A) 0.2 mm
B) 0.1 mm
C) 0.5 mm
D) 0.02 mm
Correct Answer: A
Solution :
Given \[d=1\,mm=1\times {{10}^{-3}}m\] \[D=1m\] \[\lambda =500nm=5\times {{10}^{-7}}m\] As width of central maxima = width of 10 maxima \[\therefore \] \[\frac{2D\lambda }{a}\,=10\,\left( \frac{\lambda D}{d} \right)\] \[\Rightarrow \] \[a=\frac{d}{5}\,=\frac{{{10}^{-3}}}{5}\,=0.2\,\times {{10}^{-3}}\,m\] \[a=0.2\,mm\]You need to login to perform this action.
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