A ray of light is travelling from glass to ai | vedclass

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(B) The refractive index of glass with respect to air is $\mu = 1.5$.The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{1.5} = \fra

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$80^{\circ}$

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(B) The refractive index of glass with respect to air is $\mu = 1.5$.The critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667$.$C = \sin^{-1}(0.667) \approx 41.8^{\circ} \approx 42^{\circ}$.Since the angle of incidence $i = 50^{\circ}$ is greater than the critical angle $C = 42^{\circ}$,the light ray undergoes total internal reflection.In total internal reflection,the angle of reflection $r$ is equal to the angle of incidence $i$,so $r = 50^{\circ}$.The deviation $\delta$ is the angle between the original path of the incident ray and the reflected ray.From the geometry of the reflection,the deviation is $\delta = 180^{\circ} - (i + r) = 180^{\circ} - (50^{\circ} + 50^{\circ}) = 180^{\circ} - 100^{\circ} = 80^{\circ}$.

$A$ ray of light is travelling from glass to air. (refractive index of glass $= 1.5$). The angle of incidence is $50^{\circ}$. The deviation of the ray is

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