A ray of light travelling in glass `(mu=3//2)` is incident on a horizontal glass air surface at the critical angle `theta_(C)`. If a thin layer of water `(mu=4//3)` is now poured on the glass air surface, the ray of light emerge into air at the water air surface at an angle of `pi//k`, radians find the value of k.

Answer

To solve the problem, we will follow a systematic approach using Snell's law and the concept of critical angle.

### Step-by-Step Solution:

1. **Understanding the Critical Angle**:
- The critical angle \( \theta_C \) is defined at the glass-air interface where light travels from a denser medium (glass) to a less dense medium (air).
- Using Snell's law:
\[ \mu_g \sin(\theta_C) = \mu_a \sin(90^\circ) \]
- Here, \( \mu_g = \frac{3}{2} \) (for glass) and \( \mu_a = 1 \) (for air).
- Therefore, we have:
\[ \frac{3}{2} \sin(\theta_C) = 1 \]
- Rearranging gives:
\[ \sin(\theta_C) = \frac{2}{3} \]

2. **Introducing Water Layer**:
- When a thin layer of water (with refractive index \( \mu_w = \frac{4}{3} \)) is poured on the glass-air surface, we need to analyze the new situation.
- The ray of light will now travel from glass to water and then from water to air.

3. **Applying Snell's Law at the Glass-Water Interface**:
- Let \( \theta_w \) be the angle of refraction in water.
- Using Snell's law at the glass-water interface:
\[ \mu_g \sin(\theta_C) = \mu_w \sin(\theta_w) \]
- Substituting the values:
\[ \frac{3}{2} \cdot \frac{2}{3} = \frac{4}{3} \sin(\theta_w) \]
- Simplifying gives:
\[ 1 = \frac{4}{3} \sin(\theta_w) \]
- Rearranging leads to:
\[ \sin(\theta_w) = \frac{3}{4} \]

4. **Applying Snell's Law at the Water-Air Interface**:
- Let \( \theta_a \) be the angle of refraction in air.
- Using Snell's law at the water-air interface:
\[ \mu_w \sin(\theta_w) = \mu_a \sin(\theta_a) \]
- Substituting the values:
\[ \frac{4}{3} \cdot \frac{3}{4} = 1 \cdot \sin(\theta_a) \]
- This simplifies to:
\[ 1 = \sin(\theta_a) \]
- Thus, we find:
\[ \theta_a = \frac{\pi}{2} \]

5. **Finding the Value of \( k \)**:
- The problem states that the ray emerges into air at an angle of \( \frac{\pi}{k} \).
- Since we found \( \theta_a = \frac{\pi}{2} \), we can set:
\[ \frac{\pi}{k} = \frac{\pi}{2} \]
- Solving for \( k \) gives:
\[ k = 2 \]

### Final Answer:
The value of \( k \) is \( 2 \).