A fish looking up through the water sees the | vedclass

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(B) The light from the outside world enters the water and undergoes refraction. Due to total internal reflection,the fish sees the outside world withi

Vedclass · Accepted Answer

$\frac{36}{\sqrt{7}}$

Vedclass · Answer

(B) The light from the outside world enters the water and undergoes refraction. Due to total internal reflection,the fish sees the outside world within a circular cone of light.The radius $r$ of this circular horizon is given by the formula $r = \frac{h}{\sqrt{\mu^2 - 1}}$,where $h$ is the depth of the fish and $\mu$ is the refractive index of water.Given $h = 12 \ cm$ and $\mu = \frac{4}{3}$.Substituting the values: $r = \frac{12}{\sqrt{(4/3)^2 - 1}} = \frac{12}{\sqrt{16/9 - 1}} = \frac{12}{\sqrt{7/9}} = \frac{12 	imes 3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \ cm$.

$A$ 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 water surface,the radius of this circle in $cm$ is

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