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(B) The fringe width is given by $\beta = \frac{\lambda D}{d}$.For the fringe width to be maximum,the slit separation $d$ must be minimum,and for the

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$\frac{2 \lambda D a_{0}}{d_{0}^{2}-a_{0}^{2}}$

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(B) The fringe width is given by $\beta = \frac{\lambda D}{d}$.For the fringe width to be maximum,the slit separation $d$ must be minimum,and for the fringe width to be minimum,the slit separation $d$ must be maximum.The slit separation is $d(t) = d_{0} + a_{0} \sin \omega t$.The maximum value of $d$ is $d_{\max} = d_{0} + a_{0}$ and the minimum value of $d$ is $d_{\min} = d_{0} - a_{0}$.Therefore,the minimum fringe width is $\beta_{\min} = \frac{\lambda D}{d_{0} + a_{0}}$ and the maximum fringe width is $\beta_{\max} = \frac{\lambda D}{d_{0} - a_{0}}$.The difference between the largest and smallest fringe width is $\beta_{\max} - \beta_{\min} = \frac{\lambda D}{d_{0} - a_{0}} - \frac{\lambda D}{d_{0} + a_{0}}$.Simplifying this expression: $\beta_{\max} - \beta_{\min} = \lambda D \left( \frac{(d_{0} + a_{0}) - (d_{0} - a_{0})}{d_{0}^{2} - a_{0}^{2}} \right) = \frac{2 \lambda D a_{0}}{d_{0}^{2} - a_{0}^{2}}$.

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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