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 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 \[\beta \propto D\]