Critical angle of glass is `theta_(1)` and that of water is `theta_(2)`. The critical angle for water and glass surface would be `(mu_(g)=3//2, mu_(w)=4//3)`

Options

A. less than `theta_(2)`
B. between `theta_(1)` and `theta_(2)`
C. greater than `theta_(2)`
D. less than `theta_(1)`

Answer

To find the critical angle for the interface between glass and water, we can use Snell's law and the definition of the critical angle. The critical angle is defined as the angle of incidence above which total internal reflection occurs.

### Step-by-Step Solution:

1. **Understand the Critical Angle**:
The critical angle \( \theta_c \) for a medium is given by 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, and \( n_2 \) is the refractive index of the medium into which the light is entering.

2. **Given Refractive Indices**:
- For glass, \( \mu_g = \frac{3}{2} \)
- For water, \( \mu_w = \frac{4}{3} \)

3. **Critical Angle for Glass**:
The critical angle \( \theta_1 \) for glass can be calculated when light travels from glass to air (assuming air has a refractive index of 1):
\[ \sin(\theta_1) = \frac{1}{\mu_g} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]

4. **Critical Angle for Water**:
Similarly, the critical angle \( \theta_2 \) for water can be calculated when light travels from water to air:
\[ \sin(\theta_2) = \frac{1}{\mu_w} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

5. **Critical Angle for Glass-Water Interface**:
Now, we want to find the critical angle when light travels from water to glass. Here, \( n_1 = \mu_w \) and \( n_2 = \mu_g \):
\[ \sin(\theta_c) = \frac{\mu_w}{\mu_g} = \frac{\frac{4}{3}}{\frac{3}{2}} = \frac{4}{3} \times \frac{2}{3} = \frac{8}{9} \]

6. **Finding the Critical Angle**:
To find the critical angle \( \theta_c \):
\[ \theta_c = \sin^{-1}\left(\frac{8}{9}\right) \]

7. **Conclusion**:
The critical angle for the water-glass surface is \( \theta_c = \sin^{-1}\left(\frac{8}{9}\right) \).