A) 1.5
B) 0.015
C) 2.0
D) 0.15
Correct Answer: D
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,\[d=3\text{ }mm,\text{ }\lambda =5000\text{ }\overset{o}{\mathop{\text{A}}}\,=5\times {{10}^{-7}}m\] \[=5\times {{10}^{-4}}mm,\] \[D=90\text{ }cm=900\text{ }mm.\] \[\therefore \]\[W=\frac{5\times {{10}^{-4}}\times 900}{3}=15\times {{10}^{-2}}mm\] \[=0.15\text{ }mm\]You need to login to perform this action.
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