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 4/3 . The area of the surface of water through which light from the bulb can emerge out is xπ m². The value of x is_______.

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answer is 9.

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Detailed Solution

C : Criticle angle

$\begin{array}{l} \tan C = \frac{r}{h} \\ r = htan C \\ \sin C = \frac{1}{μ} = \frac{3}{4} \\ \tan C = \frac{3}{\sqrt{7}} \\ r = \sqrt{7} \times \frac{3}{\sqrt{7}} = 3 \end{array}$
Area of surface = πr² = 9πm²

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