Answer:
Let \[{{I}_{0}}\] be the intensity of light from each slit. When the slit is not painted, \[{{I}_{\max }}={{(\sqrt{{{I}_{0}}}+\sqrt{{{I}_{0}}})}^{2}}=\mathbf{4}{{\mathbf{I}}_{\mathbf{0}}}\] \[{{I}_{\max }}={{(\sqrt{{{I}_{0}}}-\sqrt{{{I}_{0}}})}^{2}}=\mathbf{0}\] When one of the slits is painted, it transmits half of the original intensity. \[\therefore \] \[{{I}_{\max }}={{\left( \sqrt{{{I}_{0}}}+\sqrt{\frac{{{I}_{0}}}{2}} \right)}^{2}}={{I}_{0}}{{\left( 1+\frac{1}{\sqrt{2}} \right)}^{2}}\] \[\mathbf{=2}\mathbf{.914}{{\mathbf{I}}_{\mathbf{0}}}\] \[{{I}_{\min }}={{\left( {{I}_{0}}-\sqrt{\frac{{{I}_{0}}}{2}} \right)}^{2}}={{I}_{0}}{{\left( 1-\frac{1}{\sqrt{2}} \right)}^{2}}\] \[=0.086\,{{I}_{0}}.\] Hence on painting one of the two slits, the intensity of maxima decreases from \[4{{I}_{0}}\]to \[2.914{{I}_{0}}\]and that of minima increases from 0 to\[0.086{{I}_{0}}\]. The contrast between the bright and dark fringes decreases.
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