In a biprism experiment,the slit separation i | vedclass

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(A) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$,Wavelength $\lambda = 5000 	ext{ Å} = 5 	imes 10^{-7} \,m$,Change in fringe width $\Delta \bet

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Away or towards the slit by $25 \,cm$

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(A) Given: Slit separation $d = 1 \,mm = 10^{-3} \,m$,Wavelength $\lambda = 5000 	ext{ Å} = 5 	imes 10^{-7} \,m$,Change in fringe width $\Delta \beta = \beta_{2} - \beta_{1} = 12.5 	imes 10^{-5} \,m$.Fringe width formula is $\beta = \frac{\lambda D}{d}$.The change in fringe width is given by $\Delta \beta = \frac{\lambda}{d} \Delta D$,where $\Delta D = D_{2} - D_{1}$ is the shift in the screen position.Rearranging for $\Delta D$: $\Delta D = \frac{d \cdot \Delta \beta}{\lambda}$.Substituting the values: $\Delta D = \frac{10^{-3} 	imes 12.5 	imes 10^{-5}}{5 	imes 10^{-7}}$.$\Delta D = \frac{12.5 	imes 10^{-8}}{5 	imes 10^{-7}} = 2.5 	imes 10^{-1} \,m = 0.25 \,m = 25 \,cm$.Therefore,the screen should be moved away or towards the slit by $25 \,cm$.

In a biprism experiment,the slit separation is $1 \,mm$. Using monochromatic light of wavelength $5000 \text{ Å}$,an interference pattern is obtained on the screen. Where should the screen be moved,so that the change in fringe width is $12.5 \times 10^{-5} \,m$?

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