When light of wavelength `lambda` is incident on an equilateral prism, kept on its minimum deviation position, it is found that the angle of deviation equals the angle of the prism itself. The refractive index of the material of the prism for the wavelength `lambda` is

Options

A. `sqrt3`
B. `sqrt(3//2)`
C. `2`
D. `sqrt2`

Explanation

To find the refractive index of the material of an equilateral prism when light of wavelength `lambda` is incident on it in the minimum deviation position, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Prism Configuration**:
- We have an equilateral prism, which means each angle of the prism (denoted as \( A \)) is \( 60^\circ \).

2. **Minimum Deviation Condition**:
- In the minimum deviation position, the angle of deviation (\( \delta \)) is equal to the angle of the prism (\( A \)). Therefore, we have:
\[ \delta = A = 60^\circ \]

3. **Relation Between Angles**:
- The formula for deviation in terms of the angle of incidence (\( I \)), angle of emergence (\( E \)), and angle of the prism (\( A \)) is given by:
\[ \delta = I + E - A \]
- At minimum deviation, \( I = E \). Thus, we can write:
\[ \delta = 2I - A \]

4. **Substituting Values**:
- Since \( \delta = 60^\circ \) and \( A = 60^\circ \), we can substitute into the equation:
\[ 60^\circ = 2I - 60^\circ \]
- Rearranging gives:
\[ 2I = 120^\circ \implies I = 60^\circ \]

5. **Using Snell's Law**:
- According to Snell's law:
\[ \mu = \frac{\sin I}{\sin R} \]
- At minimum deviation, the angle of refraction \( R \) can be expressed as:
\[ R = \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

6. **Calculate the Refractive Index**:
- Now substituting \( I = 60^\circ \) and \( R = 30^\circ \) into Snell's law:
\[ \mu = \frac{\sin 60^\circ}{\sin 30^\circ} \]
- We know:
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin 30^\circ = \frac{1}{2} \]
- Therefore:
\[ \mu = \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3} \]

### Final Answer:
The refractive index of the material of the prism for the wavelength \( \lambda \) is \( \sqrt{3} \).

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