Question

An optical fiber has a refractive index of 1.641 for the core and 1.422 for the cladding.
The critical angle above which a ray will be totally internally reflected is

• A. \( 60^\circ \)
• B. \( 41^\circ \)
• C. \( 45^\circ \)
• D. \( 37^\circ \)

Hint:

For total internal reflection, use \( \theta_c = \sin^{-1}(n_2/n_1) \) with \( n_1>n_2 \).

Answer 1

The Correct Option is A

Critical angle \( \theta_c \) is calculated as: \[ \theta_c = \sin^{-1}\left(\frac{n_2}{n_1}\right) = \sin^{-1}\left(\frac{1.422}{1.641}\right) = \sin^{-1}(0.8665) \approx 60^\circ \]