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JEE Main 2020 (6 Sep, Shift 2) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2020 (6 Sep, Shift 2)

Q86JEE Main 2020 · 6 SEP · SHIFT 2Medium

Superposition, Coherence & Intensity

A Young's double-slit experiment is performed using monochromatic light of wavelength λ\lambda. The intensity of light at a point on the screen, where the path difference is λ\lambda, is K units. The intensity of light at a point where the path difference is λ6\frac{\lambda}{6} is given by nK12\frac{nK}{12}, where n is an integer. The value of n is __________.

Numerical answer type. Work it out, then open the answer.

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Answer: 9

  1. Path difference λ\lambda gives a maximum: K=4I0K = 4I_0.
  2. Path difference λ6\frac{\lambda}{6}: ϕ=π3\phi = \frac{\pi}{3}, I=Kcos⁡2π6=3K4=9K12I = K\cos^2\frac{\pi}{6} = \frac{3K}{4} = \frac{9K}{12}
  3. So n=9n = 9.
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