JEE Mains · Physics · STD 12 - Ray optics and optical instruments
- JEE Mains 2022
- Medium
- Numerical
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 \(x \,\pi m ^{2}\). The value of \(x\) is ...............
Pick an option to check your answer.
Answer & Solution
Correct Answer
(C) \(9\)
Step-by-step Solution
Detailed explanation
\(C\) : Criticle angle
\(\tan C=\frac{r}{h}\)
\(r=h \tan C\)
\(\sin C =\frac{1}{\mu}=\frac{3}{4}\)
\(\tan C =\frac{3}{\sqrt{7}}\)
\(r =\sqrt{7} \times \frac{3}{\sqrt{7}}=3\)
Area of surface \(=\pi r^{2}=9 \; \pi m ^{2}\)
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