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(A) Let $R$ be the radius of the circular region through which outside objects are visible.Let $	heta$ be the critical angle for the water-air interf

Vedclass · Accepted Answer

$\frac{3(H-h)}{\sqrt{7}}$

Vedclass · Answer

(A) Let $R$ be the radius of the circular region through which outside objects are visible.Let $	heta$ be the critical angle for the water-air interface.Then,$\sin 	heta = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.Now,$	an 	heta = \frac{\sin 	heta}{\cos 	heta} = \frac{\sin 	heta}{\sqrt{1 - \sin^2 	heta}}$.Substituting the value of $\sin 	heta$:$	an 	heta = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{1 - 9/16}} = \frac{3/4}{\sqrt{7/16}} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$.From the geometry of the problem,the diver's eyes are at a height $h$ from the bottom,so the distance from the water surface is $(H-h)$.Thus,$	an 	heta = \frac{R}{H-h}$.Equating the two expressions for $	an 	heta$:$\frac{R}{H-h} = \frac{3}{\sqrt{7}}$.Therefore,$R = \frac{3(H-h)}{\sqrt{7}}$.

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