In the Young's double slit experiment, the distance between the slits varies in time as d ( t)= d 0+ a 0, sin,t; where d 0, and a0 are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as:

(A) \frac{\lambda D}{d_{0}+a_{0}}

(C) \frac{2 \lambda {D}\left({d}_{0}\right)}{\left({d}_{0}^{2}-{a}_{0}^{2}\right)}

(D) \frac{\lambda D}{d_{0}^{2}} a_{0}

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JEE Mains · Physics · STD 12 - Wave optics

JEE Mains 2021
Hard
MCQ

In the Young's double slit experiment, the distance between the slits varies in time as \({d}({t})={d}_{0}+{a}_{0}\, sin\omega \,t\) ; where \({d}_{0}, \omega\) and \(a_{0}\) are constants. The difference between the largest fringe width and the smallest fringe width obtained over time is given as :

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Answer & Solution

Correct Answer

(B) \(\frac{2 \lambda \cdot D a_{0}}{\left(d_{0}^{2}-a_{0}^{2}\right)}\)

Step-by-step Solution

Detailed explanation

Fringe Width, \(\beta=\frac{\lambda D }{ d }\)

\(\beta_{\max } \Rightarrow d _{\min } \text { and } \beta_{\min } \Rightarrow d _{\max }\)

\(d = d _{0}+ a _{0} \sin \omega t\)

\(d _{\max }= d _{0}+ a _{0} \text { and } d _{\min }= d _{0}- a _{0}\)

\(\therefore \beta_{\min }=\frac{\lambda D }{ d _{0}+ a _{0}}\) and

\(\therefore \beta_{\max }=\frac{\lambda D }{ d _{0}- a _{0}}\)

\(\beta_{\max }-\beta_{\min }=\frac{\lambda D }{ d _{0}- a _{0}}-\frac{\lambda D }{ d _{0}+ a _{0}}=\frac{2 \lambda Da _{0}}{ d _{0}^{2}- a _{0}^{2}}\)

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