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

To establish the conditions for total internal reflection at the boundary of two media, with light speeds \( v_1 \) in medium '1' and \( v_2 \) in medium '2' where \( v_2>v_1 \), we use Snell's Law: \[n_1 \sin \theta_1 = n_2 \sin \theta_2\]. Here, \( n_1 \) and \( n_2 \) represent the refractive indices of media '1' and '2', respectively. The refractive index is inversely related to the speed of light in a medium: \[ n = \frac{c}{v} \], with \( c \) being the speed of light in a vacuum. Consequently, \( n_1=\frac{c}{v_1} \) and \( n_2=\frac{c}{v_2} \). Total internal reflection occurs when light travels from a denser medium to a lighter one (from medium '1' to medium '2', where \( v_2>v_1 \), implying \( n_2<n_1 \)). For this phenomenon, the angle of refraction \(\theta_2\) must be 90°, leading to: \[ n_1 \sin \theta_c = n_2 \sin 90^\circ \], where \(\theta_c\) is the critical angle. This simplifies to: \[\sin \theta_c = \frac{n_2}{n_1}\]. Substituting the refractive indices with their speed dependencies: \[\sin \theta_c = \frac{v_1}{v_2}\]. To achieve total internal reflection, the angle of incidence \(\theta_1\) must exceed this critical angle: \[\theta_1 > \sin^{-1} \left( \frac{v_1}{v_2} \right)\]. Therefore, total internal reflection requires the incident ray from medium '1' to strike the interface at an angle greater than \(\sin^{-1} \left( \frac{v_1}{v_2} \right)\).