A) 1.4
B) l5
C) 1.6
D) None of these
Correct Answer: C
Solution :
[c] Change in path difference for any point on screen is \[|\mu -1.8|t.\] For central maxima, phase difference = 0. Hence, \[d\sin \theta -|\mu -1.8|t=0\] \[\Rightarrow \,\,\,d\,\theta =|\mu -1.8|t\] [q is very small and is in radian] \[\Rightarrow \,\,\left| \mu -\left. 1.8 \right| \right.=\frac{{{10}^{-3}}\times 0.1}{0.5\times {{10}^{-3}}}=0.2\Rightarrow \mu =2\] or 1.6
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