm.
= (1.75 − 1)   ⇒ f 2= 40 cm

f2  30  1 1 1
29. [34] For L1 : − =
Image from lens L1 will be virtual and to the left of v −6 24
L1 at focal length 40 cm. So the object for lens L2 will 1 1 1
be 80 cm from L2 which is equal to 2f. Final image = − ⇒ v =−8cm

v 24 6
is formed at 80 cm from lens L2 on the right side of
the lens. 1 1 1
For L2 : − =
v (−8 − 10) 9

⇒ x = 120
1 1 1
27. [30] R=30 cm = − ⇒ v = 18
v 9 18

m1=1 m2= 1.5 f = 24 cm f = 9 cm
O I2
Object
I1
6cm
15 cm 8cm 10cm 18cm

So OI2 = 6 + 10 + 18 = 34 cm

10 JEE PYQs Physics