Question #118519
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)
1
Expert's answer
2020-05-27T10:32:20-0400

According to Snell's Law, sinθ1sinθ2=n2n1=λ1λ2\frac{sin \theta_1}{\sin \theta_2} = \frac{n_2}{n_1} = \frac{\lambda_1}{\lambda_2}, where n is a refractive index and λ\lambda is the wavelength. When passing from medium A into B, wavelength changes from λ1\lambda_1 to (1η)λ1(1-\eta)\lambda_1, therefore n2n1=λ1(1η)λ1=11η\frac{n_2}{n_1}= \frac{\lambda_1}{(1-\eta)\lambda_1} = \frac{1}{1-\eta}. Substituting η=0.22\eta = 0.22, obtain n2n11.282\frac{n_2}{n_1} \approx 1.282.


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