ExamBro
ExamBro
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.
Verified Solution

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 \(=\pi r^{2}=9 \; \pi m ^{2}\)

Same subject
Explore more questions on app