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Two slits separated by 1 mm in Young's double slit experiment are illuminated by the violet light of the wavelength 400 nm. The interference fringes are obtained on the screen placed at 1 m from the slits. Find the fringe width. If the violet light is replaced by the red light of the wavelength 700 nm, find the percentage change in fringe width. |
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Answer» Slit separation, d = 1 mm = 1 x 10-3 m wavelength \(\lambda\) = 6.5 x 10-7 m screen distance D = 1m Distance b/w 3rd dark fringe and 5th bright fringe Dark fringe yn = \(\frac{(2n-1)\lambda D}{2d}\) Bright fringe yn = \(\frac{n\lambda D}d\) y5 - y3 = \(\frac{5\times6.5\times10^{-7}\times1}{10^{-3}}\) - \(\frac{(2\times3-1)\times6.5\times10^{-7}}{2\times10^{-3}}\) = 32.5 x 10-4 - 16.25 x 10-4 = 16.25 x 10-4 m The distance b/w the third dark fringe and fifth bright fringe is 1.625 mm. |
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