A light ray is incident on the surface of a sphere of refractive index 𝑛 at an angle of incidence 𝜃₀. The ray partially refracts into the sphere with angle of refraction 𝜙₀ 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. 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 𝛼 = 180°, then all the possible values of 𝜃0 will be	(1) 30° and 0°
(Q) If 𝑛 = √3 and 𝛼 = 180°, then all the possible values of 𝜃0 will be	(2) 60° and 0°
(R) If 𝑛 = √3 and 𝛼 = 180°, then all the possible values of 𝜙0 will be	(3) 45° and 0°
(S) If 𝑛 = √2 and 𝜃0 = 45°, then all the possible values of 𝛼 will be	(4) 150°
	(5) 0°

The correct option is P → 5; Q → 2; R → 1; S → 4.

1. Formula for Angle of Deviation ($\alpha$):
When a light ray enters a sphere of refractive index $n$ at an angle of incidence ${\theta }_0$, it refracts into the sphere with an angle of refraction ${\phi }_0$.
By Snell's Law:
$\sin \left({\theta }_0\right)=n \sin \left({\phi }_0\right)$
The ray is deviated at three stages during its path:
- At the first refraction, the deviation is:
${\delta }_1={\theta }_0-{\phi }_0$ (clockwise)
- At the internal reflection on the back surface, the angle of incidence is ${\phi }_0$, so the deviation is:
${\delta }_2=180°-2{\phi }_0$ (clockwise)
- At the second refraction (emerging from the sphere), the angle of incidence is ${\phi }_0$ and the angle of emergence is ${\theta }_0$, so the deviation is:
${\delta }_3={\theta }_0-{\phi }_0$ (clockwise)
Therefore, the total angle of deviation $\alpha$ is given by:
$\alpha ={\delta }_1+{\delta }_2+{\delta }_3=({\theta }_0-{\phi }_0)+(180°-2{\phi }_0)+({\theta }_0-{\phi }_0)$
$\alpha =180°+2{\theta }_0-4{\phi }_0$

2. Analyzing List-I with the Deviation Formula:

Case (P): If $n=2$ and $\alpha =180°$:
Substituting $\alpha =180°$ into the deviation formula:
$180°=180°+2{\theta }_0-4{\phi }_0$
$\Rightarrow {\theta }_0=2{\phi }_0$
Using Snell's Law:
$\sin \left({\theta }_0\right)=n \sin \left({\phi }_0\right)$
$\sin \left(2{\phi }_0\right)=2 \sin \left({\phi }_0\right)$
$2 \sin \left({\phi }_0\right)\cos \left({\phi }_0\right)=2 \sin \left({\phi }_0\right)$
This equation gives two possibilities:
1) $\sin \left({\phi }_0\right)=0\Rightarrow {\phi }_0=0°\Rightarrow {\theta }_0=0°$
2) $\cos \left({\phi }_0\right)=1\Rightarrow {\phi }_0=0°\Rightarrow {\theta }_0=0°$
Thus, the only possible value of ${\theta }_0$ is $0°$.
Hence, P → 5.

Case (Q): If $n=\sqrt{3}$ and $\alpha =180°$:
As derived above, $\alpha =180°\Rightarrow {\theta }_0=2{\phi }_0$.
Using Snell's Law:
$\sin \left(2{\phi }_0\right)=\sqrt{3} \sin \left({\phi }_0\right)$
$2 \sin \left({\phi }_0\right)\cos \left({\phi }_0\right)=\sqrt{3} \sin \left({\phi }_0\right)$
This yields:
1) $\sin \left({\phi }_0\right)=0\Rightarrow {\phi }_0=0°\Rightarrow {\theta }_0=0°$
2) $\cos \left({\phi }_0\right)=\frac{\sqrt{3}}{2}\Rightarrow {\phi }_0=30°\Rightarrow {\theta }_0=2(30°)=60°$
Thus, the possible values of ${\theta }_0$ are $60°$ and $0°$.
Hence, Q → 2.

Case (R): If $n=\sqrt{3}$ and $\alpha =180°$:
From the derivation in Case (Q), the possible values of ${\phi }_0$ are $30°$ and $0°$.
Hence, R → 1.

Case (S): If $n=\sqrt{2}$ and ${\theta }_0=45°$:
Using Snell's Law:
$\sin (45°)=\sqrt{2} \sin \left({\phi }_0\right)$
$\frac{1}{\sqrt{2}}=\sqrt{2} \sin \left({\phi }_0\right)$
$\Rightarrow \sin \left({\phi }_0\right)=\frac{1}{2}\Rightarrow {\phi }_0=30°$
Now, substitute ${\theta }_0=45°$ and ${\phi }_0=30°$ into the deviation formula:
$\alpha =180°+2(45°)-4(30°)$
$\alpha =180°+90°-120°=150°$
Thus, the possible value of $\alpha$ is $150°$.
Hence, S → 4.