A glass-slab is immersed in water. What will be the critical angle for a light ray at glass-water interface? Where `_(a)n_(g)=1.50,_(a)n_(w)=1.33 and sin^(-1)(0.887)=62.5`

Options

A. `48.8^(@)`
B. `72.8^(@)`
C. `62.5^(@)`
D. `64.5^(@)`

Explanation

To find the critical angle for a light ray at the glass-water interface, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Values**:
- Refractive index of glass with respect to air, \( n_g = 1.50 \)
- Refractive index of water with respect to air, \( n_w = 1.33 \)

2. **Determine the Refractive Index of Glass with respect to Water**:
- The refractive index of glass with respect to water can be calculated using the formula:
\[ n_{gw} = \frac{n_g}{n_w} \]
- Substituting the values:
\[ n_{gw} = \frac{1.50}{1.33} \]

3. **Calculate the Value**:
- Performing the division:
\[ n_{gw} \approx 1.126 \]

4. **Apply the Formula for Critical Angle**:
- The critical angle \( C \) can be found using Snell's Law:
\[ \frac{1}{\sin C} = n_{gw} \]
- Rearranging gives:
\[ \sin C = \frac{1}{n_{gw}} \]

5. **Substitute the Value of \( n_{gw} \)**:
- Thus,
\[ \sin C = \frac{1}{1.126} \approx 0.887 \]

6. **Find the Critical Angle**:
- To find the critical angle, take the inverse sine:
\[ C = \sin^{-1}(0.887) \]
- From the problem statement, we know:
\[ C \approx 62.5^\circ \]

### Final Answer:
The critical angle for a light ray at the glass-water interface is approximately \( 62.5^\circ \).

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