A prism of apex angle `A=60^@` has the refractive index `mu=sqrt2.` The angle of incidence for minimum deviation is

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. None of these

Explanation

To solve the problem of finding the angle of incidence for minimum deviation in a prism with an apex angle \( A = 60^\circ \) and a refractive index \( \mu = \sqrt{2} \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Condition for Minimum Deviation**:
At minimum deviation, the angles of refraction at both faces of the prism are equal. This means:
\[ r_1 = r_2 = \frac{A}{2} \]
Given \( A = 60^\circ \):
\[ r_1 = r_2 = \frac{60^\circ}{2} = 30^\circ \]

2. **Apply Snell's Law**:
According to Snell's law at the first interface (air to prism):
\[ \mu = \frac{\sin I}{\sin r_1} \]
Here, \( \mu = \sqrt{2} \) and \( r_1 = 30^\circ \).

3. **Substitute the Values into Snell's Law**:
We can substitute the known values into the equation:
\[ \sqrt{2} = \frac{\sin I}{\sin 30^\circ} \]
Since \( \sin 30^\circ = \frac{1}{2} \), we can rewrite the equation as:
\[ \sqrt{2} = \frac{\sin I}{\frac{1}{2}} \]

4. **Rearranging the Equation**:
Rearranging gives:
\[ \sin I = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} \]

5. **Finding the Angle of Incidence**:
Now, we need to find the angle \( I \) such that:
\[ \sin I = \frac{\sqrt{2}}{2} \]
The angle \( I \) that satisfies this equation is:
\[ I = 45^\circ \]

### Final Answer:
The angle of incidence for minimum deviation is \( I = 45^\circ \).