Question 102 of 295

A small bulb is placed at the bottom of a tank containing water to a depth of $\sqrt{7}m$. The refractive index of water is $\dfrac{4}{3}$. The area of the surface of water through which light from the bulb can emerge out is $x\pi {m}^{2}$. The value of $x$ is _____ .

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Answer: 9

Apply Snell's law, $\dfrac{\sin ⁡90^{\circ}}{\sin ⁡C}=\mu$
$\Rightarrow \sin C=\dfrac{1}{\mu }\Rightarrow \dfrac{R}{\sqrt{{R}^{2}+{h}^{2}}}=\dfrac{3}{4}$



Squaring, $16{R}^{2}=9{R}^{2}+9{h}^{2}$
$\Rightarrow 7{R}^{2}=9{h}^{2} \Rightarrow R=\dfrac{3}{\sqrt{7}}h=\dfrac{3}{\sqrt{7}}\times \sqrt{7}=3m$
The area of the circle is, $A=\pi {R}^{2}=\pi {\left(3\right)}^{2}{m}^{2}\Rightarrow \alpha =9$