A ray of light propagates from glass `(""_(a)mu_(g)=1.5)` to water `("'"_(a)mu_(w)4/3)` . The value of critical angle is

Options

A. `sin_(-1)(1/2)`
B. `sin^(-1)((sqrt8)/(9))`
C. `sin^(-1)(8/9)`
D. `sin^(-1)(5/7)`

Explanation

To find the critical angle when light propagates from glass to water, we can use Snell's Law, which states that:

\[ n_1 \sin(\theta_1) = n_2 \sin(\theta_2) \]

Where:
- \(n_1\) is the refractive index of the first medium (glass),
- \(n_2\) is the refractive index of the second medium (water),
- \(\theta_1\) is the angle of incidence,
- \(\theta_2\) is the angle of refraction.

The critical angle (\(\theta_c\)) occurs when the angle of refraction (\(\theta_2\)) is 90 degrees. At this point, the light is refracted along the boundary, and any angle of incidence greater than this will result in total internal reflection.

### Step-by-Step Solution:

1. **Identify the refractive indices**:
- For glass, \(n_g = 1.5\)
- For water, \(n_w = \frac{4}{3} \approx 1.33\)

2. **Apply Snell's Law at the critical angle**:
At the critical angle, we have:
\[ n_g \sin(\theta_c) = n_w \sin(90^\circ) \]
Since \(\sin(90^\circ) = 1\), this simplifies to:
\[ n_g \sin(\theta_c) = n_w \]

3. **Substitute the values of the refractive indices**:
\[ 1.5 \sin(\theta_c) = \frac{4}{3} \]

4. **Solve for \(\sin(\theta_c)\)**:
Rearranging the equation gives:
\[ \sin(\theta_c) = \frac{4/3}{1.5} \]
\[ \sin(\theta_c) = \frac{4}{3} \times \frac{1}{1.5} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9} \]

5. **Find the critical angle \(\theta_c\)**:
To find \(\theta_c\), take the inverse sine:
\[ \theta_c = \sin^{-1}\left(\frac{8}{9}\right) \]

### Final Answer:
The value of the critical angle \(\theta_c\) is \(\sin^{-1}\left(\frac{8}{9}\right)\).