Wave Optics Part 2

Concept:When light travels from one medium to another, its frequency remains constant, but its wavelength and speed change according to the refractive index.Explanation:The relation between refractive index and wavelength in two media is: $\mu_1 \lambda_1 = \mu_2 \lambda_2$.Given: wavelength in water $\lambda_1 = 540$ nm, refractive index of water $\mu_1 = \frac{4}{3}$, and refractive index of the second medium $\mu_2 = \frac{3}{2}$.To find $\lambda_2$, rearrange: $\lambda_2 = \frac{\mu_1 \lambda_1}{\mu_2}$.Substitute: $\lambda_2 = \frac{\frac{4}{3} \times 540}{\frac{3}{2}} = \frac{4 \times 540 \times 2}{3 \times 3} = \frac{4320}{9} = 480$ nm.Answer:The wavelength of light in the second medium is $480$ nm, so the correct option is C.