Calculate the speed of light in a medium in which critical angle is \( 45^\circ \).
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 2.12 \times 10^{8}\,\mathrm{m/s} \)
At the critical angle \( C \) the refracted ray grazes the surface, so \( \mu\sin C = 1\cdot\sin 90^\circ \). The speed in the medium follows from \( \mu = c/v \).
\( \displaystyle \mu = \frac{1}{\sin C} = \frac{1}{\sin 45^\circ} = \sqrt{2} \)
\( \displaystyle v = \frac{c}{\mu} = \frac{3\times10^{8}}{\sqrt{2}} = 2.12\times10^{8}\,\mathrm{m\,s^{-1}} \)