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JEE Mains · Physics · STD 12 - Wave optics
  • JEE Mains 2024
  • Hard
  • Numerical

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

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Answer & Solution

Correct Answer

(D) \(200\)

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Detailed explanation

\(\mathrm{I}=\mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)\)

\(\frac{\mathrm{I}_0}{4}=\cos ^2\left(\frac{\Delta \phi}{2}\right)\)

\(\Delta \phi=\frac{2 \pi}{3}\)

\(\frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right)=\frac{2 \pi}{3}\)

\(\mathrm{y}=\frac{\lambda \mathrm{D}}{3 \mathrm{~d}}=\frac{600 \times 10^{-9} \times 1}{3 \times 10^{-3}}=2 \times 10^{-4} \mathrm{~m}\)

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