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MHT CET 2025 (23 Apr, Shift 2) · Superposition, Coherence & Intensity

Superposition, Coherence & Intensity question from MHT CET 2025 (23 Apr, Shift 2)

Q32MHT CET 2025 · 23 APR · SHIFT 2Medium

Superposition, Coherence & Intensity

In Young's double slit experiment, the light of wavelength ' λ\lambda ' is used. The intensity at a point on the screen is ' I ' where the path difference is λ4\frac{\lambda}{4}. If ' I0I_0 ' denotes the maximum intensity then the ratio of ' I0I_0 ' to ' I ' is (cos⁡45∘=1/2)\left(\cos 45^\circ = 1/\sqrt{2}\right)

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Answer: (a)

  1. Δx=λ/4⇒ϕ=π/2\Delta x = \lambda/4 \Rightarrow \phi = \pi/2.
  2. I=I0cos⁡245∘=I02I = I_0\cos^2 45^\circ = \dfrac{I_0}{2}
  3. So the answer is (a) I0:I=2:1I_0 : I = 2 : 1.
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