A prism has a refractive index of 1.5. If a l | vedclass

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(B) From the geometry of the prism,the angle of incidence $	heta$ at the interface $PQ$ is $60^{\circ}$.For total internal reflection to occur,the an

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$1.3$

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(B) From the geometry of the prism,the angle of incidence $	heta$ at the interface $PQ$ is $60^{\circ}$.For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$,i.e.,$	heta \geq C$.Since $	heta = 60^{\circ}$,we have $60^{\circ} \geq C$.Taking the sine on both sides,$\sin 60^{\circ} \geq \sin C$.We know that $\sin C = \frac{\mu_{liquid}}{\mu_{prism}}$.Substituting the values,$\frac{\sqrt{3}}{2} \geq \frac{\mu_{liquid}}{1.5}$.$\mu_{liquid} \leq 1.5 	imes \frac{\sqrt{3}}{2} \approx 1.5 	imes 0.866 = 1.299$.Thus,the refractive index of the liquid must be less than or equal to approximately $1.3$.

$A$ prism has a refractive index of $1.5$. If a light ray undergoes total internal reflection at the interface with a liquid,what is the maximum possible refractive index of the liquid? (Refer to the provided diagram for geometry).

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