JEE Advanced Previous Year Questions (2021 - 2024): Ray Optics and Optical Instruments | Physics for JEE Main & Advanced PDF Download

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Q1: Two equilateral-triangular prisms P 1 and P 2 are kept with their sides parallel to each other, in vacuum,
as shown in the gure. A light ray enters prism P 1 at an angle of incidence ? such that the outgoing ray
undergoes minimum deviation in prism P 2. If the respective refractive indices of P 1 and P 2 are v(3/2) and
v3, then ? = sin ?¹ [ v(3/2) sin ( p / ß) ], where the value of ß is ____. [JEE Advanced 2024 Paper 2]
Ans: 12
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Q1: Two equilateral-triangular prisms P 1 and P 2 are kept with their sides parallel to each other, in vacuum,
as shown in the gure. A light ray enters prism P 1 at an angle of incidence ? such that the outgoing ray
undergoes minimum deviation in prism P 2. If the respective refractive indices of P 1 and P 2 are v(3/2) and
v3, then ? = sin ?¹ [ v(3/2) sin ( p / ß) ], where the value of ß is ____. [JEE Advanced 2024 Paper 2]
Ans: 12
At surface BC
Q2: A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence ? 0. The
ray partially refracts into the sphere with angle of refraction ? 0 and then partly reects from the back
surface. The reected ray then emerges out of the sphere after a partial refraction. The total angle of
deviation of the emergent ray with respect to the incident ray is a.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
(a) P ? 5; Q ? 2; R ? 1; S ? 4
(b) P ? 5; Q ? 1; R ? 2; S ? 4
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Q1: Two equilateral-triangular prisms P 1 and P 2 are kept with their sides parallel to each other, in vacuum,
as shown in the gure. A light ray enters prism P 1 at an angle of incidence ? such that the outgoing ray
undergoes minimum deviation in prism P 2. If the respective refractive indices of P 1 and P 2 are v(3/2) and
v3, then ? = sin ?¹ [ v(3/2) sin ( p / ß) ], where the value of ß is ____. [JEE Advanced 2024 Paper 2]
Ans: 12
At surface BC
Q2: A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence ? 0. The
ray partially refracts into the sphere with angle of refraction ? 0 and then partly reects from the back
surface. The reected ray then emerges out of the sphere after a partial refraction. The total angle of
deviation of the emergent ray with respect to the incident ray is a.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
(a) P ? 5; Q ? 2; R ? 1; S ? 4
(b) P ? 5; Q ? 1; R ? 2; S ? 4
(c) P ? 3; Q ? 2; R ? 1; S ? 4
(d) P ? 3; Q ? 1; R ? 2; S ? 5    [JEE Advanced 2024 Paper 1]
Ans: (a)
a = ( ? - f ) + (180 - 2 f ) + ( ? - 2 f )
a = 180 + 2 ? - 4 f
(P) a = 180 + 2 ? - 4 f
180 = 180 + 2 ? - 4 f ? ? = 2 f ... (i)
sin ? = 2 sin f ... (ii)
From (i) & (ii)
sin ? = 2 sin 
?
2
 ? cos 
?
2
 = 1
?
2
 = 0
? ? = 0
(Q) ? = 2 f ... (i)
sin ? = v3 sin f ... (ii)
From (i) & (ii)
sin ? = v3 sin 
?
2
? cos 
?
2
 = 
v3
2
?
2
 = 30, 150
? = 60, 300 (Rejected)
? = 60&0
0 0 0 0 0
0 0
0 0
0 0 0 0
0 0
0
0 0
0
0
0 0
0 0
0
0
0
0
0
0
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Q1: Two equilateral-triangular prisms P 1 and P 2 are kept with their sides parallel to each other, in vacuum,
as shown in the gure. A light ray enters prism P 1 at an angle of incidence ? such that the outgoing ray
undergoes minimum deviation in prism P 2. If the respective refractive indices of P 1 and P 2 are v(3/2) and
v3, then ? = sin ?¹ [ v(3/2) sin ( p / ß) ], where the value of ß is ____. [JEE Advanced 2024 Paper 2]
Ans: 12
At surface BC
Q2: A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence ? 0. The
ray partially refracts into the sphere with angle of refraction ? 0 and then partly reects from the back
surface. The reected ray then emerges out of the sphere after a partial refraction. The total angle of
deviation of the emergent ray with respect to the incident ray is a.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
(a) P ? 5; Q ? 2; R ? 1; S ? 4
(b) P ? 5; Q ? 1; R ? 2; S ? 4
(c) P ? 3; Q ? 2; R ? 1; S ? 4
(d) P ? 3; Q ? 1; R ? 2; S ? 5    [JEE Advanced 2024 Paper 1]
Ans: (a)
a = ( ? - f ) + (180 - 2 f ) + ( ? - 2 f )
a = 180 + 2 ? - 4 f
(P) a = 180 + 2 ? - 4 f
180 = 180 + 2 ? - 4 f ? ? = 2 f ... (i)
sin ? = 2 sin f ... (ii)
From (i) & (ii)
sin ? = 2 sin 
?
2
 ? cos 
?
2
 = 1
?
2
 = 0
? ? = 0
(Q) ? = 2 f ... (i)
sin ? = v3 sin f ... (ii)
From (i) & (ii)
sin ? = v3 sin 
?
2
? cos 
?
2
 = 
v3
2
?
2
 = 30, 150
? = 60, 300 (Rejected)
? = 60&0
0 0 0 0 0
0 0
0 0
0 0 0 0
0 0
0
0 0
0
0
0 0
0 0
0
0
0
0
0
0
(R) ? = 2 f
sin ? = v3 sin f
sin 2 ? = v3 sin f
cos f = 
v3
2
f = 30, 150 (Rejected)
f = 30&0 ... (i)
(S) sin 45 = v2 cos f
cos f = 1/2
f = 60
a = 180 + 2 ? - 4 f
a = 180 + 90 - 120 ... (iv)
a = 180 - 30; a = 150°
Q3: A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the gure. The
radius of curvature of the convex surface (SPU) is 9 cm, while the planar surface (STU) acts as a mirror. This
beaker is lled with a liquid of refractive index n up to the level QPR. If the image of a point object O at a
height of h (OT in the gure) is formed onto itself, then, which of the following option(s) is(are) correct?   
(a) For n = 1.42, h = 50 cm.
(b) For n = 1.35, h = 36 cm.
(c) For n = 1.45, h = 65 cm.
(d) For n = 1.48, h = 85 cm.   [JEE Advanced 2024 Paper 1]
Ans: (a), (b)
Since STU is a plane mirror, we can take mirror image of the whole situation about it and nal image can be
assumed to be at a distance h below the base.
0 0
0 0
0 0
0
0
0
0
0
0
0 0
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Q1: Two equilateral-triangular prisms P 1 and P 2 are kept with their sides parallel to each other, in vacuum,
as shown in the gure. A light ray enters prism P 1 at an angle of incidence ? such that the outgoing ray
undergoes minimum deviation in prism P 2. If the respective refractive indices of P 1 and P 2 are v(3/2) and
v3, then ? = sin ?¹ [ v(3/2) sin ( p / ß) ], where the value of ß is ____. [JEE Advanced 2024 Paper 2]
Ans: 12
At surface BC
Q2: A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence ? 0. The
ray partially refracts into the sphere with angle of refraction ? 0 and then partly reects from the back
surface. The reected ray then emerges out of the sphere after a partial refraction. The total angle of
deviation of the emergent ray with respect to the incident ray is a.
Match the quantities mentioned in List-I with their values in List-II and choose the correct option.
(a) P ? 5; Q ? 2; R ? 1; S ? 4
(b) P ? 5; Q ? 1; R ? 2; S ? 4
(c) P ? 3; Q ? 2; R ? 1; S ? 4
(d) P ? 3; Q ? 1; R ? 2; S ? 5    [JEE Advanced 2024 Paper 1]
Ans: (a)
a = ( ? - f ) + (180 - 2 f ) + ( ? - 2 f )
a = 180 + 2 ? - 4 f
(P) a = 180 + 2 ? - 4 f
180 = 180 + 2 ? - 4 f ? ? = 2 f ... (i)
sin ? = 2 sin f ... (ii)
From (i) & (ii)
sin ? = 2 sin 
?
2
 ? cos 
?
2
 = 1
?
2
 = 0
? ? = 0
(Q) ? = 2 f ... (i)
sin ? = v3 sin f ... (ii)
From (i) & (ii)
sin ? = v3 sin 
?
2
? cos 
?
2
 = 
v3
2
?
2
 = 30, 150
? = 60, 300 (Rejected)
? = 60&0
0 0 0 0 0
0 0
0 0
0 0 0 0
0 0
0
0 0
0
0
0 0
0 0
0
0
0
0
0
0
(R) ? = 2 f
sin ? = v3 sin f
sin 2 ? = v3 sin f
cos f = 
v3
2
f = 30, 150 (Rejected)
f = 30&0 ... (i)
(S) sin 45 = v2 cos f
cos f = 1/2
f = 60
a = 180 + 2 ? - 4 f
a = 180 + 90 - 120 ... (iv)
a = 180 - 30; a = 150°
Q3: A glass beaker has a solid, plano-convex base of refractive index 1.60, as shown in the gure. The
radius of curvature of the convex surface (SPU) is 9 cm, while the planar surface (STU) acts as a mirror. This
beaker is lled with a liquid of refractive index n up to the level QPR. If the image of a point object O at a
height of h (OT in the gure) is formed onto itself, then, which of the following option(s) is(are) correct?   
(a) For n = 1.42, h = 50 cm.
(b) For n = 1.35, h = 36 cm.
(c) For n = 1.45, h = 65 cm.
(d) For n = 1.48, h = 85 cm.   [JEE Advanced 2024 Paper 1]
Ans: (a), (b)
Since STU is a plane mirror, we can take mirror image of the whole situation about it and nal image can be
assumed to be at a distance h below the base.
0 0
0 0
0 0
0
0
0
0
0
0
0 0
Since object and image are at the same distance from the equivalent lens, hence:
h = 2 F
1
F
 = 
(1.6 - 1)
1
 × 
2
9
 + 
(n - 1)
1
 × 
-2
9
1
h
 = 
1.2
9
 + 
2(1 - n)
9
1
h
 = 
3.2 - 2n
9
h = 
9
1.6 - n
 cm
(A) for n = 1.42,   h = 50 cm
(B) for n = 1.35,   h = 36 cm
(C) for n = 1.45,   h = 60 cm
(D) for n = 1.48,   h = 75 cm
qq
eq