JEE Mains · Physics · STD 12 - Wave optics
- JEE Mains 2015
- Hard
- MCQ
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
Pick an option to check your answer.
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, \(\Delta x=\frac{\lambda}{4}\)
\(\frac{d y}{D}=\frac{\lambda}{4}\) ..... \((i)\)
\({\frac{{\lambda D}}{d} = \beta }\) ..... \((ii)\)
Multiplying equation \((i)\) and \((ii)\) we get,
\(y=\frac{\beta}{4}\)
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