Glass has a refractive index mu with respect | vedclass

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(C) Given that the refractive index of glass with respect to air is $\mu$. The critical angle $	heta$ is defined by the relation $\sin 	heta = \frac

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

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(C) Given that the refractive index of glass with respect to air is $\mu$. The critical angle $	heta$ is defined by the relation $\sin 	heta = \frac{1}{\mu}$,which implies $\mu = \frac{1}{\sin 	heta}$ .....$(i)$.Now,consider the ray incident from air to glass. According to Snell's law,$\mu = \frac{\sin i}{\sin r}$.Here,the angle of incidence $i = 	heta$,so $\mu = \frac{\sin 	heta}{\sin r}$.Rearranging for $\sin r$,we get $\sin r = \frac{\sin 	heta}{\mu}$ .....$(ii)$.Substituting the value of $\mu$ from equation $(i)$ into equation $(ii)$:$\sin r = \frac{\sin 	heta}{(1/\sin 	heta)} = \sin^2 	heta$.Since $\sin 	heta = \frac{1}{\mu}$,then $\sin r = \left(\frac{1}{\mu}ight)^2 = \frac{1}{\mu^2}$.Therefore,the angle of refraction is $r = \sin^{-1}\left(\frac{1}{\mu^2}ight)$.

Glass has a refractive index $\mu$ with respect to air,and the critical angle for a ray of light going from glass to air is $\theta$. If a ray of light is incident from air on the glass with an angle of incidence $\theta$,the corresponding angle of refraction is:

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