A ray of light enters a rectangular glass slab of refractive index `sqrt(3)` at an angle of incidence `60^(@)`. It travels a distance of 5cm inside the slab and emerges out of the slab. The perpendicular distance between the incident and the emergent rays is

Options

A. `5sqrt(3)cm`
B. `(5)/(2)cm`
C. `5sqrt(3//2)cm`
D. 5cm

Explanation

To solve the problem, we need to determine the perpendicular distance between the incident ray and the emergent ray after the ray of light passes through a rectangular glass slab. Here’s a step-by-step solution:

### Step 1: Identify the given values
- Refractive index of the glass slab, \( n = \sqrt{3} \)
- Angle of incidence, \( i = 60^\circ \)
- Distance traveled inside the slab, \( d = 5 \, \text{cm} \)

### Step 2: Apply Snell's Law at the air-glass interface
Using Snell's Law:
\[ n_1 \sin(i) = n_2 \sin(r) \]
where \( n_1 = 1 \) (for air), \( n_2 = \sqrt{3} \), \( i = 60^\circ \), and \( r \) is the angle of refraction in the glass.

Substituting the values:
\[ 1 \cdot \sin(60^\circ) = \sqrt{3} \cdot \sin(r) \]
\[ \sin(60^\circ) = \frac{\sqrt{3}}{2} \]
So,
\[ \frac{\sqrt{3}}{2} = \sqrt{3} \cdot \sin(r) \]
Dividing both sides by \( \sqrt{3} \):
\[ \sin(r) = \frac{1}{2} \]
Thus, \( r = 30^\circ \).

### Step 3: Calculate the angle of emergence
When the ray exits the slab, it will again follow Snell's Law. The angle of incidence at the second interface (glass-air) will be equal to the angle of refraction \( r \) we just calculated, which is \( 30^\circ \).

Using Snell's Law again:
\[ \sqrt{3} \cdot \sin(30^\circ) = 1 \cdot \sin(e) \]
Since \( \sin(30^\circ) = \frac{1}{2} \):
\[ \sqrt{3} \cdot \frac{1}{2} = \sin(e) \]
\[ \sin(e) = \frac{\sqrt{3}}{2} \]
Thus, \( e = 60^\circ \).

### Step 4: Determine the perpendicular distance
The total angle of deviation from the normal as the ray enters and exits the slab is:
\[ \text{Total deviation} = i + e - 180^\circ = 60^\circ + 60^\circ - 180^\circ = -60^\circ \]
However, we are interested in the perpendicular distance \( d \) between the incident and emergent rays.

Using the geometry of the situation, we can relate the perpendicular distance \( D \) to the angles:
\[ D = d \cdot \sin(i - r) \]
Substituting \( d = 5 \, \text{cm} \), \( i = 60^\circ \), and \( r = 30^\circ \):
\[ D = 5 \cdot \sin(60^\circ - 30^\circ) \]
\[ D = 5 \cdot \sin(30^\circ) \]
\[ D = 5 \cdot \frac{1}{2} = 2.5 \, \text{cm} \]

### Final Answer
The perpendicular distance between the incident and emergent rays is \( 2.5 \, \text{cm} \).

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