question_answer

A fish in a water tank sees the outside world as if it is at the vertex of a cone, such that the circular base of the cone is at the surface of water. If the depth of the fish is d and the critical angle for water-air interface is\[{{i}_{c}}\], what is the radius of the circle?

Answer:

                The radius of the circular base of the cone will be \[r=d\tan {{i}_{c}}\]                 The light rays from the small bulbs S which are incident at an angle\[i>{{i}_{c}}\] are totally internally reflected and cannot emerge out of water surface. The light from the bulb S comes out through a circular patch of radius r given by                 \[\tan {{i}_{c}}=\frac{OA}{OS}=\frac{r}{h}\] or \[r=h\,\tan {{i}_{c}}\] \[\therefore \] Area of patch \[=\pi {{r}^{2}}=\pi {{h}^{2}}{{\tan }^{2}}{{i}_{c}}=\pi {{h}^{2}}\frac{{{\sin }^{2}}{{i}_{c}}}{{{\cos }^{2}}{{i}_{c}}}\] \[=\frac{\pi {{h}^{2}}{{\sin }^{2}}{{i}_{c}}}{1-{{\sin }^{2}}{{i}_{c}}}\] Now \[h=80\,cm=0.8\,m\,and\,\sin \,{{i}_{c}}=\frac{1}{\mu }=\frac{1}{1.33}=\frac{3}{4}\] \[\therefore \] Area of patch \[=\frac{3.142\times {{(0.80)}^{2}}\times {{\left( \frac{3}{4} \right)}^{2}}}{1-{{\left( \frac{3}{4} \right)}^{2}}}=2.6\,{{m}^{2}}.\]