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(D) Let the initial intensities be $I_1 = I_2 = I_0$. The maximum intensity is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (2\sqrt{I_0})^2 = 4I_0$ and th

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Dark fringes will become less dark and bright fringes will become less bright.

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(D) Let the initial intensities be $I_1 = I_2 = I_0$. The maximum intensity is $I_{max} = (\sqrt{I_1} + \sqrt{I_2})^2 = (2\sqrt{I_0})^2 = 4I_0$ and the minimum intensity is $I_{min} = (\sqrt{I_1} - \sqrt{I_2})^2 = 0$.Now,the intensity of one slit is halved,i.e.,$I_1 = I_0$ and $I_2 = I_0/2$.The new maximum intensity is $I'_{max} = (\sqrt{I_0} + \sqrt{I_0/2})^2 = I_0(1 + 1/\sqrt{2})^2 \approx 2.91 I_0$,which is less than the original $4I_0$.The new minimum intensity is $I'_{min} = (\sqrt{I_0} - \sqrt{I_0/2})^2 = I_0(1 - 1/\sqrt{2})^2 \approx 0.086 I_0$,which is greater than the original $0$.Therefore,the bright fringes will become less bright and the dark fringes will become less dark (more bright).

In Young's double-slit experiment,one of the slits is painted such that its intensity is half that of the other slit. Then:

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