A) 4mm
B) 5.6 mm
C) 14mm
D) 28mm
Correct Answer: D
Solution :
[d] At the area of total darkness minima will occur for both the wavelengths. \[\therefore \,\,\,\frac{(2n+1)}{2}{{\lambda }_{1}}=\frac{(2m+1)}{2}{{\lambda }_{2}}\] \[\Rightarrow \,\,(2n+1){{\lambda }_{1}}=(2m+1){{\lambda }_{2}}\] or\[\frac{(2n+1)}{(2m+1)}=\frac{560}{400}=\frac{7}{5}\] or \[10n=14m+2\] by inspection for m=2, n=3 and for m=7, n= 10, the distance between them will be the distance between such points. i.e., \[\Delta s=\frac{D{{\lambda }_{1}}}{d}\left\{ \frac{(2{{n}_{2}}+1)-(2{{n}_{1}}+1)}{2} \right\}\] put \[{{n}_{2}}=10\], \[{{n}_{1}}=3\] On solving we get, \[\Delta s=28mm\].
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