Light passes from air into water of refractive index 4/3. Its wavelength decreases by 120 nm. The wavelength of this light in air is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 480 nm
The frequency of light does not change when it enters water. Its speed becomes \( n \) times smaller, so its wavelength also becomes \( n \) times smaller.
\( n = \frac{4}{3} \)
\( \lambda_a - \lambda_w = 120\ \mathrm{nm} \)
\( \begin{aligned} \lambda_w &= \frac{\lambda_a}{n} \\ &= \frac{3}{4}\,\lambda_a \end{aligned} \)
\( \begin{aligned} \lambda_a - \lambda_w &= \lambda_a - \frac{3}{4}\,\lambda_a \\ &= \frac{\lambda_a}{4} \end{aligned} \)
\( \begin{aligned} \frac{\lambda_a}{4} &= 120\ \mathrm{nm} \\ \lambda_a &= 480\ \mathrm{nm} \end{aligned} \)
Hence, the wavelength of the light in air is 480 nm. In water it is 360 nm.
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 of a plane wave – “… Now, if c represents the speed of light in vacuum, then,”
\( \displaystyle n_1=\frac{c}{v_1} \)\( \qquad (10.4) \)
“and”
\( \displaystyle n_2=\frac{c}{v_2} \)\( \qquad (10.5) \)
“are known as the refractive indices of medium 1 and medium 2, respectively. …”