A light ray is incident on the surface of a sphere of refractive index $$n$$ at an angle of incidence $$\theta_0$$. 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		List-II
(P)	If $$n = 2$$ and $$\alpha = 180°$$, then all the possible values of $$\theta_0$$ will be	(1)	$$30°$$ and $$0°$$
(Q)	If $$n = \sqrt{3}$$ and $$\alpha = 180°$$, then all the possible values of $$\theta_0$$ will be	(2)	$$60°$$ and $$0°$$
(R)	If $$n = \sqrt{3}$$ and $$\alpha = 180°$$, then all the possible values of $$\phi_0$$ will be	(3)	$$45°$$ and $$0°$$
(S)	If $$n = \sqrt{2}$$ and $$\theta_0 = 45°$$, then all the possible values of $$\alpha$$ will be	(4)	$$150°$$
		(5)	$$0°$$

The ray suffers two refractions (at the first and the last surface) and one total internal reflection (at the back surface).
Let the angle of incidence be $$\theta_0$$ (in air) and the corresponding angle of refraction inside the sphere be $$\phi_0$$.
From Snell’s law (air to glass):

$$\sin \theta_0 = n \sin \phi_0 \qquad -(1)$$

For a spherical drop the total deviation produced by “refraction-reflection-refraction’’ is a standard result (used while studying rainbows):

$$\alpha \;=\; 180^\circ \;+\; 2\theta_0 \;-\; 4\phi_0 \qquad -(2)$$

Using $$(1)$$ and $$(2)$$ we answer each item.

Putting in $$(2):\; 180^\circ = 180^\circ + 2\theta_0 - 4\phi_0 \Longrightarrow \theta_0 = 2\phi_0.$$ Substituting in $$(1):\; \sin(2\phi_0) = 2\sin\phi_0.$$ Since $$\sin(2\phi_0)=2\sin\phi_0\cos\phi_0,$$ we get $$2\sin\phi_0\cos\phi_0 = 2\sin\phi_0.$$ Either $$\sin\phi_0 = 0$$ or $$\cos\phi_0 = 1.$$ Both give $$\phi_0 = 0^\circ \Longrightarrow \theta_0 = 0^\circ.$$ Hence all possible $$\theta_0$$ values = $$0^\circ$$ (List-II → 5).

Again $$\theta_0 = 2\phi_0.$$ Put in $$(1):\; \sin(2\phi_0) = \sqrt{3}\,\sin\phi_0.$$ $$2\sin\phi_0\cos\phi_0 = \sqrt{3}\,\sin\phi_0 \; \Longrightarrow\; 2\cos\phi_0 = \sqrt{3}.$$ So $$\cos\phi_0 = \dfrac{\sqrt{3}}{2} \Longrightarrow \phi_0 = 30^\circ.$$ Hence $$\theta_0 = 2\phi_0 = 60^\circ.$$ The trivial solution $$\phi_0 = 0^\circ,\; \theta_0 = 0^\circ$$ also satisfies both equations. Thus all possible $$\theta_0$$ values = $$60^\circ$$ and $$0^\circ$$ (List-II → 2).

The two $$\phi_0$$ values found above are $$30^\circ$$ and $$0^\circ$$ (List-II → 1).

From $$(1):\; \sin45^\circ = \sqrt{2}\,\sin\phi_0 \;\Longrightarrow\; \dfrac{1}{\sqrt{2}} = \sqrt{2}\,\sin\phi_0 \;\Longrightarrow\; \sin\phi_0 = \dfrac{1}{2} \;\Longrightarrow\; \phi_0 = 30^\circ.$$ Now use $$(2):\; \alpha = 180^\circ + 2(45^\circ) - 4(30^\circ) = 180^\circ + 90^\circ - 120^\circ = 150^\circ.$$ Thus all possible $$\alpha$$ values = $$150^\circ$$ (List-II → 4).

Collecting the matches:
P → 5,  Q → 2,  R → 1,  S → 4.

Hence the correct option is Option A: P → 5; Q → 2; R → 1; S → 4.