• # question_answer The intensity at the central maxima (O) in a Young's double slit experimental is ${{I}_{0}}.$ the distance OP equals one-third of fringe width of the pattern, then show that the intensity at point P would be ${{I}_{0}}/4.$   Young s double slit experiment

Intensity can be found out if we know the phase difference. which can be calculated he help of path difference. So, first of all path difference will be calculated. Given,               $OP={{Y}_{n}}$ The distance OP equals one-third of fringe width $(\beta )$ of the pattern, i.e.        ${{Y}_{n}}=\frac{\beta }{3}=\frac{1}{3}\left( \frac{D\lambda }{d} \right)=\frac{D\lambda }{3d}$          $\left[ \because \beta =\frac{D\lambda }{d} \right]$ $\Rightarrow$   $\frac{d{{Y}_{n}}}{D}=\frac{\lambda }{3}$ $\therefore$ Path difference $={{S}_{2}}P-{{S}_{1}}P=\frac{d{{Y}_{n}}}{D}=\frac{\lambda }{3}$ $\therefore$ Phase difference, $\phi =\frac{2\pi }{\lambda }\times$path difference                         $=2\pi /\lambda \times \lambda /3=2\pi /3$ If intensity at central fringe is ${{I}_{0}},$ then intensity at a point P, where phase difference is $\phi$ is given by             $I={{I}_{0}}{{\cos }^{2}}\phi$ $\therefore I={{I}_{0}}{{\left( \cos \frac{2\pi }{3} \right)}^{2}}={{I}_{0}}{{\left( -\cos \frac{\pi }{3} \right)}^{2}}={{I}_{0}}{{\left( -\frac{1}{2} \right)}^{2}}=\frac{{{I}_{0}}}{4}$Hence, the intensity at the point P would be ${{I}_{0}}/4.$