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(B) The light from the outside world reaches the fish only if the angle of incidence is less than or equal to the critical angle $C$. The light rays f

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The light from the outside world reaches the fish only if the angle of incidence is less than or equal to the critical angle $C$. The light rays forming the boundary of the circular horizon correspond to the critical angle $C$.For water with refractive index $\mu = \frac{4}{3}$,the critical angle $C$ is given by $\sin C = \frac{1}{\mu} = \frac{3}{4}$.From the geometry,the radius $r$ of the circular horizon is given by $r = h 	an C$,where $h = 12 \ cm$ is the depth of the fish.Since $\sin C = \frac{3}{4}$,we have $\cos C = \sqrt{1 - \sin^2 C} = \sqrt{1 - (\frac{3}{4})^2} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.Therefore,$	an C = \frac{\sin C}{\cos C} = \frac{3/4}{\sqrt{7}/4} = \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 water surface,the radius of this circle in $cm$ is

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