If minimum deviation = `30^(@)`, then speed of light in shown prism will be

Options

A. `3/sqrt(2)xx10^(8)m//s`
B. `1/sqrt(2) xx 10^(8)m//s`
C. `2/sqrt(3) xx 10^(8)m//s`
D. `(2KAepsilon_(0))/((K^(2)+1)d)`

Explanation

To find the speed of light in a prism when the minimum deviation is given, we can follow these steps:

### Step-by-Step Solution

1. **Identify Given Values**:
- Minimum deviation, \( \delta_m = 30^\circ \)
- Angle of the prism, \( A = 60^\circ \)

2. **Use the Prism Formula**:
The refractive index \( \mu \) of the prism can be calculated using the formula:
\[ \mu = \frac{\sin\left(\frac{A + \delta_m}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Calculate \( A + \delta_m \)**:
\[ A + \delta_m = 60^\circ + 30^\circ = 90^\circ \]
Therefore,
\[ \frac{A + \delta_m}{2} = \frac{90^\circ}{2} = 45^\circ \]

4. **Calculate \( \frac{A}{2} \)**:
\[ \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \]

5. **Substitute Values into the Prism Formula**:
Now substitute these values into the formula for \( \mu \):
\[ \mu = \frac{\sin(45^\circ)}{\sin(30^\circ)} \]

6. **Evaluate the Sine Values**:
- \( \sin(45^\circ) = \frac{1}{\sqrt{2}} \)
- \( \sin(30^\circ) = \frac{1}{2} \)

7. **Calculate \( \mu \)**:
\[ \mu = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{1}{\sqrt{2}} \times 2 = \frac{2}{\sqrt{2}} = \sqrt{2} \]

8. **Use the Refractive Index to Find Speed of Light in the Prism**:
The relationship between the speed of light in vacuum \( C \) and the speed of light in the medium \( V \) is given by:
\[ \mu = \frac{C}{V} \]
Rearranging gives:
\[ V = \frac{C}{\mu} \]

9. **Substitute Known Values**:
Given \( C = 3 \times 10^8 \, \text{m/s} \) and \( \mu = \sqrt{2} \):
\[ V = \frac{3 \times 10^8}{\sqrt{2}} \, \text{m/s} \]

10. **Final Calculation**:
\[ V \approx \frac{3 \times 10^8}{1.414} \approx 2.121 \times 10^8 \, \text{m/s} \]

### Conclusion
The speed of light in the prism is approximately \( 2.121 \times 10^8 \, \text{m/s} \).

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