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(B) Initially,the ray strikes the $AB$ interface at the critical angle $	heta_c$ with air. By Snell's law,$\mu \sin 	heta_c = 1 \cdot \sin 90^\circ

Vedclass · Accepted Answer

Lesser than $\mu$

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(B) Initially,the ray strikes the $AB$ interface at the critical angle $	heta_c$ with air. By Snell's law,$\mu \sin 	heta_c = 1 \cdot \sin 90^\circ = 1$.When the slab of refractive index $\mu_1$ is placed,the ray passes through the interface $AB$ and then strikes the interface $CD$ at an angle of incidence $r_1$. For total internal reflection to occur at $CD$,the condition is $\mu_1 \sin r_1 = 1 \cdot \sin 90^\circ = 1$.From the geometry of the slab,the angle of incidence at $CD$ $(r_1)$ is related to the angle of refraction at $AB$ $(r)$. However,since the ray is incident at the critical angle $	heta_c$ at $AB$,it will refract into the medium $\mu_1$ at an angle $r_1$ such that $\mu \sin 	heta_c = \mu_1 \sin r_1$.Since $\mu \sin 	heta_c = 1$ and we require $\mu_1 \sin r_1 = 1$ for total internal reflection at $CD$,we must have $\mu_1 \sin r_1 = \mu \sin 	heta_c = 1$.For total internal reflection to occur at $CD$,the angle of incidence $r_1$ must be greater than or equal to the critical angle for the $\mu_1$-air interface. Since the ray is already at the critical angle relative to air,and $\mu_1$ must be less than $\mu$ to allow the ray to enter and then undergo $TIR$ at a shallower angle,the condition is satisfied when $\mu_1 < \mu$.

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