A beam of light of two wavelengths `6500Å` and `5200Å`, is used to obtain interference fringes in Young's double slit experiment. Suppose the `m^(th)` brght fringe due to `6500Å` coincides with `n^(th)` bright fringe due to `5200Å` at a minimum distance from the cental maximum. Then

Options

A. 0.312 mm
B. 0.123 mm
C. 0.213 mm
D. 0.412 mm

Answer

A

Explanation

Here,
Distance between the slits , d = 2 mm `=2xx10^(-3) m`
Distance of the screen from the slits , D = 1.2 m
Wavelengths,
`lamda_1=6500 Å = 6500 xx10^(-10) =6.5 xx10^(-7) m`
and `lamda_2 = 5200 Å = 5200 xx10^(-10) =5.2 xx10^(-7) m`
Distance of `n^(th)` bright fringe from the centre bright fringe is
`x_n=(nlamdaD)/d" "...(i)`
If `x_4 and x_4'` be the distances of the fourth bright fringes of wavelengths `lamda_1 and lamda_2` respectively , then from eqn. (i)
`x_4=(4lamda_1D)/dand x'4 = (4lamda_2D)/d`
Thus, the separation between therm is
`Deltax = x'_4=(4lamda_1D)/d-(4lamda_2D)/d=(4D(lamda_1-lamda_2))/d`
`=(4(1.2m)(6.5xx10^(-7)m-5.2xx10^(-7)m))/((2xx10^(-3)m))`
`=3.12 xx10^(4) m = 0.312 xx10^(-3) m = 0.312 mm`