Question 16

Added on: Sep 19, 2026

MATRIX MATCHHARD

A light ray is incident on the surface of a sphere of refractive index $n$ at an angle of incidence $heta_0$ . The ray partially refracts into the sphere with angle of refraction $\phi_0$ 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 $\alpha$ . Match the quantities mentioned in List-I with their values in List-II and choose the correct option.

List - I

If $n = 2$ and $\alpha = 180^\circ$ , then all the possible values of $heta_0$ will be

If $n = \sqrt{3}$ and $\alpha = 180^\circ$ , then all the possible values of $heta_0$ will be

If $n = \sqrt{3}$ and $\alpha = 180^\circ$ , then all the possible values of $\phi_0$ will be

If $n = \sqrt{2}$ and $heta_0 = 45^\circ$ , then all the possible values of $\alpha$ will be

List-II

$30^\circ$ and $0^\circ$

$60^\circ$ and $0^\circ$

$45^\circ$ and $0^\circ$

$150^\circ$

$0^\circ$

(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

Detailed Solution

The total deviation for one internal reflection in a sphere is given by $\alpha = |180^\circ + 2\theta_0 - 4\phi_0|$ .

Also, by Snell's Law, $\sin \theta_0 = n \sin \phi_0$ .

(P) For $n = 2$ and $\alpha = 180^\circ$ : $180^\circ = |180^\circ + 2\theta_0 - 4\phi_0| \Rightarrow 2\theta_0 = 4\phi_0 \Rightarrow \theta_0 = 2\phi_0$ .

Then $\sin(2\phi_0) = 2 \sin\phi_0 \Rightarrow 2\sin\phi_0 \cos\phi_0 = 2\sin\phi_0 \Rightarrow \sin\phi_0 = 0$ or $\cos\phi_0 = 1$ .

Both yield $\phi_0 = 0^\circ \Rightarrow \theta_0 = 0^\circ$ . Match 5.

(Q) For $n = \sqrt{3}$ and $\alpha = 180^\circ$ : $\theta_0 = 2\phi_0$ . Then $\sin(2\phi_0) = \sqrt{3} \sin\phi_0 \Rightarrow 2\sin\phi_0 \cos\phi_0 = \sqrt{3} \sin\phi_0 \Rightarrow \cos\phi_0 = \sqrt{3}/2$ or $\sin\phi_0 = 0$ . This gives $\phi_0 = 30^\circ, 0^\circ \Rightarrow \theta_0 = 60^\circ, 0^\circ$ . Match 2.

(R) From (Q), the possible values of $\phi_0$ are $30^\circ$ and $0^\circ$ . Match 1.

(S) For $n = \sqrt{2}$ and $\theta_0 = 45^\circ$ : $\sin 45^\circ = \sqrt{2} \sin \phi_0 \Rightarrow 1/\sqrt{2} = \sqrt{2} \sin \phi_0 \Rightarrow \sin \phi_0 = 1/2 \Rightarrow \phi_0 = 30^\circ$ .

Then $\alpha = |180^\circ + 2(45^\circ) - 4(30^\circ)| = |180^\circ + 90^\circ - 120^\circ| = 150^\circ$ . Match 4.

Correct Option: A

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