JEE Mains · Physics · STD 12 - 10. Wave optics
In a Young's double slit experiment with light of wavelength \(\lambda \) the separation of slits is \(d\) and distance of screen is \(D\) such that \(D >> d >> \lambda \). If the fringe width is \(\beta \). the distance from point of maximum intensity to the point where intensity falls to half of maximum intensity on either side is
- A \(\frac{\beta }{6}\)
- B \(\frac{\beta }{3}\)
- C \(\frac{\beta }{4}\)
- D \(\frac{\beta }{2}\)
Answer & Solution
Correct Answer
(C) \(\frac{\beta }{4}\)
Step-by-step Solution
Detailed explanation
\(2 I_{0}=4 I_{0} \cos ^{2}\left(\frac{\Delta \phi}{2}\right) \quad\) here, \(\Delta \phi=\frac{\pi}{2}\) But, \(\Delta \phi=\frac{2 \pi}{\lambda} \Delta x\) so, \(\Delta x=\frac{\lambda}{4}\) \(\frac{d y}{D}=\frac{\lambda}{4}\) ..... \((i)\) \({\frac{{\lambda D}}{d} = \beta }\)…
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