When a ray is refracted from one medium to an | vedclass

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(C) The refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ in a medium,given by $\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lam

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$\sin^{-1} \left( \frac{2}{3} \right)$

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(C) The refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ in a medium,given by $\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$.For total internal reflection,the critical angle $C$ is defined by the relation $\sin C = \frac{\mu_2}{\mu_1}$,where $\mu_1$ is the denser medium and $\mu_2$ is the rarer medium.Given $\lambda_1 = 6000 \, \mathring A$ and $\lambda_2 = 4000 \, \mathring A$.Since $\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$,we have $\sin C = \frac{4000}{6000} = \frac{2}{3}$.Therefore,the critical angle $C = \sin^{-1} \left( \frac{2}{3} \right)$.

When a ray is refracted from one medium to another,the wavelength changes from $6000 \, \mathring A$ to $4000 \, \mathring A$. The critical angle for the interface will be

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