(A) 540
(B) 840
(C) 480
(D) 380
The frequency of light remains the same when it travels from one medium to another.
The relation between wavelength and refractive index is
$$ \lambda \propto \frac{1}{n} $$
or
$$ n \lambda = \text{constant} $$
For light travelling in water,
$$ n_1 \lambda_1 = \frac{4}{3} \times 540 $$
For light travelling in the second medium,
$$ n_2 \lambda_2 = \frac{3}{2} \times \lambda_2 $$
Since frequency is constant,
$$ n_1 \lambda_1 = n_2 \lambda_2 $$
$$ \frac{4}{3} \times 540 = \frac{3}{2} \times \lambda_2 $$
$$ \lambda_2 = \frac{\frac{4}{3} \times 540 \times 2}{3} $$
$$ \lambda_2 = 480\ \text{nm} $$
Hence, the wavelength of light in the second medium is
480 nm
Updated for JEE Main 2026: This PYQ is important for JEE Mains, JEE Advanced and other competitive exams. Practice more questions from this chapter.