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Physics · Geometrical Optics

JEE Advanced 2024 — Paper 1 — Question 32

A light ray is incident on the surface of a sphere of refractive index n at an angle of incidence θ0\theta_{0}. The ray partially refracts into the sphere with angle of refraction ϕ0\phi_{0} and then partly reflects from the back surface. The reflected 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 α\alpha. Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List-IList-II
(P) If n=2\mathrm{n}=2 and α=180∘\alpha=180^{\circ}, then all the possible values of θ0\theta_{0} will be(1) 30∘30^{\circ} and 000^{0}
(Q) If n=3\mathrm{n}=\sqrt{3} and α=180∘\alpha=180^{\circ}, then all the possible values of θ0\theta_{0} will be(2) 60∘60^{\circ} and 0∘0^{\circ}
(R) If n=3\mathrm{n}=\sqrt{3} and α=180∘\alpha=180^{\circ}, then all the possible values of ϕ0\phi_{0} will be(3) 45∘45^{\circ} and 0∘0^{\circ}
(S) If n=2\mathrm{n}=\sqrt{2} and θ0=45∘\theta_{0}=45^{\circ}, then all the possible values of α\alpha will be(4) 150∘150^{\circ}
(5) 0∘0^{\circ}
  1. Option A:

    P→5;Q→2;R→1;S→4\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1 ; \mathrm{S} \rightarrow 4

    Correct
  2. Option B:

    P→5;Q→1;R→2;S→4\mathrm{P} \rightarrow 5 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 4

  3. Option C:

    P→3;Q→2;R→1\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 2 ; \mathrm{R} \rightarrow 1; S→4\mathrm{S} \rightarrow 4

  4. Option D:

    P→3;Q→1;R→2;S→5\mathrm{P} \rightarrow 3 ; \mathrm{Q} \rightarrow 1 ; \mathrm{R} \rightarrow 2 ; \mathrm{S} \rightarrow 5

Answer: A

Step-by-step solution

(P) sin⁡θ0=2sin⁡ϕ0\sin \theta_{0}=2 \sin \phi_{0}

When, α=180\alpha=180

2θ0−4ϕ0=0\begin{gathered} 2 \theta_{0}-4 \phi_{0}=0 \end{gathered}

θ0=2ϕ0\theta_{0}=2 \phi_{0}

sin⁡2ϕ0=2sin⁡ϕ0\sin 2 \phi_{0}=2 \sin \phi_{0}

2sin⁡ϕ0cos⁡ϕ0=2sin⁡ϕ02 \sin \phi_{0} \cos \phi_{0}=2 \sin \phi_{0}

cos⁡ϕ0=1\cos \phi_{0}=1

ϕ0=0\phi_{0}=0,

hence θ0=0\theta_{0}=0 (Q) sin⁡θ0=3sin⁡ϕ0\sin \theta_{0}=\sqrt{3} \sin \phi_{0}

cos⁡ϕ0=32\cos \phi_{0}=\frac{\sqrt{3}}{2}

ϕ0=30∘\phi_{0}=30^{\circ} θ0=60∘\theta_{0}=60^{\circ}

Hence, possible value of θ0\theta_{0} is zero and 60∘60^{\circ} (R) ϕ0=0\phi_{0}=0 and 30∘30^{\circ} (S)

sin⁡θ0=2sin⁡ϕ0\sin \theta_{0}=\sqrt{2} \sin \phi_{0}

sin⁡ϕ0=12,ϕ0=30∘\sin \phi_{0}=\frac{1}{2}, \phi_{0}=30^{\circ} (given θ0=45∘\theta_{0}=45^{\circ} )

Hence, α=180+2θ0−4ϕ0=180∘+90∘−120∘=150∘\alpha=180+2 \theta_{0}-4 \phi_{0}=180^{\circ}+90^{\circ}-120^{\circ}=150^{\circ}

Solution figure

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Physics
Chapter
Geometrical Optics
Topic
Refraction at Curved Surface and Glass Sphere

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