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JEE Main 2022 (26 Jun, Shift 2) · Total Internal Reflection

Total Internal Reflection question from JEE Main 2022 (26 Jun, Shift 2)

Q28JEE Main 2022 · 26 JUN · SHIFT 2Medium

Total Internal Reflection

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 $\frac{4}{3}$ . The area of the surface of water through which light from the bulb can emerge out is x $\pi$ m2. The value of x is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 9

Light escapes only through a circle of radius $r = h\tan C = \frac{h}{\sqrt{\mu^2 - 1}}$ , with $\sin C = \frac{1}{\mu}$ .
$\mu^2 - 1 = \frac{16}{9} - 1 = \frac{7}{9}$ , so $r = \frac{\sqrt{7}}{\sqrt{7}/3} = 3\text{ m}$
Area $= \pi r^2 = 9\pi\text{ m}^2$
So $x = 9$ .

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