In two different Young's double-slit experime | vedclass

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(A) 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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$4:1$

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(A) 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 between the slits and the screen,and $d$ is the distance between the slits.Given that the fringe width $\beta$ is the same in both experiments,we have $\beta_1 = \beta_2$.Therefore,$\frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2}$.Rearranging for the ratio of distances $\frac{D_1}{D_2}$,we get $\frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} 	imes \frac{d_1}{d_2}$.Given $\frac{\lambda_1}{\lambda_2} = \frac{1}{2}$ (so $\frac{\lambda_2}{\lambda_1} = 2$) and $\frac{d_1}{d_2} = \frac{2}{1} = 2$.Substituting these values: $\frac{D_1}{D_2} = 2 	imes 2 = 4$.Thus,the ratio is $4:1$.

In two different Young's double-slit experiments,the fringe width is the same when the ratio of wavelengths is $1:2$. If the ratio of the distance between the slits in the two cases is $2:1$,then the ratio of the distance between the slits and the screen in the two experiments is:

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