A light source, which emits two wavelength `lamda_(1)=400nm` and `lamda_(2)=600nm`, is used in a Young's double slit experiment. If recorded fringe width for `lamda_(1)` and `lamda_(2)` are `beta_(1)` and `beta_(2)` and the number of fringes for them within a distance y on one side of the central maximum are `m_(1)` and `m_(2)` respectively, then

Options

A. `beta_(2) gt beta_(1)`
B. `m_(1) gt m_(2)`
C. Form the central maximum, `3^(rd)` maximum of `lambda_(2)` overlaps with `5^(th)` minimum of `lambda_(1)`
D. The angular separation of fringes for `lambda_(1)` is greater than `lambda_(2)`

Answer

A, B, C

Explanation

`beta=(lambdaD)/(d) therefore lambda_(2) gt lambda_(1)` so `beta_(2) gt beta_(1)`
No of fringes in a given width `(m)=(y)/(beta)rArrm_(2)ltm_(1)`
`3^(rd)` maximum of `lambda_(2)=(3lambda_(2)D)/(d)=(1800D)/(d)`
`5^(th)` maximum of `lambda_(1)=(9lambda_(1)D)/(2d)=(1800D)/(d)`
So, `3^(rd)` maxima of `lambda_(2)` will meet with `5^(th)` minimum of `lambda_(1)`
Angular sepration `=(lambda)/(d) rArr` Angular seperation for `lambda_(1)` will be lesser
`beta=(lambdaD)/(d) therefore lambda_(2) gt lambda_(1)` So `beta_(2) gt beta_(1)`