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NEET 2016 (Phase 1) · YDSE: Film, Medium & Two Wavelengths

YDSE: Film, Medium & Two Wavelengths question from NEET 2016 (Phase 1)

Q65NEET 2016 · PHASE 1Medium

YDSE: Film, Medium & Two Wavelengths

The intensity at the maximum in a Young's double slit experiment is II0. Distance between two slits is d = 5λ\lambda, where λ\lambda is the wavelength of light used in the expreriment. What will be the intensity in front of one of the slits on the screen placed at a distance D = 10d?

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Answer: (b)

  1. In front of a slit, Δ=d22D=d20=5λ20=λ4\Delta = \dfrac{d^2}{2D} = \dfrac{d}{20} = \dfrac{5\lambda}{20} = \dfrac{\lambda}{4}.
  2. Phase ϕ=2πλΔ=π2\phi = \dfrac{2\pi}{\lambda}\Delta = \dfrac{\pi}{2}.
  3. I=I0cos⁡2(ϕ/2)=I0cos⁡2π4=I02I = I_0\cos^2(\phi/2) = I_0\cos^2\tfrac{\pi}{4} = \dfrac{I_0}{2}
  4. So the answer is (b) I02\frac{I_0}{2}.
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