Angle of Deviation for Light Ray Refracting and Reflecting Inside Sphere

To determine the correct matching between List-I and List-II, we first establish the expression for the total deviation $\alpha$ of a light ray undergoing two refractions and one internal reflection inside a sphere of refractive index $n$.

1. General Derivation for Total Deviation

Let:

• $\theta_0$ be the angle of incidence at the first surface.
• $\phi_0$ be the angle of refraction inside the sphere.

By Snell's law at the first surface (air to sphere): $\sin\theta_0 = n \sin\phi_0$

Due to the spherical symmetry of the sphere, the chord representing the ray inside the sphere forms an isosceles triangle with the center of the sphere. Thus:

1. The deviation at the first refraction (entry) is: $\delta_1 = \theta_0 - \phi_0$
2. The angle of incidence at the back surface inside the sphere is also $\phi_0$. The deviation at internal reflection is: $\delta_2 = 180^\circ - 2\phi_0$
3. The angle of incidence at the exit surface inside the sphere is $\phi_0$, and the angle of refraction into air is $\theta_0$. The deviation at the second refraction (exit) is: $\delta_3 = \theta_0 - \phi_0$

The total angle of deviation $\alpha$ of the emergent ray with respect to the incident ray is the sum of these three deviations: $\alpha = \delta_1 + \delta_2 + \delta_3$ $\alpha = (\theta_0 - \phi_0) + (180^\circ - 2\phi_0) + (\theta_0 - \phi_0) = 180^\circ + 2\theta_0 - 4\phi_0$

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2. Analysis of List-I Items

(P) For $n = 2$ and $\alpha = 180^\circ$:

Substitute $\alpha = 180^\circ$ into the total deviation formula: $180^\circ = 180^\circ + 2\theta_0 - 4\phi_0 \implies \theta_0 = 2\phi_0$

Using Snell's law with $n = 2$: $\sin\theta_0 = 2\sin\phi_0$ $\sin(2\phi_0) = 2\sin\phi_0$ $2\sin\phi_0 \cos\phi_0 = 2\sin\phi_0$ $2\sin\phi_0(\cos\phi_0 - 1) = 0$

This yields two possibilities:

• $\sin\phi_0 = 0 \implies \phi_0 = 0^\circ \implies \theta_0 = 0^\circ$
• $\cos\phi_0 = 1 \implies \phi_0 = 0^\circ \implies \theta_0 = 0^\circ$

Thus, the only possible value of $\theta_0$ is $0^\circ$. $\text{P} \rightarrow (5)$

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(Q) For $n = \sqrt{3}$ and $\alpha = 180^\circ$:

From $\alpha = 180^\circ$, we have $\theta_0 = 2\phi_0$.

Using Snell's law with $n = \sqrt{3}$: $\sin\theta_0 = \sqrt{3}\sin\phi_0$ $\sin(2\phi_0) = \sqrt{3}\sin\phi_0$ $2\sin\phi_0 \cos\phi_0 = \sqrt{3}\sin\phi_0$ $\sin\phi_0 \left(2\cos\phi_0 - \sqrt{3}\right) = 0$

This yields two solutions:

1. $\sin\phi_0 = 0 \implies \phi_0 = 0^\circ \implies \theta_0 = 0^\circ$
2. $2\cos\phi_0 - \sqrt{3} = 0 \implies \cos\phi_0 = \frac{\sqrt{3}}{2} \implies \phi_0 = 30^\circ \implies \theta_0 = 2(30^\circ) = 60^\circ$

Thus, the possible values of $\theta_0$ are $60^\circ$ and $0^\circ$. $\text{Q} \rightarrow (2)$

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(R) For $n = \sqrt{3}$ and $\alpha = 180^\circ$:

From the derivation in part (Q), the corresponding possible values for the angle of refraction $\phi_0$ are: $\phi_0 = 30^\circ \text{ and } 0^\circ$

$\text{R} \rightarrow (1)$

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(S) For $n = \sqrt{2}$ and $\theta_0 = 45^\circ$:

By Snell's law: $\sin(45^\circ) = \sqrt{2}\sin\phi_0$ $\frac{1}{\sqrt{2}} = \sqrt{2}\sin\phi_0 \implies \sin\phi_0 = \frac{1}{2} \implies \phi_0 = 30^\circ$

Now, calculate the total deviation $\alpha$: $\alpha = 180^\circ + 2\theta_0 - 4\phi_0 = 180^\circ + 2(45^\circ) - 4(30^\circ) = 180^\circ + 90^\circ - 120^\circ = 150^\circ$

$\text{S} \rightarrow (4)$

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Conclusion

The correct matching is: $\text{P} \rightarrow 5; \quad \text{Q} \rightarrow 2; \quad \text{R} \rightarrow 1; \quad \text{S} \rightarrow 4$

This corresponds to Option A.