A ray of light falls on a transparent glass slab of refractive index `sqrt (2)`. If the reflected and refracted rays are mutually perpendicular, then the angle of incidence is

Options

A. `tan^-1(sqrt(3/2))`
B. `sin^(-1)(sqrt(2/3))`
C. `tan^-1(sqrt(2))`
D. `sin^(-1)(sqrt(3))`

Answer

To solve the problem, we need to find the angle of incidence when a ray of light falls on a transparent glass slab with a refractive index of \( \sqrt{2} \), and the reflected and refracted rays are mutually perpendicular.

### Step-by-Step Solution:

1. **Understanding the Geometry**:
- When the reflected and refracted rays are mutually perpendicular, it means that the angle between the reflected ray and the refracted ray is \( 90^\circ \).
- Let \( i \) be the angle of incidence, \( r \) be the angle of refraction, and \( R \) be the angle of reflection. According to the law of reflection, \( R = i \).

2. **Setting Up the Relationship**:
- Since the reflected and refracted rays are perpendicular, we can write:
\[ R + r = 90^\circ \]
- Substituting \( R = i \) into the equation gives us:
\[ i + r = 90^\circ \]
- Rearranging this, we find:
\[ r = 90^\circ - i \]

3. **Applying Snell's Law**:
- Snell's law states:
\[ n_1 \sin(i) = n_2 \sin(r) \]
- Here, \( n_1 \) (refractive index of air) is \( 1 \) and \( n_2 \) (refractive index of glass) is \( \sqrt{2} \). Thus, we can write:
\[ 1 \cdot \sin(i) = \sqrt{2} \cdot \sin(r) \]

4. **Substituting for \( r \)**:
- We know from our earlier relationship that \( r = 90^\circ - i \). Therefore, we can substitute this into Snell's law:
\[ \sin(i) = \sqrt{2} \cdot \sin(90^\circ - i) \]
- Using the identity \( \sin(90^\circ - x) = \cos(x) \), we have:
\[ \sin(i) = \sqrt{2} \cdot \cos(i) \]

5. **Rearranging the Equation**:
- Dividing both sides by \( \cos(i) \) (assuming \( \cos(i) \neq 0 \)):
\[ \tan(i) = \sqrt{2} \]

6. **Finding the Angle of Incidence**:
- To find \( i \), we take the inverse tangent:
\[ i = \tan^{-1}(\sqrt{2}) \]

### Final Answer:
The angle of incidence \( i \) is \( \tan^{-1}(\sqrt{2}) \).