Question

If the critical angle for total internal reflection from any medium to vacuum is 30°, then the velocity of light in the medium is.

• A. \( 3 \times 10^8 \, \text{m/sec} \)
• B. \( 1.5 \times 10^8 \, \text{m/sec} \)
• C. \( 6 \times 10^8 \, \text{m/sec} \)
• D. \( 4.5 \times 10^8 \, \text{m/sec} \)

Hint:

Critical angle \( \theta_c \) and velocity of light in the medium are inversely related through refractive indices.

Answer 1

The Correct Option is B

Step 1: Formula for critical angle.
The critical angle (\( \theta_c \)) is related to the refractive indices of the two media by the formula: \[ \sin(\theta_c) = \frac{{n_2}}{{n_1}} \] where \( n_1 \) is the refractive index of the medium and \( n_2 \) is the refractive index of vacuum (\( n_2 = 1 \)). The refractive index \( n_1 \) is related to the velocity of light in the medium by: \[ n_1 = \frac{{c}}{{v}} \] where \( c = 3 \times 10^8 \, \text{m/sec} \) is the speed of light in vacuum and \( v \) is the velocity of light in the medium.
Step 2: Using given information.
We know the critical angle is \( 30^\circ \), so \[ \sin(30^\circ) = \frac{1}{2} = \frac{1}{\frac{c}{v}} \Rightarrow v = 1.5 \times 10^8 \, \text{m/sec} \] Conclusion: The velocity of light in the medium is \( 1.5 \times 10^8 \, \text{m/sec} \).