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JEE Main 2013 (9 Apr, Shift 1) · Polarisation

Polarisation question from JEE Main 2013 (9 Apr, Shift 1)

Q61JEE Main 2013 · 9 APR · SHIFT 1Difficult

Polarisation

A person lives in a high-rise building on the bank of a river 50 m wide. Across the river is a well lit tower of height 40 m. When the person, who is at a height of 10 m, looks through a polarizer at an appropriate angle at light of the tower reflecting from the river surface, he notes that intensity of light coming from distance X from his building is the least and this corresponds to the light coming from light bulbs at height 'Y' on the tower. The values of X and Y are respectively close to (refractive index of water ∼43\sim \frac{4}{3})

Figure

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Answer: (b)

10 mX ≈ 13 mθBY ≈ 27 m40 mriver 50 mtan θB = 4/3
  1. Reflected light is least (fully polarised) at Brewster's angle: tan⁡θB=μ=43\tan\theta_B = \mu = \frac{4}{3} (angle from the vertical).
  2. From the person (height 10 m): tan⁡θB=X10⇒X=403≈13 m\tan\theta_B = \frac{X}{10} \Rightarrow X = \frac{40}{3} \approx 13\ \mathrm{m}
  3. From the tower: tan⁡θB=50−XY⇒Y=34(50−13.3)≈27 m\tan\theta_B = \frac{50 - X}{Y} \Rightarrow Y = \frac{3}{4}(50 - 13.3) \approx 27\ \mathrm{m}
  4. So the answer is (b) 13 m, 27 m.
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