The wavelengths of light in two liquids x and | vedclass

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(C) The refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ of light in a medium,given by $\mu = \frac{c}{v} = \frac{\lambda_

Vedclass · Accepted Answer

$30$

Vedclass · Answer

(C) The refractive index $\mu$ is inversely proportional to the wavelength $\lambda$ of light in a medium,given by $\mu = \frac{c}{v} = \frac{\lambda_0}{\lambda}$.For a critical angle $C$ between two media with refractive indices $\mu_1$ and $\mu_2$ (where $\mu_1 > \mu_2$),the condition is $\sin C = \frac{\mu_2}{\mu_1}$.Since $\mu \propto \frac{1}{\lambda}$,we have $\frac{\mu_2}{\mu_1} = \frac{\lambda_1}{\lambda_2}$.Given $\lambda_x = 3500 \, \mathring{A}$ and $\lambda_y = 7000 \, \mathring{A}$.Here,$\mu_x > \mu_y$ because $\lambda_x $\sin C = \frac{\lambda_x}{\lambda_y} = \frac{3500}{7000} = \frac{1}{2}$.Therefore,$C = \arcsin(0.5) = 30^o$.

The wavelengths of light in two liquids $x$ and $y$ are $3500 \, \mathring{A}$ and $7000 \, \mathring{A}$ respectively. The critical angle of $x$ with respect to $y$ is .......... $^o$.

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