question_answer

One of the two slits in Young's double slit experiment is so painted that it transmits half the intensity of the other. What is the effect on interference fringes?

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.