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 find the minimum deviation produced by a prism with a given refractive index and refractive angle, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Refractive index (μ) = √(3/2)
- Refractive angle (A) = 90°

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

3. **Substitute the Known Values:**
Substitute A = 90° into the formula:
\[ \mu = \frac{\sin\left(\frac{90° + \delta_m}{2}\right)}{\sin\left(\frac{90°}{2}\right)} \]
This simplifies to:
\[ \mu = \frac{\sin\left(45° + \frac{\delta_m}{2}\right)}{\sin(45°)} \]

4. **Calculate sin(45°):**
We know that:
\[ \sin(45°) = \frac{\sqrt{2}}{2} \]

5. **Set Up the Equation:**
Substitute the value of μ:
\[ \frac{\sqrt{3}}{2} = \frac{\sin\left(45° + \frac{\delta_m}{2}\right)}{\frac{\sqrt{2}}{2}} \]
This can be rearranged to:
\[ \sqrt{3} = \sin\left(45° + \frac{\delta_m}{2}\right) \cdot \sqrt{2} \]

6. **Solve for sin(45° + δm/2):**
Rearranging gives:
\[ \sin\left(45° + \frac{\delta_m}{2}\right) = \frac{\sqrt{3}}{\sqrt{2}} \]
We can simplify this to:
\[ \sin\left(45° + \frac{\delta_m}{2}\right) = \sin(60°) \]

7. **Set Angles Equal:**
Since sine is positive in the first quadrant, we can equate the angles:
\[ 45° + \frac{\delta_m}{2} = 60° \]

8. **Solve for δm:**
Rearranging gives:
\[ \frac{\delta_m}{2} = 60° - 45° = 15° \]
Therefore:
\[ \delta_m = 30° \]

### Final Answer:
The minimum deviation produced by the prism is **30°**.

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