Refractive index of a rectangular glass slab is `mu = sqrt(3)`. Alight ray incident at an angle 60° is displaced laterally through 2.5 cm. Distance travelled by light in the slab is

Options

A. 4 cm/s towards the mirror
B. 5cm
C. `2.5sqrt(3)` cm
D. `3cm`

Explanation

To solve the problem, we need to find the distance traveled by light in a rectangular glass slab given the refractive index and the lateral displacement of the light ray. Let's break down the solution step by step.

### Step 1: Understand the Given Information
- Refractive index of the glass slab, \( \mu = \sqrt{3} \)
- Angle of incidence, \( \theta_i = 60^\circ \)
- Lateral displacement, \( d = 2.5 \, \text{cm} \)

### Step 2: Apply Snell's Law
Using Snell's Law, which states that \( \mu_1 \sin \theta_1 = \mu_2 \sin \theta_2 \), we can find the angle of refraction \( \theta_r \) in the glass slab.

Here, \( \mu_1 = 1 \) (for air), \( \theta_1 = 60^\circ \), and \( \mu_2 = \sqrt{3} \).

\[ \sin \theta_r = \frac{\mu_1}{\mu_2} \sin \theta_i = \frac{1}{\sqrt{3}} \sin 60^\circ \]
\[ \sin 60^\circ = \frac{\sqrt{3}}{2} \]
\[ \sin \theta_r = \frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{2} = \frac{1}{2} \]

### Step 3: Determine the Angle of Refraction
From the above calculation, we find:
\[ \sin \theta_r = \frac{1}{2} \implies \theta_r = 30^\circ \]

### Step 4: Analyze the Geometry of the Problem
The light ray travels through the slab and is laterally displaced. We can visualize this situation as a right triangle formed by the path of the light ray.

- The angle of refraction \( \theta_r = 30^\circ \)
- The lateral displacement \( d = 2.5 \, \text{cm} \)

### Step 5: Use Trigonometric Relationships
In the right triangle formed, we can use the sine function to relate the lateral displacement and the distance traveled in the slab.

Let \( L \) be the distance traveled by light in the slab (the hypotenuse of the triangle).

Using the sine of the angle:
\[ \sin \theta_r = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{d}{L} \]
\[ \sin 30^\circ = \frac{2.5}{L} \]
Since \( \sin 30^\circ = \frac{1}{2} \):
\[ \frac{1}{2} = \frac{2.5}{L} \]

### Step 6: Solve for \( L \)
Rearranging the equation gives:
\[ L = 2.5 \times 2 = 5 \, \text{cm} \]

### Final Answer
The distance traveled by light in the slab is \( L = 5 \, \text{cm} \).

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