• # question_answer A small bulb is placed at the bottom of a tank, containing water at a depth of 80 cm. What is the area of the surface of water, through which light from the bulb can emerge out? Refractive index of water is 4/3.

The light rays coming from bulb can pass through the surface, if angle of incidence at surface is less than or equal to critical angle (c) for water-air interface. If h is depth of bulb from the surface, then light will emerge only through a circle of radius r given by Critical angle for water-air interface From the diagram,          $r=h\,\,tan\,c$   ...(i) where depth,                  h = 80 cm = 0.80 m but        $\sin c=\frac{1}{{}_{a}{{n}_{w}}}=\frac{3}{4}$ Right angle triangle $\therefore$      $\tan \,c=\frac{3}{\sqrt{7}}$ $\therefore$ Radius,    $t=h\tan c$       [from Eq.(i)]                         $=0.80\times 3/\sqrt{7}$ $\therefore$ Area of circular surface of water, $A=\pi {{r}^{2}}$ $=3.14\times {{\left( 0.8\times \frac{3}{\sqrt{7}} \right)}^{2}}=3.14\times 0.64\times \frac{9}{7}$ $=2.6\,{{m}^{2}}$