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JEE Mains · Physics · STD 12 - Ray optics and optical instruments
  • JEE Mains 2023
  • Hard
  • Numerical

A pole is vertically submerged in swimming pool, such that it gives a length of shadow \(2.15 \,m\) within water when sunlight is incident at an angle of \(30^{\circ}\) with the surface of water. If swimming pool is filled to a height of \(1.5\,m\), then the height of the pole above the water surface in centimetres is \(\left(\eta_W=4 / 3\right)........\)

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Answer & Solution

Correct Answer

(D) \(50\)

Step-by-step Solution

Detailed explanation

By Snell's law

\(1 \sin 60^{\circ}=\frac{4}{3} \sin r \rightarrow \sin r=\frac{3 \sqrt{3}}{8} \rightarrow \tan r=\frac{3 \sqrt{3}}{\sqrt{37}}\)

By the diagram

\(x \sqrt{3}+1.5 \tan r =2.15\)

\(x \sqrt{3}=2.15-1.5 \times \frac{3 \sqrt{3}}{\sqrt{37}}\)

\(x =\frac{2.15}{\sqrt{3}}-\frac{1.5 \times 3}{\sqrt{37}}\)

\(=1.241-0.739\)

\(=0.502\)

\(\approx 0.50 \text { meter }\)

\(x =50\,cm\)

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