Calculate the speed of light in a medium in which critical angle is 45°.

$2.12 × 10^8 m/s$

The correct answer is Option (1) → $2.12 × 10^8 m/s$

Refractive index of medium w.r.t. air:

$\mu = \frac{1}{\sin C} = \frac{1}{\sin 45^\circ} = \frac{1}{\frac{1}{\sqrt{2}}} = \sqrt{2}$

Speed of light in medium:

$v = \frac{c}{\mu} = \frac{3 \times 10^8}{\sqrt{2}}$

$v = 2.12 \times 10^8 \, \text{m/s}$