Light of wavelength 500 nm in air has a wavelength of 375 nm in a liquid. A plane wave of this light falls from air on the surface of the liquid at an angle of incidence of \(30^\circ\). The angle of refraction is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\sin^{-1}\left(\frac{3}{8}\right)\)
The frequency does not change on refraction. So the ratio of the wavelengths in the two media equals the ratio of the speeds. The same ratio of speeds gives \( \sin i/\sin r \).
\( \begin{aligned} \frac{\sin i}{\sin r} &= \frac{v_1}{v_2} \\ &= \frac{\lambda_1}{\lambda_2} \\ &= \frac{500}{375} \\ &= \frac{4}{3} \end{aligned} \)
\( \begin{aligned} \sin r &= \frac{3}{4}\sin i \\ &= \frac{3}{4}\times\frac{1}{2} \\ &= \frac{3}{8} \end{aligned} \)
Hence, the angle of refraction is \( \sin^{-1}(3/8) \). It is less than \(30^\circ\), so the wave bends towards the normal.
NCERT (2026-27):
Refraction of a plane wave – “… where i and r are the angles of incidence and refraction, respectively. Thus we obtain”
\( \displaystyle \frac{\sin i}{\sin r}=\frac{v_1}{v_2} \)\( \qquad (10.3) \)
Refraction of a plane wave – “… Further, if λ1 and λ2 denote the wavelengths of light in medium 1 and medium 2, respectively and if the distance BC is equal to λ1 then the distance AE will be equal to λ2 … thus,”
\( \displaystyle \frac{\lambda_1}{\lambda_2}=\frac{BC}{AE}=\frac{v_1}{v_2}\quad \text{or}\quad \frac{v_1}{\lambda_1}=\frac{v_2}{\lambda_2} \)\( \qquad (10.7) \)