Question

If a ray of light takes \(t_1\) and \(t_2\) times in two media of absolute refractive indices \(\mu_1\) and \(\mu_2\) respectively to travel same distance, then the relation between the times and refractive indices is:

• A. \(\mu_1 t_1 = \mu_2 t_2\)
• B. \(\mu_1 t_2 = \mu_2 t_1\)
• C. \(t_1 \sqrt{\mu_1} = t_2 \sqrt{\mu_2}\)
• D. \(\sqrt{\mu_1} t_1 = \sqrt{\mu_2} t_2\)

Hint:

For light traveling the same distance in different media, the product of the refractive index and the time taken is a constant, illustrating the inverse relationship between speed and refractive index.

Answer 1

The Correct Option is B

Step 1: Relate the times to the velocities in the media. \[ v = \frac{c}{\mu} \] \[ t = \frac{d}{v} = \frac{d \mu}{c} \] where \(d\) is the distance, \(c\) is the speed of light, and \(v\) is the velocity of light in the medium. 
Step 2: Derive the relationship using the indices and times. Since \( v_1 = \frac{c}{\mu_1} \) and \( v_2 = \frac{c}{\mu_2} \), \[ t_1 = \frac{d}{v_1} = \frac{d \mu_1}{c} \] \[ t_2 = \frac{d}{v_2} = \frac{d \mu_2}{c} \] Equating and rearranging gives: \[ \mu_1 t_2 = \mu_2 t_1 \]