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

 

Answer:

            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 \[{{I}_{0}}/4.\]