If the value of the refractive index of a denser medium with respect to the rarer medium is 1.414, then the critical angle is:

45°

The correct answer is Option (2) → 45°

Given:

Refractive index $n = 1.414$ (denser to rarer medium)

Critical angle $\theta_c$ is given by:

$\sin \theta_c = \frac{n_{\text{rarer}}}{n_{\text{denser}}} = \frac{1}{n}$

$\sin \theta_c = \frac{1}{1.414} = 0.707$

Therefore, $\theta_c = \arcsin(0.707) \approx 45^\circ$

Answer: Critical angle $\theta_c = 45^\circ$