A beam of light has a wavelength of 480 nm in liquid A and 400 nm in liquid B. The refractive index of liquid B with respect to liquid A is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 1.2
The frequency of the light is the same in both liquids. So the wavelength is proportional to the speed, while the refractive index is inversely proportional to it.
\( \begin{aligned} n_{BA} &= \frac{n_B}{n_A} \\ &= \frac{v_A}{v_B} \\ &= \frac{\lambda_A}{\lambda_B} \\ &= \frac{480}{400} \\ &= 1.2 \end{aligned} \)
Hence, the refractive index of liquid B with respect to liquid A is 1.2.
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. …”