Unpolarized light is incident on a plane glass surface having refractive index . The angle of incidence at which reflected and refracted rays would become perpendicular to each other is :

Options

A. `15^(@)`
B. `30^(@)`
C. `45^(@)`
D. `60^(@)`

Answer

To solve the problem, we need to find the angle of incidence at which the reflected and refracted rays become perpendicular to each other. Let's break down the solution step by step.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- When unpolarized light strikes a plane glass surface, we have an incident ray, a reflected ray, and a refracted ray.
- Let the angle of incidence be \( I \), the angle of reflection also be \( I \), and the angle of refraction be \( R \).

2. **Setting Up the Relationship**:
- According to the problem, the reflected and refracted rays are perpendicular to each other. This means that the angle between the reflected ray and the refracted ray is \( 90^\circ \).
- Therefore, we can express this relationship as:
\[ I + 90^\circ + R = 180^\circ \]
- Rearranging this gives:
\[ R = 90^\circ - I \]

3. **Applying Snell's Law**:
- Snell's Law states that:
\[ \frac{\sin I}{\sin R} = n \]
- Here, \( n \) is the refractive index of the glass, which is given as \( \sqrt{3} \).

4. **Substituting for \( R \)**:
- We substitute \( R \) from the previous step into Snell's Law:
\[ \frac{\sin I}{\sin(90^\circ - I)} = \sqrt{3} \]
- Since \( \sin(90^\circ - I) = \cos I \), we can rewrite the equation as:
\[ \frac{\sin I}{\cos I} = \sqrt{3} \]

5. **Simplifying the Equation**:
- The left-hand side can be expressed as:
\[ \tan I = \sqrt{3} \]

6. **Finding the Angle**:
- To find \( I \), we take the inverse tangent:
\[ I = \tan^{-1}(\sqrt{3}) \]
- We know that \( \tan 60^\circ = \sqrt{3} \), so:
\[ I = 60^\circ \]

### Final Answer:
The angle of incidence at which the reflected and refracted rays become perpendicular is \( 60^\circ \).

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