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Topic: Interference and Young's double-slit experiment· Optics

The intensity at the maximum in a Young's double slit experiment is I_0. Distance between two slits is d = 5λ, where λ is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at…

The intensity at the maximum in a Young's double slit experiment is \(I_0\). Distance between two slits is \(d = 5\lambda\), where \(\lambda\) is the wavelength of light used in the experiment. What will be the intensity in front of one of the slits on the screen placed at a distance \(D = 10\,d\) ?

1. \(I_0\)

2. \(\frac{I_0}{4}\)

3. \(\frac{3}{4}I_0\)

4. \(\frac{I_0}{2}\)

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Answer: 4. \(\frac{I_0}{2}\)

Key idea: find the phase difference at the point in front of one slit, then use \(I=I_0\cos^2(\phi/2)\).

  1. The point is \(d/2\) from the central axis, so path difference \(\Delta=\dfrac{d\cdot(d/2)}{D}=\dfrac{d^2}{2D}=\dfrac{d}{20}=\dfrac{\lambda}{4}\).
  2. Phase difference \(\phi=\dfrac{2\pi}{\lambda}\cdot\dfrac{\lambda}{4}=\dfrac{\pi}{2}\).
  3. \(I=I_0\cos^2\dfrac{\pi}{4}=\dfrac{I_0}{2}\).
Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report
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