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 1m away from the slits. |
(i) Find the distance of the second |
(a) bright fringe and |
(b) dark fringe from the central maximum. |
(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 = 1 m (i) (a) The distance of nth order bright fringe from central fringe is given by \[{{Y}_{n}}=nD\lambda /d\] For second bright fringe, \[{{Y}_{2}}=2D\lambda /d\] \[\frac{=2\times 1\times 4.5\times {{10}^{-7}}}{1.5\times {{10}^{-4}}}=6\times {{10}^{-3}}m\] The distance of the second bright fringe, \[{{Y}_{2}}=6mm\] (b) The distance of nth order dark fringe from central fringe is given by \[Y{{'}_{n}}=(2n-1)\frac{D\lambda }{2d}\] For second dark fringe n = 2 \[Y{{'}_{n}}=(2\times 2-1)\frac{D\lambda }{2d}=\frac{3D\lambda }{2d}\] \[Y{{'}_{n}}=\frac{3}{2}\times \frac{1\times 4.5\times {{10}^{-7}}}{1.5\times {{10}^{-4}}}=4.5\times {{10}^{-3}}m\] The distance of the second dark fringe, \[Y{{'}_{n}}=4.5mm\] (ii) With increase of D, fringe width also increases as \[\beta =\frac{D\lambda }{d}\] or \[\beta \propto D\]
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