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A beam of light propagating at an angle $\alpha_1$ from a medium 1 through to another 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 $2,\left(\lambda_2\right)$, is
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Verified Answer
The correct answer is:
$\frac{\sin \alpha_2}{\sin \alpha_1} \lambda_1$
We know that,
$$
\lambda=\frac{\lambda_0}{\mu}
$$
According to the question,
$$
\begin{array}{r}
\frac{\sin \alpha_1}{\sin \alpha_2}=\frac{\mu_2}{\mu_1}=\frac{\lambda_1}{\lambda_2} \\
\lambda_2=\lambda_1 \frac{\sin \alpha_2}{\sin \alpha_1}
\end{array}
$$
$$
\lambda=\frac{\lambda_0}{\mu}
$$
According to the question,
$$
\begin{array}{r}
\frac{\sin \alpha_1}{\sin \alpha_2}=\frac{\mu_2}{\mu_1}=\frac{\lambda_1}{\lambda_2} \\
\lambda_2=\lambda_1 \frac{\sin \alpha_2}{\sin \alpha_1}
\end{array}
$$
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