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A point source of light is placed 4 m below the surface of water of refractive index $5 / 3$. The minimum diameter of a disc which should be placed over the source on the surface of water to cut-off all light coming out of water is
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The correct answer is:
6 m
Given, $\sin i=\frac{1}{\mu}=\frac{3}{5}$
So, $\quad \tan i=\frac{3}{4}=\frac{r}{4}$
It gives $\quad r=3 \mathrm{~m}$
So, the diameter $=6 \mathrm{~m}$.
So, $\quad \tan i=\frac{3}{4}=\frac{r}{4}$
It gives $\quad r=3 \mathrm{~m}$
So, the diameter $=6 \mathrm{~m}$.
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