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^(@)`

Explanation

To solve the problem, we need to find the angle of incidence (i) for a prism with a refractive index (μ) of √2 and a refracting angle (A) of 60° at minimum deviation.

### Step-by-Step Solution:

1. **Identify Given Values**:
- Refractive index (μ) = √2
- Refracting angle (A) = 60°

2. **Use the Relation for Minimum Deviation**:
In the case of minimum deviation (D), the relationship between the angles of the prism and the angles of refraction is given by:
\[ r_1 + r_2 = A \]
Since at minimum deviation, \( r_1 = r_2 \), we can write:
\[ 2r = A \]
Therefore:
\[ r = \frac{A}{2} = \frac{60°}{2} = 30° \]

3. **Apply Snell's Law**:
According to Snell's law:
\[ \mu = \frac{\sin i}{\sin r} \]
Substituting the known values:
\[ \sqrt{2} = \frac{\sin i}{\sin 30°} \]
Since \( \sin 30° = \frac{1}{2} \), we can substitute this value:
\[ \sqrt{2} = \frac{\sin i}{\frac{1}{2}} \]

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

5. **Finding the Angle of Incidence**:
We know that:
\[ \sin 45° = \frac{\sqrt{2}}{2} \]
Therefore:
\[ i = 45° \]

### Final Answer:
The angle of incidence (i) is **45°**.

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