A light wave travels from glass to water. The refractive index for glass and water are `3/2 ` and `4/3` respectively. The value of the critical angle will be:

Options

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

Answer

To find the critical angle when a light wave travels from glass to water, we can use Snell's law and the formula for the critical angle. Here are the steps to solve the problem:

### Step-by-Step Solution:

1. **Identify the Refractive Indices**:
- The refractive index of glass, \( n_g = \frac{3}{2} \).
- The refractive index of water, \( n_w = \frac{4}{3} \).

2. **Use the Formula for Critical Angle**:
The critical angle \( c \) can be found using the formula:
\[ \sin c = \frac{n_w}{n_g} \]
where \( n_g \) is the refractive index of the medium from which light is coming (glass), and \( n_w \) is the refractive index of the medium into which light is entering (water).

3. **Substitute the Values**:
Substitute the values of the refractive indices into the formula:
\[ \sin c = \frac{n_w}{n_g} = \frac{\frac{4}{3}}{\frac{3}{2}} \]

4. **Simplify the Expression**:
To simplify \( \frac{\frac{4}{3}}{\frac{3}{2}} \):
\[ \sin c = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9} \]

5. **Find the Critical Angle**:
Now, to find the critical angle \( c \), we take the inverse sine:
\[ c = \sin^{-1}\left(\frac{8}{9}\right) \]

6. **Calculate the Value**:
Using a calculator, find \( c \):
\[ c \approx 62.73^\circ \]

Thus, the value of the critical angle is approximately \( 62.73^\circ \).