Q. In Youngs double slit experiment, the slits are 3 mm apart. The wavelength of light used is $ 5000\text{ }\overset{o}{\mathop{\text{A}}}\, $ and the distance between the slits and the screen is 90 cm. The fringe width in mm is
JamiaJamia 2006
Solution:
Let $ \lambda $ be wavelength of monochromatic light, used to illuminate the slit 5, and d be the distance between coherent sources, then width of slits is given by
where D is distance between screen and source. Given, $ d=3\text{ }mm,\text{ }\lambda =5000\text{ }\overset{o}{\mathop{\text{A}}}\,=5\times {{10}^{-7}}m $ $ =5\times {{10}^{-4}}mm, $ $ D=90\text{ }cm=900\text{ }mm. $ $ \therefore $ $ W=\frac{5\times {{10}^{-4}}\times 900}{3}=15\times {{10}^{-2}}mm $ $ =0.15\text{ }mm $
