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

The fish receives light coming from above only along those directions for which the rays, after refraction at the water-air interface, can actually emerge into air. Inside the water this set of directions forms a cone whose semi-vertical angle equals the critical angle $$\theta_c$$ for the water-air interface. Rays making a larger angle with the normal suffer total internal reflection and are not part of the “visible world”. Hence the circular horizon on the water surface is the base of this cone.

Step 1 Critical angle at a water-air surface
Snell’s law for the limiting (critical) ray is $$n_{\text{water}}\;\sin\theta_c \;=\; n_{\text{air}}\;\sin 90^{\circ}$$ With $$n_{\text{water}} = \tfrac{4}{3}$$ and $$n_{\text{air}} = 1$$, $$\sin\theta_c \;=\;\frac{1}{\,\tfrac{4}{3}} \;=\;\frac{3}{4}$$

Step 2 Find $$\tan\theta_c$$
For $$\sin\theta_c = \tfrac{3}{4}$$, $$\cos\theta_c \;=\;\sqrt{1-\sin^2\theta_c} \;=\; \sqrt{1-\tfrac{9}{16}} \;=\; \sqrt{\tfrac{7}{16}} \;=\; \tfrac{\sqrt{7}}{4}$$ Therefore, $$\tan\theta_c \;=\;\frac{\sin\theta_c}{\cos\theta_c} \;=\;\frac{\,\tfrac{3}{4}}{\,\tfrac{\sqrt{7}}{4}} \;=\;\frac{3}{\sqrt{7}}$$

Step 3 Relate the geometry to the fish’s depth
Let the fish be at depth $$h = 12\text{ cm}$$ below the surface. For the limiting ray, the horizontal (radial) distance $$r$$ on the surface and the depth $$h$$ form a right triangle inside the water: $$\tan\theta_c = \frac{r}{h} \implies r = h\,\tan\theta_c$$ Substituting, $$r = 12 \times \frac{3}{\sqrt{7}} = \frac{36}{\sqrt{7}}\;\text{cm}$$

Thus, the radius of the circular horizon seen by the fish is $$\boxed{\dfrac{36}{\sqrt{7}}\text{ cm}}$$.

Option B which is: $$36/\sqrt{7}$$