The refractive index of a medium is `sqrt(3)`. What is the angle of refraction, if unpolarizing light is incident on the polarising angle of the medium ?

Options

A. `60^(@)`
B. `45^(@)`
C. `30^(@)`
D. `0^(@)`

Explanation

To solve the problem, we need to find the angle of refraction when unpolarized light is incident on the polarizing angle of a medium with a refractive index of \(\sqrt{3}\).

### Step-by-Step Solution:

1. **Identify the given data**:
- Refractive index \( n = \sqrt{3} \).

2. **Use Brewster's Law to find the polarizing angle (\(\theta_P\))**:
- According to Brewster's Law, the refractive index \( n \) is given by:
\[ n = \tan(\theta_P) \]
- Substituting the value of \( n \):
\[ \sqrt{3} = \tan(\theta_P) \]

3. **Calculate the polarizing angle (\(\theta_P\))**:
- From trigonometric values, we know that:
\[ \tan(60^\circ) = \sqrt{3} \]
- Therefore, we can conclude:
\[ \theta_P = 60^\circ \]

4. **Apply Snell's Law to find the angle of refraction (\(R\))**:
- Snell's Law states:
\[ n = \frac{\sin(i)}{\sin(r)} \]
- Here, \( i = \theta_P = 60^\circ \) and \( n = \sqrt{3} \):
\[ \sqrt{3} = \frac{\sin(60^\circ)}{\sin(R)} \]

5. **Substitute the value of \(\sin(60^\circ)\)**:
- We know that:
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
- Therefore, substituting this into the equation gives:
\[ \sqrt{3} = \frac{\frac{\sqrt{3}}{2}}{\sin(R)} \]

6. **Rearranging to find \(\sin(R)\)**:
- Cross-multiplying gives:
\[ \sqrt{3} \cdot \sin(R) = \frac{\sqrt{3}}{2} \]
- Dividing both sides by \(\sqrt{3}\):
\[ \sin(R) = \frac{1}{2} \]

7. **Determine the angle \(R\)**:
- We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]
- Therefore, we conclude:
\[ R = 30^\circ \]

### Final Answer:
The angle of refraction \( R \) is \( 30^\circ \).

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