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A small bulb is placed at the bottom of a tank containing water to a depth of $80 \mathrm{~cm}$. 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.)
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When the angle of incidence is equal to the critical angle the ray of light grazes the surface. Hence the light will appear to emerge out from a circle whose edges makes angle $i_c$ with the vertical.
The area of surface of water through which light will come out, $\mathrm{a}=\pi \mathrm{r}^2$
$$
\begin{aligned}
&\mu=\frac{1}{\sin \mathrm{i}_{\mathrm{c}}} \quad \therefore \sin \mathrm{i}_{\mathrm{c}}=\frac{1}{\mu}=\frac{1}{1.33}=0.752 \\
&\therefore \mathrm{i}_{\mathrm{c}}=48.76^{\circ}, \tan \mathrm{i}_{\mathrm{c}}=\frac{\mathrm{AB}}{\mathrm{AO}} \Rightarrow \mathrm{AB}=\mathrm{AO} \tan \mathrm{i}_{\mathrm{c}} \\
&\mathrm{r}=80 \times \tan 48.76^{\circ}=91.2 \mathrm{~cm} . \\
&\therefore \text { Area, } a=\pi \mathrm{r}^2=\pi \times(0.912)^2=2.61 \mathrm{~m}^2
\end{aligned}
$$
The area of surface of water through which light will come out, $\mathrm{a}=\pi \mathrm{r}^2$
$$
\begin{aligned}
&\mu=\frac{1}{\sin \mathrm{i}_{\mathrm{c}}} \quad \therefore \sin \mathrm{i}_{\mathrm{c}}=\frac{1}{\mu}=\frac{1}{1.33}=0.752 \\
&\therefore \mathrm{i}_{\mathrm{c}}=48.76^{\circ}, \tan \mathrm{i}_{\mathrm{c}}=\frac{\mathrm{AB}}{\mathrm{AO}} \Rightarrow \mathrm{AB}=\mathrm{AO} \tan \mathrm{i}_{\mathrm{c}} \\
&\mathrm{r}=80 \times \tan 48.76^{\circ}=91.2 \mathrm{~cm} . \\
&\therefore \text { Area, } a=\pi \mathrm{r}^2=\pi \times(0.912)^2=2.61 \mathrm{~m}^2
\end{aligned}
$$
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