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(A) The light from the outside world reaches the fish only if the angle of incidence is less than or equal to the critical angle $	heta_{c}$.For wate

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) The light from the outside world reaches the fish only if the angle of incidence is less than or equal to the critical angle $	heta_{c}$.For water with refractive index $\mu = \frac{4}{3}$,the critical angle $	heta_{c}$ is given by $\sin 	heta_{c} = \frac{1}{\mu} = \frac{3}{4}$.From the geometry of the problem,the radius $R$ of the circular horizon is related to the depth $h = 12 \, cm$ by $R = h 	an 	heta_{c}$.Since $\sin 	heta_{c} = \frac{3}{4}$,we have $\cos 	heta_{c} = \sqrt{1 - \sin^2 	heta_{c}} = \sqrt{1 - (\frac{3}{4})^2} = \sqrt{1 - \frac{9}{16}} = \sqrt{\frac{7}{16}} = \frac{\sqrt{7}}{4}$.Therefore,$	an 	heta_{c} = \frac{\sin 	heta_{c}}{\cos 	heta_{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 surface,the radius of this circle in $cm$ is

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