When a glass prism of refracting angle `60^(@)` is immersed in a liquid its angle of minimum deviation is `30^(@)`. The critical angle of glass with respect to the liquid medium is

Options

A. `42^(@)`
B. `45^(@)`
C. `50^(@)`
D. `52^(@)`

Explanation

To solve the problem, we need to find the critical angle of glass with respect to the liquid medium when a glass prism of refracting angle \( A = 60^\circ \) is immersed in a liquid and the angle of minimum deviation \( D = 30^\circ \).

### Step-by-Step Solution:

1. **Identify the Given Values:**
- Refracting angle of the prism, \( A = 60^\circ \)
- Angle of minimum deviation, \( D = 30^\circ \)

2. **Use the Formula for Refractive Index of the Prism:**
The refractive index \( \mu \) of the prism with respect to the surrounding medium (liquid in this case) can be calculated using the formula:
\[ \mu = \frac{\sin\left(\frac{A + D}{2}\right)}{\sin\left(\frac{A}{2}\right)} \]

3. **Calculate \( \frac{A + D}{2} \) and \( \frac{A}{2} \):**
- \( A + D = 60^\circ + 30^\circ = 90^\circ \)
- \( \frac{A + D}{2} = \frac{90^\circ}{2} = 45^\circ \)
- \( \frac{A}{2} = \frac{60^\circ}{2} = 30^\circ \)

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

5. **Substitute the Sine Values into the Refractive Index Formula:**
\[ \mu = \frac{\sin(45^\circ)}{\sin(30^\circ)} = \frac{\frac{1}{\sqrt{2}}}{\frac{1}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \]

6. **Relate the Refractive Index to the Critical Angle:**
The critical angle \( C \) can be found using the formula:
\[ \sin C = \frac{1}{\mu} \]
Therefore:
\[ C = \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) \]

7. **Calculate the Critical Angle:**
- Since \( \sin^{-1}\left(\frac{1}{\sqrt{2}}\right) = 45^\circ \), we find that:
\[ C = 45^\circ \]

### Final Answer:
The critical angle of glass with respect to the liquid medium is \( 45^\circ \).

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