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

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$4$

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(C) The fringe width $\beta$ in Young's double slit experiment is given by the formula: $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance between the slits and the screen,and $d$ is the distance between the two slits.Given that the new wavelength $\lambda' = 2\lambda$ and the new slit separation $d' = \frac{d}{2}$.The new fringe width $\beta'$ is given by: $\beta' = \frac{\lambda' D}{d'} = \frac{(2\lambda) D}{(d/2)} = 4 	imes \frac{\lambda D}{d} = 4\beta$.Therefore,the resultant fringe width becomes $4$ times the initial fringe width. Hence,$n = 4$.

In Young's double slit experiment,if the wavelength of the light used is doubled and the distance between the two slits is halved,the resultant fringe width becomes $n$ times the initial fringe width. Find the value of $n$.

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