A glass-slab is immersed in water. What will be the crirtical 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 1: Understand the Concept of Critical Angle
The critical angle is the angle of incidence above which total internal reflection occurs when light travels from a denser medium to a less dense medium. In this case, we are dealing with light moving from glass (denser) to water (less dense).

### Step 2: Identify the Refractive Indices
Given:
- Refractive index of glass (n_g) = 1.50
- Refractive index of water (n_w) = 1.33

### Step 3: Calculate the Relative Refractive Index
The relative refractive index of glass with respect to water can be calculated using the formula:
\[ \mu_{g/w} = \frac{n_g}{n_w} \]
Substituting the values:
\[ \mu_{g/w} = \frac{1.50}{1.33} \approx 1.126 \]

### Step 4: Use the Critical Angle Formula
The critical angle (C) can be calculated using the formula:
\[ \sin C = \frac{n_w}{n_g} \]
Substituting the values:
\[ \sin C = \frac{1.33}{1.50} \approx 0.8867 \]

### Step 5: Calculate the Critical Angle
Now, we find the critical angle by taking the inverse sine (arcsin) of the value obtained:
\[ C = \sin^{-1}(0.8867) \]
Using the provided information that \(\sin^{-1}(0.887) = 62.5^\circ\), we can conclude:
\[ C \approx 62.5^\circ \]

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

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