question_answer

In Young's double slit experiment, the two slits 0.15 mm apart are illuminated by monochromatic light of wavelength 450 nm.

The screen is 1.0 m away from the slits.

(i) Find the distance of the second

(a) bright fringe.

(b) dark fringe from the central maxima.

(ii) How will the fringe pattern change, if the screen is moved away from the slits?

Answer:

Distance between the two sources, \[d=0.15\,mm=1.5\times {{10}^{-\,4}}m\]             Wavelength, \[\lambda =450\,nm=4.5\times {{10}^{-7}}m\]             Distance of screen from source, \[D=1m\]             Given, \[OP={{y}_{n}}\] The distance OP equals one-third of fringe width of the pattern. i.e.        \[{{y}_{n}}=\frac{\beta }{3}=\frac{1}{3}\left( \frac{D\lambda }{d} \right)=\frac{d\lambda }{3d}\Rightarrow \frac{d{{y}_{n}}}{D}=\frac{\lambda }{3}\] Path difference \[={{S}_{2}}P-{{S}_{1}}P=\frac{d{{y}_{n}}}{D}=\frac{\lambda }{3}S\] \[\therefore \] Phase difference, \[\phi =\frac{2\pi }{\lambda }\times \] path difference             \[=\frac{2\pi }{\lambda }\times \frac{\lambda }{3}=\frac{2\pi }{3}\] If intensity at central fringe is \[{{I}_{0}},\] then intensity at a point P, where phase difference \[\phi ,\] is given by \[I={{I}_{0}}{{\cos }^{2}}\phi \Rightarrow I={{I}_{0}}{{\left( \cos \frac{2\pi }{3} \right)}^{2}}\] \[={{I}_{0}}{{\left( -\cos \frac{\pi }{3} \right)}^{2}}={{I}_{0}}{{\left( -\frac{1}{2} \right)}^{2}}=\frac{{{I}_{0}}}{4}\] Hence, the intensity at point P would be \[\frac{{{I}_{0}}}{4}.\]