A) 6000\[\overset{0}{\mathop{A}}\,\]
B) 4000 \[\overset{0}{\mathop{A}}\,\]
C) 1200 \[\overset{0}{\mathop{A}}\,\]
D) 2400 \[\overset{0}{\mathop{A}}\,\]
Correct Answer: A
Solution :
When distance between screen and source is D, and d the distance between coherent sources, then fringe width (W) is given by \[W=\frac{D\lambda }{d}\] where\[\lambda \]is wavelength of monochromatic light. \[\therefore \] \[\lambda =\frac{Wd}{D}\] Given, D = 1 m, \[d=1\text{ }mm={{10}^{-3}}m,\] \[W=0.06\text{ }cm,\] \[=0.06\times {{10}^{-2}}m\] \[\therefore \] \[\lambda =\frac{0.06\times {{10}^{-2}}\times {{10}^{-3}}}{1}\] \[=6\times {{10}^{-7}}m=6000{\AA}\]You need to login to perform this action.
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