Question

At the interface between two materials having refractive indices \( n_1 \) and \( n_2 \), the critical angle for reflection of an EM wave is \( \theta_c \). The \( n_1 \) material is replaced by another material having refractive index \( n_3 \), such that the critical angle at the interface between \( n_1 \) and \( n_3 \) materials is \( \theta_{c3} \). If \( n_1>n_2>n_3 \), \( \frac{n_2}{n_3} = \frac{2}{5} \), and \( \sin\theta_{c2} - \sin\theta_{c1} = \frac{1}{2} \), then \( \theta_{c1} \) is:

• A. \( \sin^{-1} \left( \frac{5}{6n_1} \right) \)
• B. \( \sin^{-1} \left( \frac{2}{3n_1} \right) \)
• C. \( \sin^{-1} \left( \frac{1}{3n_1} \right) \)
• D. \( \sin^{-1} \left( \frac{1}{6n_1} \right) \)

Hint:

To solve problems involving critical angles, use Snell's law and the relationships between the refractive indices and the angles of incidence and refraction.

Answer 1

The Correct Option is A

Using the relationship between critical angles and refractive indices, and applying Snell's law at the interface, we can derive the necessary expression for \( \theta_{c1} \). The given condition \( \sin\theta_{c2} - \sin\theta_{c1} = \frac{1}{2} \) helps in finding the value of \( \theta_{c1} \). Thus, the correct expression for \( \theta_{c1} \) is \( \sin^{-1} \left( \frac{5}{6n_1} \right) \).