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  • A small bulb is placed at the bottom of a tank containing water to a depth of <img alt="\sqrt{7} \mathrm{~m}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Csqrt%7B7%7D%20%5Cmathrm%7B%7Em%

A small bulb is placed at the bottom of a tank containing water to a depth of \sqrt{7} \mathrm{~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 \mathrm{x a\, m^{2}}. The value of  \mathrm{x}is _____________.

Option: 1

9


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer

Circle of illumination


\mathrm{\sin i_{c}=\frac{1}{u}}

\mathrm{\sin i_{c}=\frac{3}{4}}
\mathrm{\tan i_{c}=\frac{r}{n}=\frac{3}{\sqrt{7}} }

\mathrm{r=3 \mathrm{~m} }
\mathrm{\text { Area }=\pi r^{2}=9 \pi=x \pi }
\mathrm{\therefore x=9 }

Posted by

avinash.dongre

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