Question

A small bulb is placed at the bottom of a tank containing water to 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 $1.33$. (Consider the bulb to be a point source)

• A. $2.6\,m^2$
• B. $3.6\,m^2$
• C. $4.2\,m^2$
• D. $5.8 \,m^2$

Answer 1

Answer: (A)

As shown in the figure all those light rays which are incident on the surface at angle of incidence more than critical angle, does total internal reflection and are reflected back in water only. All those light rays which are incident below critical angle emerges out of surface bending away from normal. All those light beams which are incident at critical angle grazes the surface of water.

We know
$sin \,C = \frac{1}{^{a}\mu_{w}} $
$C = sin^{-1}\left(\frac{1}{^{a}\mu_{w}}\right) $
$ C = sin^{-1}\left(\frac{1}{1.33}\right) $
$ \Rightarrow sin \,C = \frac{1}{1.33} = \frac{3}{4}$
$ tan \,C = \frac{R}{OP}\left(radius\right) \left[\because\left(0.80\right)^{2} = 0.6400\right] $
$ R = tan\, C \times OP = tan \,C \left(0.80\right) $
Area $= \pi R^{2} = \pi \times tan^{2}C \left(0.64\right) $
$A = \pi\left(0.64\right)\times tan^{2}C$
$ = \pi\left(0.64\right)\times\frac{sin^{2}C}{cos^{2}C} = \pi \left(0.64\right)\times \frac{sin^{2}C}{1-sin^{2}C } $
$ = \pi\left(0.64\right)\times\frac{9}{16}\times\frac{16}{7}$
$=\frac{22}{7}\times0.64\times\frac{9}{7} = 2.6 m^{2}$