JEE Mains · Physics · STD 12 - 9. Ray optics and optical instruments
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 \(\left(\eta_W=4 / 3\right)........\)
- A \(49\)
- B \(48\)
- C \(47\)
- D \(50\)
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 \(x \sqrt{3}+1.5 \tan r =2.15\) \(x \sqrt{3}=2.15-1.5 \times \frac{3 \sqrt{3}}{\sqrt{37}}\)…
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