The refractive index of a prism for a monochromatic wave is `sqrt(2)` and its refracting angle is `60^(@)` for minimum deviation, the angle of indidence will be

Options

A. `30^@`
B. `45^@`
C. `60^@`
D. `75^@`

Answer

To find the angle of incidence for a prism with a refractive index of \(\sqrt{2}\) and a refracting angle of \(60^\circ\) at minimum deviation, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values:**
- Refractive index (\(\mu\)) = \(\sqrt{2}\)
- Refracting angle of the prism (\(A\)) = \(60^\circ\)

2. **Determine the Angle of Refraction (\(R\)) at Minimum Deviation:**
- For minimum deviation, the angle of refraction \(R\) is given by:
\[ R = \frac{A}{2} \]
- Substituting the value of \(A\):
\[ R = \frac{60^\circ}{2} = 30^\circ \]

3. **Use Snell's Law:**
- According to Snell's law at the prism interface:
\[ \mu = \frac{\sin I}{\sin R} \]
- Substituting the known values:
\[ \sqrt{2} = \frac{\sin I}{\sin 30^\circ} \]

4. **Calculate \(\sin 30^\circ\):**
- We know that:
\[ \sin 30^\circ = \frac{1}{2} \]

5. **Substitute \(\sin 30^\circ\) into the equation:**
- The equation becomes:
\[ \sqrt{2} = \frac{\sin I}{\frac{1}{2}} \]

6. **Rearranging the equation to find \(\sin I\):**
- Multiply both sides by \(\frac{1}{2}\):
\[ \sin I = \sqrt{2} \times \frac{1}{2} = \frac{\sqrt{2}}{2} \]

7. **Determine the Angle of Incidence (\(I\)):**
- The value of \(\sin I = \frac{\sqrt{2}}{2}\) corresponds to:
\[ I = 45^\circ \]

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

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