A beam of light consisting of two wavelength `6500Å&5200Å`is used to obtain interferance fringes in a young's double slit experiment .The distance between the slits is `2.0mm` and the distance between the plane of the slits and thescreen is `120cm`,.what is the least distance from the central maximum where the bright fringes due to both the wave length coincides?

Options

A. 1.56 mm
B. 2.56 mm
C. 3.12 mm
D. 0.86 mm

Answer

A

Explanation

Suppose that `n_(1)`th bright fringe of `lamda_(1)` coincides with `n_(2)`th bright fringe of `lamda_(2)`.
`therefore n_(1)X_(1)=n_(2)X_(2)` where X is the fringe width
`therefore n_(1)(lamda_(1)D)/(d)=n_(2)(lamda_(2)D)/(d)`
`therefore n_(1)/n_(2)=(lamda_(2))/(lamda_(1))=(5200)/(6500)=4/5`
`therefore ` Fourth bright fringe of `lamda_(1)` coincides with 5th bright fringe of `lamda_(2)` [or 8th bright of `lamda_(1)` and 10th bright of `lamda_(2)`]
thus the least distance of coincidence from the central maximum will correspond to 4th bright fringe of `lamda_(1)` [or 5th bright fringe of `lamda_(2)`].
`therefore Y_("min")=(4lamda_(1)D)/(d)=(4xx6500xx10^(-10)xx1.2)/((2xx10^(-3)))`
`=2.4xx65xx10^(-5)`
`Y_("min")=156xx10^(-5)m=1.56mm`