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Question

(B) The refractive index of glass with respect to air is given as $\mu$.The critical angle for a ray of light traveling from glass to air is $	heta$.

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The refractive index of glass with respect to air is given as $\mu$.The critical angle for a ray of light traveling from glass to air is $	heta$.According to the definition of critical angle,$\mu = \frac{1}{\sin 	heta}$,which implies $\sin 	heta = \frac{1}{\mu}$.Now,consider a ray of light incident from air onto the glass surface with an angle of incidence $i = 	heta$.Let $r$ be the corresponding angle of refraction in the glass.Applying Snell's Law at the interface: $n_1 \sin i = n_2 \sin r$.Here,$n_1 = 1$ (air) and $n_2 = \mu$ (glass).$1 \cdot \sin 	heta = \mu \cdot \sin r$.Substituting $\sin 	heta = \frac{1}{\mu}$ into the equation:$\frac{1}{\mu} = \mu \cdot \sin r$.$\sin r = \frac{1}{\mu^2}$.Therefore,the angle of refraction is $r = \sin^{-1}(\frac{1}{\mu^2})$.

Glass has 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 angle of incidence $\theta$,the corresponding angle of refraction is:

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