Angle of prism is `A` and its one surface is silvered. Light rays falling at an angle of incidence `2 A` on first surface return back through the same path after suffering reflection at second silvered surface. Refraction index of the material of prism is

Options

A. `2 sin A`
B. `2cos A`
C. `(1)/(2) cos A`
D. tan A

Explanation

To find the refractive index of the material of the prism when one surface is silvered, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a prism with an angle \( A \).
- One surface of the prism is silvered (reflective).
- A light ray is incident on the first surface at an angle of incidence \( 2A \).

2. **Refraction at the First Surface**:
- When the light ray strikes the first surface of the prism, it undergoes refraction.
- Let the angle of refraction be \( r \).
- According to Snell's law, we have:
\[ \sin(2A) = \mu \sin(r) \]
- Here, \( \mu \) is the refractive index of the prism.

3. **Geometry of the Prism**:
- The internal angles of the prism can be analyzed. The angle of the prism is \( A \), and the angles at the interface can be expressed as:
\[ A + (90^\circ - r) + 90^\circ = 180^\circ \]
- Simplifying this gives:
\[ A + 90^\circ - r + 90^\circ = 180^\circ \implies A + 180^\circ - r = 180^\circ \implies r = A \]

4. **Substituting the Angle of Refraction**:
- Now substituting \( r = A \) back into Snell's law:
\[ \sin(2A) = \mu \sin(A) \]

5. **Expressing the Refractive Index**:
- Rearranging the equation gives:
\[ \mu = \frac{\sin(2A)}{\sin(A)} \]
- Using the double angle identity for sine, we have:
\[ \sin(2A) = 2 \sin(A) \cos(A) \]
- Therefore, substituting this into the equation for \( \mu \):
\[ \mu = \frac{2 \sin(A) \cos(A)}{\sin(A)} \]
- The \( \sin(A) \) terms cancel out:
\[ \mu = 2 \cos(A) \]

6. **Final Result**:
- The refractive index of the material of the prism is:
\[ \mu = 2 \cos(A) \]