Question

The speed of light in two media '1' and '2' are \( v_1 \) and \( v_2 \) (\( v_2>v_1 \)) respectively. For a ray of light to undergo total internal reflection at the interface of these two media, it must be incident from:

• A. medium '1' and at an angle greater than \( \sin^{-1} \left( \frac{v_1}{v_2} \right) \)
• B. medium '1' and at an angle greater than \( \cos^{-1} \left( \frac{v_1}{v_2} \right) \)
• C. medium '2' and at an angle greater than \( \sin^{-1} \left( \frac{v_1}{v_2} \right) \)
• D. medium '2' and at an angle greater than \( \cos^{-1} \left( \frac{v_1}{v_2} \right) \)

Hint:

For total internal reflection to occur, the angle of incidence must be greater than the critical angle, which is \( \sin^{-1} \left( \frac{v_1}{v_2} \right) \) when light is moving from the slower medium.

Answer 1

The Correct Option is A

For total internal reflection to occur at the interface between two media, the angle of incidence must be greater than the critical angle \( \theta_c \), where the critical angle is given by: \[ \sin \theta_c = \frac{v_1}{v_2} \] Thus, the angle of incidence must be greater than \( \theta_c \), or equivalently, greater than \( \sin^{-1} \left( \frac{v_1}{v_2} \right) \), for total internal reflection to occur. This condition applies when light is incident from the medium with the lower speed of light (i.e., medium '1'). Therefore, the correct option is (A).