Unpolarised light is incident from air on a plane surface of a material of refractive index `mu`. At a particular angle of incidence `i`, it is found that the reflected and refracted rays are perpendicular to each other. Which of the following options is correct for this situation?

Options

A. `i=sin^(-1)(1/mu)`
B. Reflected light is polarised with its electric vector perpendicular to the plane of incidence.
C. Reflected light is polarised with its electric vector parallel to the plane of incidence.
D. `i=tan^(-1)(1/mu)`

Explanation

To solve the problem, we need to understand the relationship between the angle of incidence, the reflected ray, and the refracted ray when they are perpendicular to each other. This situation occurs at Brewster's angle.

### Step-by-Step Solution:

1. **Understanding Brewster's Angle**:
- Brewster's angle (θ_B) is the angle of incidence at which light with a particular polarization is perfectly transmitted through a transparent dielectric surface, with no reflection. At this angle, the reflected and refracted rays are perpendicular to each other.

2. **Setting Up the Relationship**:
- When the reflected ray and refracted ray are perpendicular, we can express this mathematically:
\[ \theta_r + \theta_t = 90^\circ \]
- Here, θ_r is the angle of reflection (which equals the angle of incidence, i) and θ_t is the angle of refraction.

3. **Applying Snell's Law**:
- According to Snell's Law:
\[ n_1 \sin(\theta_i) = n_2 \sin(\theta_t) \]
- For our case, \( n_1 = 1 \) (for air) and \( n_2 = \mu \) (for the material). Thus:
\[ \sin(i) = \mu \sin(\theta_t) \]

4. **Relating Angles**:
- Since the reflected and refracted rays are perpendicular:
\[ \theta_t = 90^\circ - i \]
- Therefore, substituting this into Snell's Law gives:
\[ \sin(i) = \mu \sin(90^\circ - i) = \mu \cos(i) \]

5. **Rearranging the Equation**:
- We can rearrange the equation:
\[ \sin(i) = \mu \cos(i) \]
- Dividing both sides by cos(i):
\[ \tan(i) = \mu \]

6. **Finding Brewster's Angle**:
- The Brewster's angle (θ_B) can be expressed as:
\[ \theta_B = \tan^{-1}(\mu) \]

7. **Conclusion**:
- Therefore, the correct option for the situation described in the problem is:
\[ \theta_B = \tan^{-1}(\mu) \]