In Young's double slit experiment, the two slits `0.15 mm` apart are illuminated by light of wavelength `450 nm`. The screen is `1.0 m` away from the slits. Find the distance of second bright fringe and second dark fringe from the central maximum. How will the fringe pattern change if the screen is moved away from the slits ?

Explanation

Here ,
`d = 0.15` mm = `0.15 xx 10^(-3)` m = `15 xx 10^(-5)` m
`lambda = 450` nm = `450 xx 10^(-9) m = 4.5 xx 10^(-7) m , D = 1.0` m
(a) (i) Distance of the second bright fringe
`x_(2) = (2 lambda D)/(d) " " [ because x_(n) = (n lambda D)/(d)]`
`= (2 xx 4.5 xx 10^(-7) xx 1.0)/(15 xx 10^(-5)) = (2 xx 4.5)/(15) xx 10^(-2) = 0.6 xx 10^(-2)` m = 6mm
(ii) Distance of the second dark fringe
`x_(2) = (3 lambda D)/(2d) " " [ because x_(n) = ((2n-1) lambda D)/(2d)]`
`= (3 xx 4.5 xx 10^(-7) xx 1.0)/(2 xx 15 xx 10^(-5)) = (3 xx 4.5)/(30) xx 10^(-2) = 0.45 xx 10^(-2) m = 4.5` mm
(b) Since fringe width `beta = (lambda D)/(d)`
When screen is moved away, D increases . Therefore , width of the fringe increases but the angular separation `(lambda//d)` remains the same .