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

Alright, let's tackle this problem step by step. It's about light refraction and reflection inside a sphere with a given refractive index, and we need to match certain conditions to the correct angles. Here's how I'll approach it:

Understanding the Problem

We have a light ray incident on a spherical surface with refractive index $n$. The incident angle is ${\theta }_0$, and the angle of refraction is ${\varphi }_0$. The ray refracts into the sphere, reflects off the back surface, and then refracts out again. The total deviation angle between the incident and emergent ray is $\alpha$.

We're given different scenarios in List-I and possible angle values in List-II. Our goal is to match each scenario (P, Q, R, S) with the correct angles from List-II (1, 2, 3, 4, 5).

Key Concepts

1. Snell's Law: At the first refraction, $\sin {\theta }_0=n\sin {\varphi }_0$.

2. Internal Reflection: After refraction, the ray hits the back of the sphere. The angle of incidence here is ${\varphi }_0$ (since the radius is normal to the surface).

3. Total Deviation ($\alpha$): This is the angle between the initial incident ray and the final emergent ray. For the ray to emerge with $\alpha ={180}^{\circ }$, the ray must retrace its path, meaning it reflects back exactly opposite to the incident direction.

Analyzing $\alpha ={180}^{\circ }$

For $\alpha ={180}^{\circ }$, the emergent ray is opposite to the incident ray. This happens if:

• The ray reflects back along the incident path, which occurs if the refracted ray is along the diameter (i.e., ${\varphi }_0=0^{\circ }$).

• Then, by Snell's law, $\sin {\theta }_0=n\sin 0^{\circ }=0$, so ${\theta }_0=0^{\circ }$.

But wait, the problem mentions "all possible values," and ${\theta }_0=0^{\circ }$ is always a solution for $\alpha ={180}^{\circ }$. Are there others?

Actually, if the ray refracts, reflects once, and refracts out such that the net deviation is ${180}^{\circ }$, another way is if the ray undergoes total internal reflection, but that's more complex. However, the simplest case is ${\theta }_0=0^{\circ }$, which gives $\alpha ={180}^{\circ }$.

But looking at the options, some have ${\theta }_0=0^{\circ }$ and another angle. Maybe there are other angles where after refraction and reflection, the deviation is ${180}^{\circ }$.

Let me think differently: The total deviation $\alpha$ for one reflection is given by:

$$\alpha =2({\theta }_0-{\varphi }_0)+{180}^{\circ }-2{\varphi }_0={180}^{\circ }+2{\theta }_0-4{\varphi }_0$$

But setting $\alpha ={180}^{\circ }$:

$${180}^{\circ }={180}^{\circ }+2{\theta }_0-4{\varphi }_0$$$$2{\theta }_0=4{\varphi }_0$$$${\theta }_0=2{\varphi }_0$$

And from Snell's law:

$$\sin {\theta }_0=n\sin {\varphi }_0$$$$\sin (2{\varphi }_0)=n\sin {\varphi }_0$$$$2\sin {\varphi }_0\cos {\varphi }_0=n\sin {\varphi }_0$$

If $\sin {\varphi }_0\mathrm{\ne }0$, then:

$$2\cos {\varphi }_0=n$$$$\cos {\varphi }_0=\frac{n}{2}$$$${\varphi }_0={\cos }^{-1}(\frac{n}{2})$$

Then ${\theta }_0=2{\varphi }_0$.

If $\sin {\varphi }_0=0$, then ${\varphi }_0=0^{\circ }$, ${\theta }_0=0^{\circ }$.

So possible solutions are:

1. ${\theta }_0=0^{\circ }$, ${\varphi }_0=0^{\circ }$.

2. ${\theta }_0=2{\cos }^{-1}(\frac{n}{2})$, ${\varphi }_0={\cos }^{-1}(\frac{n}{2})$, if $\frac{n}{2}\le 1$.

Now let's apply this to each case.

Case P: $n=2$, $\alpha ={180}^{\circ }$

From above:

1. ${\theta }_0=0^{\circ }$.

2. $\cos {\varphi }_0=\frac{2}{2}=1$, so ${\varphi }_0=0^{\circ }$, ${\theta }_0=0^{\circ }$. No other solution.

Thus, only ${\theta }_0=0^{\circ }$.

Looking at List-II, this matches option (5) $0^{\circ }$.

Case Q: $n=\sqrt{3}$, $\alpha ={180}^{\circ }$

From above:

1. ${\theta }_0=0^{\circ }$.

2. $\cos {\varphi }_0=\frac{\sqrt{3}}{2}$, so ${\varphi }_0={30}^{\circ }$, ${\theta }_0={60}^{\circ }$.

Thus, possible ${\theta }_0$ are ${60}^{\circ }$ and $0^{\circ }$.

Looking at List-II, this matches option (2) ${60}^{\circ }$ and $0^{\circ }$.

Case R: $n=\sqrt{3}$, $\alpha ={180}^{\circ }$, find ${\varphi }_0$

From above:

1. ${\varphi }_0=0^{\circ }$.

2. ${\varphi }_0={30}^{\circ }$.

Thus, possible ${\varphi }_0$ are ${30}^{\circ }$ and $0^{\circ }$.

Looking at List-II, this matches option (1) ${30}^{\circ }$ and $0^{\circ }$.

Case S: $n=\sqrt{2}$, ${\theta }_0={45}^{\circ }$, find $\alpha$

First, find ${\varphi }_0$ using Snell's law:

$$\sin {45}^{\circ }=\sqrt{2}\sin {\varphi }_0$$$$\frac{1}{\sqrt{2}}=\sqrt{2}\sin {\varphi }_0$$$$\sin {\varphi }_0=\frac{1}{2}$$$${\varphi }_0={30}^{\circ }$$

Now, calculate deviation $\alpha$:
After first refraction, deviation is ${\theta }_0-{\varphi }_0={15}^{\circ }$.
At reflection, angle of incidence is ${\varphi }_0={30}^{\circ }$, deviation is ${180}^{\circ }-2\times {30}^{\circ }={120}^{\circ }$.
At second refraction, deviation is again ${\theta }_0-{\varphi }_0={15}^{\circ }$.
Total deviation:

$$\alpha ={15}^{\circ }+{120}^{\circ }+{15}^{\circ }={150}^{\circ }$$

Looking at List-II, this matches option (4) ${150}^{\circ }$.

Matching the Options

Now, let's match:

• P → 5

• Q → 2

• R → 1

• S → 4

Looking at the given choices:
(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

Our matches correspond to option (A).

Final Answer

$\boxed{{\displaystyle A}}$