In Young's double slit experiment with wavelength 5890 `Å`, there are 60 fringe in the field of vision, How many fringes will be observed in the same field of vision id wavelength used is 5460 `Å`?

Explanation

The fringe width is given by
`beta = (lambda D)/(d)`
If the extent of field of vision is l.
`l = N_(1) beta_(1)`
where `N_(1)` is the number of fringes formed with wavelength l.
As the field of vision is fixed, `l = N_(1) beta_(1) = N(2) beta_(2)`
`N_(1) ((lambda_(1) D)/(d)) = N_(2) ((lambda_(2) D)/(d))` ltbgt `N_(1) lambda_(1) = N_(2) lambda_(2)`
Thus, `N_(2) = (N_(1) lambda_(1))/(lambda_(2)) = (60 xx 5890)/(5460) = 64.72 ~~ 65`