Question

A ray of light travelling in air is incident on one face of a parallel glass slab of thickness \( t \) and refractive index \( \mu \) at an angle of incidence \( i \). Total time spent by the ray inside the slab is

• A. \( \frac{\mu^2 t}{c\sqrt{1 - \mu^2 \sin^2 i}} \)
• B. \( \frac{\mu t}{c\sqrt{\mu^2 - \sin^2 i}} \)
• C. \( \frac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}} \)
• D. \( \frac{t}{c\sqrt{\mu^2 - \sin^2 i}} \)

Hint:

To easily check the dimensional consistency of the options, notice that \( T \) must have the unit of time (\( \frac{\text{length}}{\text{velocity}} = \frac{t}{c} \)). Since \( \mu \) is a dimensionless refractive index, any expression where \( \frac{t}{c} \) is multiplied by dimensionless functions of \( \mu \) and \( \sin i \) is dimensionally correct. Thus, analyzing the boundary condition at normal incidence (\( i = 0 \)), the time must reduce to \( T = \frac{\mu t}{c} \). Substituting \( i = 0 \) into option (C) gives \( \frac{\mu^2 t}{c\sqrt{\mu^2}} = \frac{\mu t}{c} \), which confirms its correctness.

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
A light ray enters a parallel glass slab of thickness \( t \) and refractive index \( \mu \). We need to determine the total time \( T \) the ray spends traveling inside the slab in terms of the angle of incidence \( i \), the slab thickness \( t \), and the speed of light in vacuum \( c \).

Step 2: Key Formula or Approach:
1. Snell's Law:
\[ 1 \cdot \sin i = \mu \cdot \sin r \implies \sin r = \frac{\sin i}{\mu} \]
2. Geometry of the path: The path length \( d \) traveled by the ray inside the slab is related to the thickness \( t \) by:
\[ d = \frac{t}{\cos r} \]
3. Velocity of light in the medium:
\[ v = \frac{c}{\mu} \]
4. Time calculation:
\[ T = \frac{\text{distance}}{\text{velocity}} = \frac{d}{v} \]

Step 3: Detailed Explanation:
Substitute the expressions for \( d \) and \( v \) into the time equation:
\[ T = \frac{t / \cos r}{c / \mu} = \frac{\mu t}{c \cos r} \]
Using the trigonometric identity \( \cos r = \sqrt{1 - \sin^2 r} \):
\[ \cos r = \sqrt{1 - \left(\frac{\sin i}{\mu}\right)^2} = \sqrt{1 - \frac{\sin^2 i}{\mu^2}} = \frac{\sqrt{\mu^2 - \sin^2 i}}{\mu} \]
Substitute \( \cos r \) back into the expression for \( T \):
\[ T = \frac{\mu t}{c \left(\frac{\sqrt{\mu^2 - \sin^2 i}}{\mu}\right)} = \frac{\mu^2 t}{c \sqrt{\mu^2 - \sin^2 i}} \]

Step 4: Final Answer:
The total time spent by the ray inside the slab is \( \frac{\mu^2 t}{c\sqrt{\mu^2 - \sin^2 i}} \).