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

Answer

To solve the problem, we need to analyze the situation where unpolarized light is incident on a surface and the reflected and refracted rays are perpendicular to each other. We will use Brewster's Law to derive the necessary conditions.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have unpolarized light incident from air (refractive index \( n_1 = 1 \)) onto a plane surface of a material with refractive index \( \mu \).
- The angle of incidence is \( I \).

**Hint**: Identify the refractive indices of the two media involved.

2. **Condition of Perpendicular Rays**:
- The problem states that the reflected and refracted rays are perpendicular to each other. This means that the angle between the reflected ray and the refracted ray is \( 90^\circ \).

**Hint**: Recall that if two angles are complementary, their sum is \( 90^\circ \).

3. **Using Snell's Law**:
- According to Snell's Law:
\[ n_1 \sin I = n_2 \sin r \]
where \( r \) is the angle of refraction.

**Hint**: Remember that \( n_1 \) is the refractive index of air (1) and \( n_2 \) is \( \mu \).

4. **Relationship Between Angles**:
- Since the reflected and refracted rays are perpendicular, we have:
\[ r = 90^\circ - I \]
- Substituting this into Snell's Law gives:
\[ \sin r = \sin(90^\circ - I) = \cos I \]
- Therefore, Snell's Law becomes:
\[ \sin I = \mu \cos I \]

**Hint**: Use the trigonometric identity for sine and cosine.

5. **Rearranging the Equation**:
- Rearranging the equation gives:
\[ \tan I = \mu \]

**Hint**: This relationship indicates that the angle of incidence is related to the refractive index.

6. **Applying Brewster's Law**:
- Brewster's Law states that at the Brewster angle \( I_B \):
\[ I_B = \tan^{-1}(\mu) \]
- This means that at this angle, the reflected light is polarized with its electric field vector perpendicular to the plane of incidence.

**Hint**: Recognize that Brewster's angle is critical for understanding polarization.

7. **Conclusion**:
- The reflected ray is polarized with its electric vector perpendicular to the plane of incidence. Thus, the correct option is that the reflected ray is polarized with its electric vector perpendicular to the plane of incidence.

### Final Answer:
The correct option is that the reflected ray is polarized with its electric vector perpendicular to the plane of incidence.