Question 312: Ray Optics

Question

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 __________.

Accepted Answer

So $$r = h{{\sin {i_c}} \over {\sqrt {1 - {{\sin }^2}{i_c}} }}$$
So $$A = \pi {r^2}$$
$$ = {{\pi {h^2}{{\sin }^2}{i_c}} \over {1 - {{\sin }^2}{i_c}}}$$
$$ = {{\pi 7 	imes {9 \over {16}}} \over {1 - {9 \over {16}}}} = {{\pi  	imes 7 	imes 9} \over 7} = 9\pi $$

Ray Optics

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 __________.

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