In Young's double slit experiment, we get `60` fringes in the field of view of monochromatic light of wavelength `4000Å`. If we use monochromatic light of wavelength `6000Å`, then the number of fringes obtained in the same field of view is

Options

A. 60
B. 90
C. 40
D. 1.5

Answer

C

Explanation

If the width of fringe of wavelength `lambda` is `beta` than for `n_(1)` fringes in the field of view `x=n beta=n(D lambda)/(d)` If the width of fringes of monochromatic light `beta'` than for `n_(2)` fringes in the field of view
`x=n_(1)beta'=n_(1) (Dlambda_(2))/(d)`
`therefore n_(1)lambda_(1)=n_(2)lambda_(2)`
Number of fringes `n_(2)=(n_(1)lambda_(1))/(lambda_(2))=(60xx4000)/(6000)=40`