On the hypotenuse of a right-angled prism (30 | vedclass

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(A) The light enters the prism normally through the short face. By geometry,the angle of incidence $i$ at the hypotenuse face is $60^{\circ}$.For tota

Vedclass · Accepted Answer

$1.30$

Vedclass · Answer

(A) The light enters the prism normally through the short face. By geometry,the angle of incidence $i$ at the hypotenuse face is $60^{\circ}$.For total internal reflection to occur,the angle of incidence must be greater than or equal to the critical angle $C$ for the interface between the prism and the liquid.Thus,$i \geq C$,which implies $\sin i \geq \sin C$.Here,$i = 60^{\circ}$,so $\sin 60^{\circ} \geq \frac{\mu_{	ext{liquid}}}{\mu_{	ext{prism}}}$.Substituting the values: $\frac{\sqrt{3}}{2} \geq \frac{\mu_{	ext{liquid}}}{1.50}$.$\mu_{	ext{liquid}} \leq 1.50 	imes \frac{\sqrt{3}}{2} = 1.50 	imes 0.866 = 1.299 \approx 1.30$.Therefore,the maximum refractive index of the liquid is $1.30$.

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