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. The critical angle is defined as the angle of incidence above which total internal reflection occurs.

### Step-by-step solution:

1. **Identify the refractive indices**:
- The refractive index of glass (μ1) = 3/2
- The refractive index of water (μ2) = 4/3

2. **Apply Snell's Law**:
Snell's Law states that:
\[ \mu_1 \sin(\theta_1) = \mu_2 \sin(\theta_2) \]
For the critical angle (θc), the angle of refraction (θ2) is 90 degrees. Thus, sin(θ2) = sin(90°) = 1.

3. **Set up the equation for the critical angle**:
\[ \mu_1 \sin(\theta_c) = \mu_2 \cdot 1 \]
Substituting the values of μ1 and μ2:
\[ \frac{3}{2} \sin(\theta_c) = \frac{4}{3} \]

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

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

6. **Conclusion**:
The critical angle θc is the angle whose sine is 8/9.

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