Two slits separated by a distance of 1 mm are illuminated with red light of wavelength 6.5×10-7 m. The interference fringes are observed on a screen placed at 1 m from the slits. The distance between the third dark fringe and the fifth bright fringe is equal to

Here, d=1mm=10−3m                          λ=6.5×10−7mD=1mNow,    x5=nλDd=5×6.5×10−7×110−3                            =32.5×10−4mand      x3=(2n−1)λ2Dd             =(2×3−1)×6.5×10−72×10−3=16.25×10−4m∴ x5−x3=(32.5−16.25)10−4m                     =16.25×10−4m=1.63mm