In two separate set-ups of the Young's double | vedclass

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(A) The fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance between the slits an

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$4:1$

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(A) The fringe width $\beta$ in Young's double slit experiment is given by $\beta = \frac{D\lambda}{d}$,where $D$ is the distance between the slits and the screen,$\lambda$ is the wavelength of light,and $d$ is the slit separation.Given that the fringe widths are equal,$\beta_1 = \beta_2$,so $\frac{\beta_1}{\beta_2} = 1$.We are given the ratios $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$ and $\frac{d_1}{d_2} = \frac{2}{1}$.Using the formula $\frac{\beta_1}{\beta_2} = \left(\frac{D_1}{D_2}ight) \left(\frac{\lambda_1}{\lambda_2}ight) \left(\frac{d_2}{d_1}ight)$,we substitute the known values:$1 = \left(\frac{D_1}{D_2}ight) 	imes \left(\frac{1}{2}ight) 	imes \left(\frac{1}{2}ight)$.$1 = \left(\frac{D_1}{D_2}ight) 	imes \frac{1}{4}$.Therefore,$\frac{D_1}{D_2} = \frac{4}{1}$.

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