A) 1.2 m
B) 3.5 mm
C) 2.8 cm
D) 8.1 mm
Correct Answer: B
Solution :
[b] At the place where maxima for both the wavelengths coincide, y will be same for both the maxima, i.e., \[\frac{{{n}_{1}}{{\lambda }_{1}}D}{d}=\frac{{{n}_{2}}{{\lambda }_{2}}D}{d}\Rightarrow \frac{{{n}_{1}}}{{{n}_{2}}}=\frac{{{\lambda }_{1}}}{{{\lambda }_{2}}}=\frac{700}{500}=\frac{7}{5}\] Minimum integral value of \[{{n}_{2}}\]is 5. Minimum distance of maxima of the two Wave lengths fr 1 om central fringe \[\frac{{{n}_{1}}{{\lambda }_{2}}D}{d}=5\times 700\times {{10}^{-9}}\times {{10}^{3}}=3.5mm.\]
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