A fish inside water sees the outside world th | vedclass

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Question

(B) The light rays from the outside world reach the fish only if the angle of incidence at the water-air interface is less than or equal to the critic

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The light rays from the outside world reach the fish only if the angle of incidence at the water-air interface is less than or equal to the critical angle $i_c$.From the geometry of the problem,the radius $r$ of the circular horizon is given by $r = h 	an(i_c)$,where $h$ is the depth of the fish.We know that $\sin(i_c) = \frac{1}{\mu}$,where $\mu$ is the refractive index of water.Thus,$	an(i_c) = \frac{\sin(i_c)}{\cos(i_c)} = \frac{1/\mu}{\sqrt{1 - (1/\mu)^2}} = \frac{1}{\sqrt{\mu^2 - 1}}$.Substituting this into the expression for $r$,we get $r = \frac{h}{\sqrt{\mu^2 - 1}}$.Given $h = 12 \, cm$ and $\mu = 4/3$,we have:$r = \frac{12}{\sqrt{(4/3)^2 - 1}} = \frac{12}{\sqrt{16/9 - 1}} = \frac{12}{\sqrt{7/9}} = \frac{12 	imes 3}{\sqrt{7}} = \frac{36}{\sqrt{7}} \, cm$.

$A$ fish inside water sees the outside world through a circular horizon. If the refractive index of water is $4/3$ and the fish is $12 \, cm$ below the water surface,then the radius of this circle is $= ..... cm$.

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