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(D) Given,refractive index $\mu = \frac{4}{3}$.Depth of the diver $h = 15 \, cm$.Let $R$ be the radius of the circular horizon.The light from the outs

Vedclass · Accepted Answer

$\frac{15 	imes 3}{\sqrt{7}} \, cm$

Vedclass · Answer

(D) Given,refractive index $\mu = \frac{4}{3}$.Depth of the diver $h = 15 \, cm$.Let $R$ be the radius of the circular horizon.The light from the outside world enters the water and reaches the diver's eyes only if the angle of incidence is less than or equal to the critical angle $C$.From Snell's law,$\sin C = \frac{1}{\mu} = \frac{1}{4/3} = \frac{3}{4}$.From the geometry of the problem,$	an C = \frac{R}{h}$.Since $\sin C = \frac{3}{4}$,we have $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - (3/4)^2} = \sqrt{1 - 9/16} = \sqrt{7/16} = \frac{\sqrt{7}}{4}$.Therefore,$	an C = \frac{\sin C}{\cos C} = \frac{3/4}{\sqrt{7}/4} = \frac{3}{\sqrt{7}}$.Thus,$R = h 	an C = 15 	imes \frac{3}{\sqrt{7}} \, cm = \frac{15 	imes 3}{\sqrt{7}} \, cm$.

$A$ diver looking up through the water sees the outside world contained in a circular horizon. The refractive index of water is $\frac{4}{3}$,and the diver's eyes are $15 \, cm$ below the surface of water. Then the radius of the circle is

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