A glass slab has a critical angle of `30^(@)` when placed in air. What will be the criticle angle when it is placed in liquid of refractive index 6/5 index?

Options

A. `45^(@)`
B. `37^(@)`
C. `53^(@)`
D. `60^(@)`

Explanation

To solve the problem, we need to determine the critical angle of a glass slab when it is placed in a liquid with a refractive index of \( \frac{6}{5} \). We will use Snell's law and the concept of critical angle.

### Step-by-Step Solution:

1. **Understanding Critical Angle**:
The critical angle \( \theta_c \) is defined as the angle of incidence above which total internal reflection occurs. It can be calculated using the formula:
\[ \sin \theta_c = \frac{n_2}{n_1} \]
where \( n_1 \) is the refractive index of the medium in which the light is coming from (glass in this case) and \( n_2 \) is the refractive index of the medium into which the light is entering (air or liquid).

2. **Finding the Refractive Index of Glass**:
From the problem, we know the critical angle of the glass slab when placed in air is \( 30^\circ \). The refractive index of air \( n_a \) is approximately \( 1 \). Using the critical angle formula:
\[ \sin 30^\circ = \frac{n_a}{n_g} \]
Since \( \sin 30^\circ = \frac{1}{2} \), we can substitute:
\[ \frac{1}{2} = \frac{1}{n_g} \]
Rearranging gives:
\[ n_g = 2 \]
So, the refractive index of glass \( n_g \) is \( 2 \).

3. **Applying Snell's Law for the Liquid**:
Now, we need to find the critical angle when the glass slab is placed in a liquid with a refractive index \( n_l = \frac{6}{5} \). The critical angle \( \theta_c' \) for the glass in the liquid can be calculated as:
\[ \sin \theta_c' = \frac{n_l}{n_g} \]
Substituting the values we have:
\[ \sin \theta_c' = \frac{\frac{6}{5}}{2} = \frac{6}{10} = \frac{3}{5} \]

4. **Calculating the Critical Angle**:
Now we need to find \( \theta_c' \) using the inverse sine function:
\[ \theta_c' = \sin^{-1}\left(\frac{3}{5}\right) \]
Using a calculator or trigonometric tables, we find:
\[ \theta_c' \approx 36.87^\circ \]
Rounding this gives approximately \( 37^\circ \).

### Final Answer:
The critical angle when the glass slab is placed in a liquid with a refractive index of \( \frac{6}{5} \) is approximately \( 37^\circ \).