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 `lambda_(1)` coincides with `n_(2)^(th)` number of bright fringe with wavelength `lambda_(2)`
`:. n_(1) beta_(1)= n_(2) beta_(2)`
`rArr n_(1)=(lambda_(1)D)/(d)=n_(2) (lambda_(2)D)/(d)`
`rArr n_(1) lambda_(1)=n_(2)lambda_(2)`..... (i)
Also at first position, the `n^(th)` bright fringe of one will coincide with `(n+1)^(th)` bright of other.
If `lambda_(2) lt lambda_(1)`
So then `n_(2) t 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))(650xx10^(-9))`
`=(n_(1)+1)(520xx10^(-9))`
`rArr 54n_(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[((6.5xx10^(-7))(1.5))/(3xx10^(-3))]`
`=1.3xx10^(-3)m`
`=1.3mm`