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(D) The light from the outside world enters the water and undergoes refraction. The fish sees the outside world within a circular cone of light,which

Vedclass · Accepted Answer

$36/\sqrt{7}$

Vedclass · Answer

(D) The light from the outside world enters the water and undergoes refraction. The fish sees the outside world within a circular cone of light,which is defined by the critical angle $	heta_c$.The radius $r$ of the circular horizon is given by the formula $r = h 	an(	heta_c)$,where $h$ is the depth of the fish.Given the refractive index of water $\mu = \frac{4}{3}$,the critical angle $	heta_c$ satisfies $\sin(	heta_c) = \frac{1}{\mu} = \frac{3}{4}$.Using the identity $	an(	heta_c) = \frac{\sin(	heta_c)}{\cos(	heta_c)} = \frac{\sin(	heta_c)}{\sqrt{1 - \sin^2(	heta_c)}}$,we get $	an(	heta_c) = \frac{3/4}{\sqrt{1 - (3/4)^2}} = \frac{3/4}{\sqrt{7/16}} = \frac{3}{\sqrt{7}}$.Substituting the values,$r = 12 	imes \frac{3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \ cm$.

$A$ fish looking up through the water sees the outside world contained in a circular horizon. If the refractive index of water is $\frac{4}{3}$ and the fish is $12 \ cm$ below the surface,the radius of this circle in $cm$ is

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