A) 0.2 mm
B) 0.1 mm
C) 0.5 mm
D) 0.02 mm
Correct Answer: A
Solution :
Given distance between two slits, \[d=1\text{ }mm={{10}^{-3}}m\] And distance of screen from slits, D = 1 m The wavelength of monochromatic light used, \[\lambda =500\,nm\,=500\times {{10}^{-9}}\,m\] Width of each slit a=? Width of central maxima in single slit pattern \[\beta '=\frac{2\lambda D}{a}\] Fringe width in double-slit experiment \[\beta =\frac{\lambda D}{d}\] So, required condition\[\frac{10\lambda D}{d}=\frac{2\lambda D}{a}\] \[\Rightarrow \]\[a=\frac{d}{5D}=\frac{1}{5}\times {{10}^{-3}}\,m=0.2\,mm\]
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