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Question

(A) 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

Vedclass · Accepted Answer

$\frac{1}{3}$

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(A) 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 of light,$D$ is the distance between the slits and the screen,and $d$ is the slit separation.According to the problem,the slit separation $d$ is made $3$ folds,so the new slit separation $d' = 3d$.The new fringe width $\beta'$ is given by: $\beta' = \frac{\lambda D}{d'} = \frac{\lambda D}{3d}$.Substituting the original formula: $\beta' = \frac{1}{3} \left( \frac{\lambda D}{d} \right) = \frac{\beta}{3}$.Therefore,the fringe width becomes $\frac{1}{3}$ times the original fringe width.

In Young's double slit interference experiment,the slit separation is made $3$ folds. The fringe width becomes ......... times.

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