Question

If velocity of light in air is \(3 \times 10^8\) m/s and that in water is \(2 \times 10^8\) m/s, then what would be the critical angle?

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

Hint:

The critical angle is the angle of incidence beyond which total internal reflection occurs. It can be calculated using the ratio of velocities in the two media.

Answer 1

The Correct Option is B

The critical angle \(\theta_c\) can be calculated as follows: \[\n\sin \theta_c = \frac{v_{\text{2}}}{v_{\text{1}}}\n\] where \(v_{\text{2}}\) represents the light velocity in water and \(v_{\text{1}}\) is the light velocity in air. By substituting the given values: \[\n\sin \theta_c = \frac{2 \times 10^8}{3 \times 10^8} = \frac{2}{3}\n\] Therefore, the critical angle is \(\sin^{-1}\left(\frac{2}{3}\right)\).