A plane wave in air falls on the surface of a transparent medium at an angle of incidence of \(45^\circ\). The angle of refraction is \(30^\circ\). The speed of light in the medium is (speed of light in air \(=3\times10^{8}\ \mathrm{m\,s^{-1}}\))
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → \(2.12\times10^{8}\ \mathrm{m\,s^{-1}}\)
In the Huygens construction, the incident and refracted wavefronts cover \( v_1\tau \) and \( v_2\tau \) in the same time. This gives the ratio of the sines of the angles as the ratio of the speeds.
\( \begin{aligned} \frac{\sin i}{\sin r} &= \frac{v_1}{v_2} \end{aligned} \)
\( \begin{aligned} v_2 &= v_1\,\frac{\sin r}{\sin i} \\ &= 3\times10^{8}\times\frac{1/2}{1/\sqrt{2}} \\ &= \frac{3\times10^{8}}{\sqrt{2}} \\ &= 2.12\times10^{8}\ \mathrm{m\,s^{-1}} \end{aligned} \)
Hence, the speed of light in the medium is \( 2.12\times10^{8} \) m s−1.
NCERT (2026-27):
Refraction of a plane wave – “… where i and r are the angles of incidence and refraction, respectively. Thus we obtain”
\( \displaystyle \frac{\sin i}{\sin r}=\frac{v_1}{v_2} \)\( \qquad (10.3) \)