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(C) For the coin at $O$ to be visible from point $E$,the light ray must emerge from the liquid surface at the edge $B$. Let $R$ be the radius of the h

Vedclass · Accepted Answer

$\sqrt{2}$

Vedclass · Answer

(C) For the coin at $O$ to be visible from point $E$,the light ray must emerge from the liquid surface at the edge $B$. Let $R$ be the radius of the hemispherical vessel. The ray travels from $O$ to $B$. The angle of incidence $	heta$ at the surface is the angle between the normal at $B$ and the ray $OB$. Since the triangle formed by the center of the top surface,$O$,and $B$ is an isosceles right triangle,the angle $	heta = 45^{\circ}$. For the ray to emerge at the surface,the angle of incidence must be less than or equal to the critical angle $c$. Thus,$	heta \leq c$,which implies $\sin 	heta \leq \sin c$. Since $\sin c = \frac{1}{\mu}$,we have $\sin 45^{\circ} \leq \frac{1}{\mu}$. $\frac{1}{\sqrt{2}} \leq \frac{1}{\mu} \implies \mu \leq \sqrt{2}$. However,for the ray to graze the surface and reach $E$,we require the critical condition $\sin c = \sin 45^{\circ} = \frac{1}{\mu}$. Therefore,$\mu = \sqrt{2}$.

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