A light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes in a Young's double slit experiment. Find the least distance (in mm) from a central maximum where the bright fringes due to both the wavelengths coincide? It is given that the distance between the slits is 3 mm and the distance between the plane of the slits and 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 `lambda_(1)` coincides with `n_(2)^(th)` number of bright fringe with wavelength `lambda_(2)`
`therefore n_(1)beta_(1) = n_(2)beta_(2)`
`rArr n_(1) (lambda_(1)D)/d = n_(2) (lambda_(2))/d`
`rArr n_(1)lambda_(1) = n_(2) lambda_(2)`.......... (i)
Also at first position, the nth bright fringe of one will coincide with (n + 1)th bright of other.
If `lambda_(2) lt lambda_(1)`
So, then `n_(2) gt n_(1)`
`rArr n_(2) = n_(1) + 1`........ (ii)
Using equation (ii) and equation (i)
`n_(1)lambda_(1) = (n_(1) + 1)lambda_(2)`
`rArr (n_(1)) (650 xx 10^(-9)) = (n_(1) +1) (52 xx 10^(-9))`
`rArr 65n_(1) = 52n_(1) + 52`
`rArr 13n_(1) = 52`
`rArr n_(1) = 4`
Thus, `y = n_(1)beta_(1)`
` = n_(1)(lambda_(1)D)/d`
`= 4[((6.5 xx 10^(-7))(1.5))/(3 xx 10^(-3))]`
`= 1.3 xx 10^(-3) m`
= 1.3 mm