In Young 's double slit experiment th two slits are illuminated by a light beam consisting of wavelenghts 4200 Å and 5040 Å. If the distance between the slits is 2.4 mm and the distance between the slits and the screen is 200 cm the minimum distance from the central bright fringe to teh point where the bright fringes due to both the wavelengths coincide is

Options

A. 0.7 mm
B. 1.4 mm
C. 2.1 mm
D. 2.8 mm

Answer

C

Explanation

Let `n_(1)` fringes of `lambda_(1)= 4200 Å` and `n_(2)` fringes of `lambda_(2)= 5040 Å` wavelength are formed in a fixed distance on screen. So `nlambda` = constant or `n_(1)lambda_(1)= n_(2)lambda_(2)`
For minimum distance to be coincide
`n_(1)= n+1, n_(2)= n`
`(n+1) 4200 = n(5040 ) `
4200 n+4200= 5040 n
`implies n = 5 `
Required distance , `x=(n_(1)lambdaD)/(d)`
Given D = 200 cm = 2m d= `2.4xx 10^(-3) `m
`n_(1)= 5+1=6 lambda_(1)= 4200 Å `
`x = (6xx4200 xx 10^(10)xx2)/(2.4xx 10^(-3))`
x`=2.1 xx10^(-3)m = 2.1 mm`