A) 1.5 mm
B) 3 mm
C) 4.5 mm
D) 6 mm
Correct Answer: C
Solution :
From the given data, note that the fringe width (b1) for \[{{\lambda }_{1}}=900\,nm\] is greater than fringe width (b2) for \[{{\lambda }_{2}}=750\,nm.\] This means that at though the central maxima of the two coincide, but first maximum for \[{{\lambda }_{1}}=900\,nm\] will be further away from the first maxima for \[{{\lambda }_{2}}=750\,nm,\] and so on. A stage may come when this mismatch equals b2, then again maxima of \[{{\lambda }_{1}}=900\,nm,\] will coincide with a maxima of \[{{\lambda }_{2}}=750\,nm,\] let this correspond to nth order fringe for l1. Then it will correspond to \[{{(n+1)}^{th}}\] order fringe for l2. Therefore \[\frac{n{{\lambda }_{1}}D}{d}=\frac{(n+1){{\lambda }_{2}}D}{d}\] \[\Rightarrow n\times 900\times {{10}^{-9}}=(n+1)750\times {{10}^{-9}}\Rightarrow n=5\] Minimum distance from Central maxima \[=\frac{n{{\lambda }_{1}}D}{d}=\frac{5\times 900\times {{10}^{-9}}\times 2}{2\times {{10}^{-3}}}\] \[=45\times {{10}^{-4}}m=4.5\,mm\]
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