JEE MAIN - Physics (2022 - 26th June Evening Shift - No. 24)
A small bulb is placed at the bottom of a tank containing water to a depth of $$\sqrt7$$ m. The refractive index of water is $${4 \over 3}$$. 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 __________.
Answer
9
Explanation
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So $$r = h{{\sin {i_c}} \over {\sqrt {1 - {{\sin }^2}{i_c}} }}$$
So $$A = \pi {r^2}$$
$$ = {{\pi {h^2}{{\sin }^2}{i_c}} \over {1 - {{\sin }^2}{i_c}}}$$
$$ = {{\pi 7 \times {9 \over {16}}} \over {1 - {9 \over {16}}}} = {{\pi \times 7 \times 9} \over 7} = 9\pi $$
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