Question

A ray of light is incident at an angle of incidence \(60º\) on the glass slab of refractive index \(\sqrt 3.\) After refraction, the light ray emerges out from other parallel faces and lateral shift between incident ray and emergent ray is \(4\sqrt 3.\) cm. The thickness of the glass slab is ______ cm.

Answer 1

Correct Answer: 12

The phenomenon described involves refraction through a parallel-sided glass slab. We will use Snell's Law and the geometry of the situation to find the thickness \( t \) of the slab. The angle of incidence is \( 60^\circ \) and the refractive index \( n = \sqrt{3} \).

1. **Snell's Law**: \( n_1 \sin \theta_1 = n_2 \sin \theta_2 \), where \( n_1 = 1 \) (refractive index of air), \( \theta_1 = 60^\circ \) and \( n_2 = \sqrt{3} \).
\[ \sin 60^\circ = \sqrt{3}/2 \]
\[ \sin \theta_2 = \frac{\sin 60^\circ}{\sqrt{3}} = \frac{\sqrt{3}/2}{\sqrt{3}} = 1/2 \]
The angle of refraction \( \theta_2 \) is \( 30^\circ \).

2. **Lateral Shift Formula**: \( \Delta = \frac{t \sin(\theta_1 - \theta_2)}{\cos \theta_2} \).
Given \( \Delta = 4\sqrt{3} \) cm.

3. **Substitution**:
\[ \Delta = \frac{t \cdot \sin(60^\circ - 30^\circ)}{\cos 30^\circ} \]
\[ \Delta = \frac{t \sin 30^\circ}{\sqrt{3}/2} \]
\[ \Delta = \frac{t \cdot \frac{1}{2}}{\sqrt{3}/2} = \frac{t}{\sqrt{3}} \]
Equating and solving for \( t \):
\[ 4\sqrt{3} = \frac{t}{\sqrt{3}} \]
\[ t = 4\sqrt{3} \times \sqrt{3} = 12 \text{ cm} \]

Thus, the thickness of the glass slab is 12 cm, which falls within the specified range of 12,12 cm.