A beam of light consisting of two wavelengths, 650 nm and 520 nm is used to obtain interference fringes in Young's double-slit experiment. What is the least distance (in m) from a central maximum where the bright fringes due to both the wavelengths coincide ? The distance between the slits is 3 mm and the distance between the plane of the slits and the screen is 150 cm.

Answer

`1.3`

Explanation

Let y be the linear distance between the centre of screen and the point at which the bright fringes due to both
Wavelength coincides. Let `n _(1) ^(th)` number of bright fringe with wavelength `lamda _(1)` coincides with `n _(2) ^(th)` number of bright fringe with wavelength `lamda _(2).`
`therefore n _(1) beta _(1) = n _(2) beta _(2)`
`implies n _(1) ( lamda _(1D))/( d ) = n _(2) ( lamda _(2 D))/( d)`
`implies n _(1) lamda _(1) = n _(2) lamda _(2) ...(i)`
Also at first position , the `n^(th)` bright fringe of one will coincide with
`(n + 1) ^(th)` bright of other.
So then `n _(2) gt n _(1)`
`implies n _(2) = n _(1) + 1 ...(ii)`
Using equation (ii) and equation (i),
`n _(1) lamda _(1) = ( n _(1) + 1 ) lamda _(2)`
`implies ( n _(1) ) ( 650 xx 10 ^(-9))`
`implies ( n _(1) +1) ( 520 xx 10 ^(-9))`
`implies 65 n _(1) = 52 n _(1) + 52`
`implies 13 n _(1) = 52`
`implies n _(1) = 4`
Thus, `y = n _(1) beta _(1)`
`= n _(1) (lamda _(1) D)/(d)`
`= 4 [ (( 6.5 xx 10 ^(-7) ) (1.5))/(3 xx 10 ^(-3))]= 1. 3 xx 10 ^(-3) m`
`= 1. 3 m`