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(A) According to Snell's Law, the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the

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$\frac{\sin \alpha_2}{\sin \alpha_1} \lambda_1$

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(A) According to Snell's Law, the ratio of the sines of the angles of incidence and refraction is equal to the ratio of the refractive indices of the two media:$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{\mu_2}{\mu_1}$We also know that the refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ in the medium, given by $\mu = \frac{c}{v} = \frac{\lambda_0}{\lambda}$, where $\lambda_0$ is the wavelength in vacuum.Therefore, $\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$.Equating the two expressions, we get:$\frac{\sin \alpha_1}{\sin \alpha_2} = \frac{\lambda_1}{\lambda_2}$Rearranging for $\lambda_2$, we find:$\lambda_2 = \lambda_1 \frac{\sin \alpha_2}{\sin \alpha_1}$

$A$ beam of light is propagating at an angle $\alpha_1$ from medium $1$ into medium $2$ at an angle $\alpha_2$. If the wavelength of light in medium $1$ is $\lambda_1$, then the wavelength of light in medium $2$, $\lambda_2$, is:

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