JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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 ...............
- A \(5\)
- B \(7\)
- C \(9\)
- D \(11\)
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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