A prism of refractive index sqrt2 has refractive angle `60^@.` In the order that a ray suffers minimum deviation it should be incident at an angle of

Options

A. `45^@`
B. `90^@`
C. `30^@`
D. None

Explanation

To solve the problem of finding the angle of incidence at which a ray suffers minimum deviation when passing through a prism with a refractive index of \(\sqrt{2}\) and a refracting angle of \(60^\circ\), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the given values:**
- Refractive index of the prism, \(\mu = \sqrt{2}\)
- Angle of the prism, \(A = 60^\circ\)

2. **Determine the angle of refraction for minimum deviation:**
- For minimum deviation, the angles of refraction at both faces of the prism are equal. Let this angle be \(r_1\).
- According to the property of the prism, we have:
\[ r_1 = r_2 = \frac{A}{2} \]
- Substituting the value of \(A\):
\[ r_1 = r_2 = \frac{60^\circ}{2} = 30^\circ \]

3. **Apply Snell's Law at the first surface of the prism:**
- According to Snell's Law:
\[ n_1 \sin I = \mu \sin r_1 \]
- Here, \(n_1\) (the refractive index of air) is \(1\), and we have:
\[ 1 \cdot \sin I = \sqrt{2} \cdot \sin(30^\circ) \]

4. **Calculate \(\sin(30^\circ)\):**
- We know that:
\[ \sin(30^\circ) = \frac{1}{2} \]
- Therefore, substituting this value into the equation gives:
\[ \sin I = \sqrt{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2} \]

5. **Find the angle of incidence \(I\):**
- To find \(I\), we take the inverse sine:
\[ I = \sin^{-1}\left(\frac{\sqrt{2}}{2}\right) \]
- This corresponds to:
\[ I = 45^\circ \]

6. **Conclusion:**
- The angle of incidence at which the ray suffers minimum deviation is \(45^\circ\).

### Final Answer:
The angle of incidence must be \(45^\circ\).

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