A glass prism of refractive index 1.5 is imme | vedclass

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(A) For total internal reflection to occur at the interface between the glass prism and water,the angle of incidence $	heta$ must be greater than or

Vedclass · Accepted Answer

$\sin 	heta \geqslant \frac{8}{9}$

Vedclass · Answer

(A) For total internal reflection to occur at the interface between the glass prism and water,the angle of incidence $	heta$ must be greater than or equal to the critical angle $C$.The condition for total internal reflection is $\sin 	heta \geqslant \sin C$.The critical angle $C$ is given by $\sin C = \frac{1}{{}_w\mu_g}$,where ${}_w\mu_g$ is the refractive index of glass with respect to water.Given,refractive index of glass $\mu_g = 1.5 = 3/2$ and refractive index of water $\mu_w = 4/3$.Therefore,${}_w\mu_g = \frac{\mu_g}{\mu_w} = \frac{3/2}{4/3} = \frac{9}{8}$.Thus,$\sin C = \frac{1}{9/8} = \frac{8}{9}$.For total internal reflection,the angle of incidence $	heta$ must satisfy $\sin 	heta \geqslant \frac{8}{9}$.

$A$ glass prism of refractive index $1.5$ is immersed in water (refractive index $4/3$). $A$ light beam incident normally on the face $AB$ is totally reflected to reach the face $BC$ if

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