A monochromatic light has a wavelength of 600 nm in air, 400 nm in glass and 450 nm in water. For this light going from glass into water, the sine of the critical angle is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\frac{8}{9}\)
The frequency is the same in every medium, so the wavelength is proportional to the speed. Since \( n=c/v \), the ratio of refractive indices equals the inverse ratio of the wavelengths. Here glass is medium 1 and water is medium 2.
\( \begin{aligned} \sin i_c &= \frac{n_2}{n_1} \\ &= \frac{v_1}{v_2} \\ &= \frac{\lambda_1}{\lambda_2} \\ &= \frac{400}{450} \\ &= \frac{8}{9} \end{aligned} \)
Hence, the sine of the critical angle for the glass–water surface is \( 8/9 \). The wavelength in air is not needed here.
NCERT (2026-27):
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,”
\( \displaystyle \frac{\lambda_1}{\lambda_2}=\frac{BC}{AE}=\frac{v_1}{v_2}\quad \text{or}\quad \frac{v_1}{\lambda_1}=\frac{v_2}{\lambda_2} \)\( \qquad (10.7) \)
Refraction at a rarer medium – “… We define an angle ic by the following equation”
\( \displaystyle \sin i_c=\frac{n_2}{n_1} \)\( \qquad (10.8) \)