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JEE Main 2024 (9 Apr, Shift 1) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from JEE Main 2024 (9 Apr, Shift 1)

Q54JEE Main 2024 · 9 APR · SHIFT 1Difficult

Superposition, Coherence & Intensity

In a Young's double slit experiment, the intensity at a point is (14)\left(\frac{1}{4}\right)th of the maximum intensity, the minimum distance of the point from the central maximum is _________ μm\mu\text{m}. (Given: λ=600 nm, d=1.0 mm,D=1.0 m\lambda = 600\text{ nm},\ d = 1.0\text{ mm}, D = 1.0\text{ m})

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

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Answer: 200

  1. I=Imax4⇒ϕ=2π3⇒Δx=λ3I = \frac{I_{max}}{4} \Rightarrow \phi = \frac{2\pi}{3} \Rightarrow \Delta x = \frac{\lambda}{3}
  2. ydD=λ3⇒y=λD3d\frac{yd}{D} = \frac{\lambda}{3} \Rightarrow y = \frac{\lambda D}{3d}
  3. y=600×10−9×13×10−3=2×10−4 my = \frac{600\times10^{-9}\times 1}{3\times 10^{-3}} = 2\times10^{-4}\text{ m}
  4. So y=200 μmy = 200\ \mu\text{m}.
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