JEE Mains · Physics · STD 12 - 10. Wave optics
In a double slit experiment shown in figure, when light of wavelength \(400 \mathrm{~nm}\) is used, dark fringe is observed at \(P\). If \(D=0.2 \mathrm{~m}\). the minimum distance between the slits \(S_1\) and \(S_2\) is ____________ \(mm\).
- A \(0.26\)
- B \(0.20\)
- C \(0.54\)
- D \(45\)
Answer & Solution
Correct Answer
(B) \(0.20\)
Step-by-step Solution
Detailed explanation
Path difference for minima at \(\mathrm{P}\) \( 2 \sqrt{\mathrm{D}^2+\mathrm{d}^2}-2 \mathrm{D}=\frac{\lambda}{2} \) \( \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}-\mathrm{D}=\frac{\lambda}{4} \) \( \therefore \sqrt{\mathrm{D}^2+\mathrm{d}^2}=\frac{\lambda}{4}+\mathrm{D} \)…
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