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(D) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is

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(D) The formula for fringe width in Young's double slit experiment is given by $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance of the screen from the sources,and $d$ is the separation between the sources.Let the initial fringe width be $\beta = \frac{\lambda D}{d}$.According to the problem,the new distance $D^{\prime} = 2D$ and the new separation $d^{\prime} = \frac{d}{2}$.The new fringe width $\beta^{\prime}$ is given by $\beta^{\prime} = \frac{\lambda D^{\prime}}{d^{\prime}}$.Substituting the new values: $\beta^{\prime} = \frac{\lambda (2D)}{d/2} = 4 	imes \frac{\lambda D}{d}$.Therefore,$\beta^{\prime} = 4\beta$.The fringe width becomes $4$ times the original value.

In Young's double slit experiment,if the separation between coherent sources is halved and the distance of the screen from the coherent sources is doubled,then the fringe width becomes:

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