Answer to Question #118519 in Atomic and Nuclear Physics for H

Question

Light passes from medium A into medium B.
As it does so the wavelength of the light is reduced by 22.0%.
What is the ratio of the refractive indices nB/nA? (to 3 s.f)

Answer 1

According to Snell's Law, $\frac{sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2}$, where n is a refractive index and $\lambda$ is the wavelength. When passing from medium A into B, wavelength changes from $\lambda_1$ to $(1-\eta)\lambda_1$, therefore $\frac{n_2}{n_1}= \frac{\lambda_1}{(1-\eta)\lambda_1} = \frac{1}{1-\eta}$. Substituting $\eta = 0.22$, obtain $\frac{n_2}{n_1} \approx 1.282$.