A ray of light travelling in glass having refractive index `_(a)mu_(g)=3//2` is incident at a critical angle C on the glass air interface. If a thin layer of water is poured on glass air interface, then what will be the angle of emergence of this ray in air when it emerges from water air inteface?

Options

A. `180^(@)`
B. `0^(@)`
C. `90^(@)`
D. `45^(@)`

Answer

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 in glass-air interface
The critical angle \( C \) is defined as the angle of incidence in the denser medium (glass) for which the angle of refraction in the less dense medium (air) is \( 90^\circ \).

Using Snell's law:
\[ \mu_g \sin C = \mu_a \sin 90^\circ \]
Where:
- \( \mu_g = \frac{3}{2} \) (refractive index of glass)
- \( \mu_a = 1 \) (refractive index of air)

Since \( \sin 90^\circ = 1 \), we can rewrite the equation as:
\[ \mu_g \sin C = 1 \]
From this, we can derive:
\[ \sin C = \frac{1}{\mu_g} = \frac{1}{\frac{3}{2}} = \frac{2}{3} \]
Thus, the critical angle \( C \) is:
\[ C = \sin^{-1}\left(\frac{2}{3}\right) \]

### Step 2: Ray incident at the critical angle on the glass-water interface
When a thin layer of water is poured on the glass surface, the ray of light will still be incident at the critical angle \( C \) at the glass-water interface.

Applying Snell's law at the glass-water interface:
\[ \mu_g \sin C = \mu_w \sin \theta_1 \]
Where \( \mu_w \) is the refractive index of water (approximately \( \frac{4}{3} \)).

Substituting \( \mu_g \) and \( \sin C \):
\[ \frac{3}{2} \cdot \frac{2}{3} = \frac{4}{3} \sin \theta_1 \]
Simplifying the left side:
\[ 1 = \frac{4}{3} \sin \theta_1 \]
Thus:
\[ \sin \theta_1 = \frac{3}{4} \]

### Step 3: Ray emerging from the water-air interface
Now, we apply Snell's law at the water-air interface:
\[ \mu_w \sin \theta_1 = \mu_a \sin \theta_2 \]
Substituting the known values:
\[ \frac{4}{3} \cdot \frac{3}{4} = 1 \cdot \sin \theta_2 \]
This simplifies to:
\[ 1 = \sin \theta_2 \]
Thus:
\[ \theta_2 = \sin^{-1}(1) = 90^\circ \]

### Conclusion
The angle of emergence of the ray in air when it emerges from the water-air interface is \( 90^\circ \).

---