A ray of light travelling in glass `(mu_g=3//2)` is incident on a horizontal glass-air surface at the critical angle `theta_C.` If a thin layer of water `(mu_w=4//3)` is now poured on the glass-air surface. At what angle will the ray of light emerges into water at glass-water surface?

Explanation

To solve the problem step by step, we will use Snell's Law and the concept of critical angle.

### Step 1: Understand the Critical Angle
The critical angle \( \theta_C \) is the angle of incidence in the denser medium (glass) at which the angle of refraction in the less dense medium (air) is \( 90^\circ \). This can be expressed using Snell's Law:
\[ \mu_g \sin(\theta_C) = \mu_{air} \sin(90^\circ) \]
Given that \( \mu_{air} = 1 \) (the refractive index of air), we can simplify this to:
\[ \mu_g \sin(\theta_C) = 1 \]

### Step 2: Calculate the Critical Angle
We know the refractive index of glass \( \mu_g = \frac{3}{2} \). Thus, we can find the critical angle \( \theta_C \):
\[ \frac{3}{2} \sin(\theta_C) = 1 \implies \sin(\theta_C) = \frac{2}{3} \]
Now, we can find \( \theta_C \) using the inverse sine function:
\[ \theta_C = \sin^{-1}\left(\frac{2}{3}\right) \]

### Step 3: Introduce Water Layer
Now, when a thin layer of water (with refractive index \( \mu_w = \frac{4}{3} \)) is poured on the glass-air surface, we need to find the angle at which the ray of light emerges into water at the glass-water surface.

### Step 4: Apply Snell's Law at Glass-Water Interface
At the glass-water interface, we can use Snell's Law again:
\[ \mu_g \sin(\theta_C) = \mu_w \sin(\theta_1) \]
Where \( \theta_1 \) is the angle of refraction in water. Since we already established that \( \mu_g \sin(\theta_C) = 1 \):
\[ 1 = \mu_w \sin(\theta_1) \implies \sin(\theta_1) = \frac{1}{\mu_w} = \frac{1}{\frac{4}{3}} = \frac{3}{4} \]

### Step 5: Calculate Angle \( \theta_1 \)
Now, we can find \( \theta_1 \):
\[ \theta_1 = \sin^{-1}\left(\frac{3}{4}\right) \]

### Step 6: Apply Snell's Law at Water-Air Interface
Next, we need to find the angle \( \theta_2 \) at which the ray of light emerges into the air from water. Applying Snell's Law again:
\[ \mu_w \sin(\theta_1) = \mu_{air} \sin(\theta_2) \]
Since \( \mu_{air} = 1 \):
\[ \mu_w \sin(\theta_1) = \sin(\theta_2) \]
Substituting \( \mu_w = \frac{4}{3} \) and \( \sin(\theta_1) = \frac{3}{4} \):
\[ \frac{4}{3} \cdot \frac{3}{4} = \sin(\theta_2) \implies \sin(\theta_2) = 1 \]
Thus, \( \theta_2 = 90^\circ \).

### Conclusion
The angle at which the ray of light emerges into the water at the glass-water surface is \( 90^\circ \).

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