The refractive index of a material of a prism | vedclass

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Question

(A) The critical angle $i_c$ for the prism material is given by $\sin(i_c) = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667$.Since $\sin(45

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(A) The critical angle $i_c$ for the prism material is given by $\sin(i_c) = \frac{1}{\mu} = \frac{1}{1.5} = \frac{2}{3} \approx 0.667$.Since $\sin(45^o) = \frac{1}{\sqrt{2}} \approx 0.707$,we have $\sin(45^o) > \sin(i_c)$,which implies $45^o > i_c$.When a light ray is incident normally on the hypotenuse $BC$,it enters the prism without deviation and strikes the face $AB$ at an angle of incidence of $45^o$.Since the angle of incidence $45^o$ is greater than the critical angle $i_c$,the ray undergoes Total Internal Reflection $(TIR)$ at face $AB$.After reflection,the ray travels parallel to the base $BC$ and strikes the face $AC$ at an angle of incidence of $45^o$.Again,since $45^o > i_c$,the ray undergoes $TIR$ at face $AC$ and emerges normally from the hypotenuse $BC$.This path corresponds to the diagram shown in option $A$.

The refractive index of a material of a prism of angles $45^o -45^o -90^o$ is $1.5$. The path of the ray of light incident normally on the hypotenuse side is shown in

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