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 \[\frac{{{I}_{0}}}{4}.\]
You need to login to perform this action.
You will be redirected in
3 sec