A prism has refractive index `sqrt((3)/(2))` and refractive angle `90^@`. Find the minimum deviation produced by prism

Options

A. `60^@`
B. `45^@`
C. `30^@`
D. `15^@`

Explanation

To solve the problem of finding the minimum deviation produced by a prism with a refractive index of \( \sqrt{\frac{3}{2}} \) and a refractive angle of \( 90^\circ \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refractive index \( n = \sqrt{\frac{3}{2}} \)
- Refractive angle \( A = 90^\circ \)

2. **Use the Formula for Minimum Deviation:**
The formula relating the refractive index \( n \), the angle of the prism \( A \), and the minimum deviation \( \delta_m \) is given by:
\[ n = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Substitute the Values into the Formula:**
- Since \( A = 90^\circ \), we have:
\[ n = \frac{\sin\left(\frac{90^\circ + \delta_m}{2}\right)}{\sin\left(\frac{90^\circ}{2}\right)} \]
- This simplifies to:
\[ n = \frac{\sin\left(\frac{90^\circ + \delta_m}{2}\right)}{\sin(45^\circ)} \]

4. **Calculate \( \sin(45^\circ) \):**
- We know that \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \).

5. **Substituting \( n \) and \( \sin(45^\circ) \) into the Equation:**
\[ \sqrt{\frac{3}{2}} = \frac{\sin\left(\frac{90^\circ + \delta_m}{2}\right)}{\frac{1}{\sqrt{2}}} \]
- Rearranging gives:
\[ \sin\left(\frac{90^\circ + \delta_m}{2}\right) = \sqrt{\frac{3}{2}} \cdot \frac{1}{\sqrt{2}} = \sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2} \]

6. **Determine the Angle Corresponding to \( \sin\left(\frac{90^\circ + \delta_m}{2}\right) = \frac{\sqrt{3}}{2} \):**
- The angle whose sine is \( \frac{\sqrt{3}}{2} \) is \( 60^\circ \):
\[ \frac{90^\circ + \delta_m}{2} = 60^\circ \]

7. **Solve for \( \delta_m \):**
- Multiply both sides by 2:
\[ 90^\circ + \delta_m = 120^\circ \]
- Subtract \( 90^\circ \) from both sides:
\[ \delta_m = 120^\circ - 90^\circ = 30^\circ \]

### Final Answer:
The minimum deviation \( \delta_m \) produced by the prism is \( 30^\circ \).