Question

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 \(\phi\) 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 \(a\). 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	I	\(30\degree\) or \(0\degree\)
Q	If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\theta_0\) will be	II	\(60\degree\) or \(0\degree\)
R	If \(n = √3\) and \(\alpha= 180°\), then all the possible values of \(\phi_0\) will be	III	\(45\degree\) or \( 0\degree\)
S	If \(n = \sqrt2\) and \(\theta_0 = 45°\), then all the possible values of  \(\alpha\) will be	IV	\(150\degree\)
			\[0\degree\]

• 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

Answer 1

The Correct Option is A

Step 1: Calculate for P 

Given \( n = 2 \) and \( \alpha = 180^\circ \),

\[ \delta = 2(\theta_0 - \phi_0) + \pi - 2\phi_0 = 2. \]

From the above equation, \( \phi_0 = 0^\circ \).

\[ \theta_0 = 2\phi_0 = 0^\circ. \]

Thus, \( \theta_0 = 0^\circ \) for P.

Step 2: Calculate for Q

Given \( n = \sqrt{3} \) and \( \alpha = 180^\circ \),

\[ 1 \cdot \sin \theta_0 = \sqrt{3} \cdot \sin \phi_0. \]

For \( \phi_0 = 0^\circ \):

\[ \theta_0 = 0^\circ \quad \text{or} \quad \theta_0 = 60^\circ. \]

Step 3: Calculate for R

Given \( n = \sqrt{3} \) and \( \alpha = 180^\circ \), using Snell’s law:

\[ n \cdot \sin \phi_0 = \sin \theta_0. \]

Solving gives \( \phi_0 = 0^\circ \) or \( \phi_0 = 30^\circ \).

Step 4: Calculate for S

Given \( n = \sqrt{2} \) and \( \theta_0 = 45^\circ \):

\[ \sin \phi_0 = \frac{\sin 45^\circ}{\sqrt{2}} = \frac{1}{2}. \]

The total angle \( \alpha \) is:

\[ \alpha = 150^\circ. \]

Conclusion

Based on the calculations:

• P → 5
• Q → 2
• R → 1
• S → 4

Final Answer:

The correct option is (A).

Answer 2

To solve the problem, we need to analyze the refraction and reflection of light rays inside a sphere with given refractive indices and angles, then match the values from List-I with List-II.

1. Understanding the problem:
A light ray incident on the surface of a sphere partially refracts inside, reflects from the back surface, and then emerges out. Given the refractive index \(n\), angle of incidence \(\theta_0\), angle of refraction \(\phi\), and total deviation angle \(\alpha\), we find possible values of these angles under different conditions.

2. Using Snell's Law and Geometry:
Snell's Law states:
\[ n_1 \sin \theta_0 = n_2 \sin \phi \] For air to sphere interface, \(n_1 = 1\) and \(n_2 = n\).
Hence, \[ \sin \theta_0 = n \sin \phi \] Given the total angle of deviation \(\alpha = 180^\circ\), geometrical relations help us find possible values of \(\theta_0\) and \(\phi\).

3. Matching List-I and List-II for each case:

• P: \(n=2, \alpha=180^\circ\). Using Snell's law and geometry, possible \(\theta_0\) values are \(30^\circ\) or \(0^\circ\). Matches List-II I.
• Q: \(n=\sqrt{3}, \alpha=180^\circ\). Possible \(\theta_0\) values are \(60^\circ\) or \(0^\circ\). Matches List-II II.
• R: \(n=\sqrt{3}, \alpha=180^\circ\). Possible \(\phi_0\) values are \(45^\circ\) or \(0^\circ\). Matches List-II I.
• S: \(n=\sqrt{2}, \theta_0=45^\circ\). Possible \(\alpha\) values are \(150^\circ\) or \(0^\circ\). Matches List-II IV and V.

4. Conclusion:
The correct matching is:
P → 5 (0° or 30°), Q → 2 (0° or 60°), R → 1 (0° or 45°), S → 4 (150° or 0°)

Final Answer:
Option A: P → 5; Q → 2; R → 1; S → 4