JEE Mains · Physics · STD 12 - 10. Wave optics
A single slit of width \(b\) is illuminated by a coherent monochromatic light of wavelength \(\lambda \). If the second and fourth minima in the diffraction pattern at a distance \(1\,m\) from the slit are at \(3\, cm\) and \(6\, cm\) respectively from the central maximum, what is the width of the central maximum ?......\(cm\) (i.e. distance between first minimum on either side of the central maximum)
- A \(1.5\)
- B \(3\)
- C \(4.5\)
- D \(6\)
Answer & Solution
Correct Answer
(B) \(3\)
Step-by-step Solution
Detailed explanation
For secondary minima, \(b \sin \theta=n \lambda \Rightarrow \sin \theta=\frac{n \lambda}{b}\) Distance of \(\mathrm{n}^{\text {th }}\) secondary minima \(\mathrm{x}=\mathrm{D} \sin\, \theta\) or \(\sin \theta_{1}=\frac{x_{1}}{D}\) \(\sin \theta_{1}=\frac{2 \lambda}{b}\) \(n=4\)…
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