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 a ray of light strikes a glass slab, it can be reflected and refracted. According to the problem, the reflected ray and the refracted ray are at right angles to each other. This means that if we denote the angle of incidence as \( i \) and the angle of refraction as \( r \), we have:
\[ i + r = 90^\circ \]
This is because the two angles add up to \( 90^\circ \) when they are perpendicular.

2. **Applying Snell's Law**:
Snell's Law relates the angles of incidence and refraction to the refractive indices of the two media:
\[ \frac{\sin i}{\sin r} = \frac{n_2}{n_1} \]
Here, \( n_1 = 1 \) (the refractive index of air) and \( n_2 = \sqrt{2} \) (the refractive index of the glass slab). Thus, we can write:
\[ \frac{\sin i}{\sin r} = \sqrt{2} \]

3. **Substituting for \( r \)**:
From the first equation \( i + r = 90^\circ \), we can express \( r \) in terms of \( i \):
\[ r = 90^\circ - i \]

4. **Using the Sine Relationship**:
Substitute \( r \) into Snell's Law:
\[ \frac{\sin i}{\sin(90^\circ - i)} = \sqrt{2} \]
Since \( \sin(90^\circ - i) = \cos i \), we have:
\[ \frac{\sin i}{\cos i} = \sqrt{2} \]

5. **Simplifying the Equation**:
The left-hand side of the equation \( \frac{\sin i}{\cos i} \) is equal to \( \tan i \):
\[ \tan i = \sqrt{2} \]

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

### Final Result:
The angle of incidence \( i \) is:
\[ i = \tan^{-1}(\sqrt{2}) \]