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

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becomes $\left(\frac{1}{20}\right)$ times the original

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(A) The formula for fringe width in a Young's double slit experiment is $\beta = \frac{\lambda D}{d}$,where $\lambda$ is the wavelength,$D$ is the distance from the slits to the screen,and $d$ is the distance between the slits.Let the initial fringe width be $\beta_1 = \frac{\lambda D_1}{d_1}$.According to the problem,the new distance between slits is $d_2 = 10 d_1$ and the new distance from the screen is $D_2 = \frac{D_1}{2}$.The new fringe width is $\beta_2 = \frac{\lambda D_2}{d_2} = \frac{\lambda (D_1 / 2)}{10 d_1} = \frac{1}{20} \left( \frac{\lambda D_1}{d_1} \right)$.Therefore,$\beta_2 = \frac{1}{20} \beta_1$.

In a double slit experiment,when the distance between slits is increased $10$ times,while their distance from the screen is halved,then the fringe width

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