JEE Mains · Physics · STD 12 - 10. Wave optics
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 : \(\lambda=600 \mathrm{~nm}, d=1.0 \mathrm{~mm}, \mathrm{D}=1.0 \mathrm{~m}\) )
- A \(197\)
- B \(198\)
- C \(199\)
- D \(200\)
Answer & Solution
Correct Answer
(D) \(200\)
Step-by-step Solution
Detailed explanation
\(\mathrm{I}=\mathrm{I}_0 \cos ^2\left(\frac{\Delta \phi}{2}\right)\) \(\frac{\mathrm{I}_0}{4}=\cos ^2\left(\frac{\Delta \phi}{2}\right)\) \(\Delta \phi=\frac{2 \pi}{3}\) \(\frac{2 \pi}{\lambda}\left(\frac{\mathrm{yd}}{\mathrm{D}}\right)=\frac{2 \pi}{3}\)…
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