Question

In vacuum, light takes time \( t \) to travel a distance \( d \) and it takes time \( T \) to travel a distance \( 5d \) in a denser medium. The critical angle of the given pair of media is

• A. \( \sin^{-1} \left( \frac{3t}{T} \right) \)
• B. \( \sin^{-1} \left( \frac{5t}{T} \right) \)
• C. \( \sin^{-1} \left( \frac{t}{T} \right) \)
• D. \( \sin^{-1} \left( \frac{2t}{T} \right) \)

Hint:

The critical angle depends on the ratio of the speeds of light in the two media, which is inversely related to the refractive indices of the media.

Answer 1

The Correct Option is B

Step 1: Use the relationship between speed, distance, and time.
The speed of light in vacuum is \( v_{\text{vacuum}} = \frac{d}{t} \), and the speed of light in the denser medium is \( v_{\text{medium}} = \frac{5d}{T} \). The critical angle \( \theta_c \) is related to the refractive indices of the two media by the equation: \[ \sin(\theta_c) = \frac{v_{\text{vacuum}}}{v_{\text{medium}}} \] Step 2: Substitute the values.
Substitute the expressions for the speeds into the equation: \[ \sin(\theta_c) = \frac{\frac{d}{t}}{\frac{5d}{T}} = \frac{T}{5t} \] Thus, the critical angle is: \[ \theta_c = \sin^{-1} \left( \frac{5t}{T} \right) \] Step 3: Conclusion.
Thus, the critical angle of the given pair of media is \( \sin^{-1} \left( \frac{5t}{T} \right) \), which corresponds to option (B).