Using light of wavelength 600 mm in Young's double slit experiment 12 bands are found on one part of the screen. If the wavelength of light is changed to 400 nm , then what will be the number of bands on that part of the screen?

Explanation

In first case wavelength `lambda_(1)` =600 nm =`600 xx 10^(-9)` m, number of fringes `n_(1)=12 ` fringe width `y_(1)` . In second case wavelength `lambda_(2)`=400 nm = `400 xx 10^(-9)`m, number of fringes `n_(2)` and fringe width `y_(2)`.
Let the distance between the slits and the screen be D separation between the slits be 2d and length of the entire fringe pattern be x.
Now , `y_(1)=(lambda_(1) D )/(2d) =(600 xx 10^(-9)D)/(2d)`
and `n_(1)=(x)/(y_(1))`
or `12= (x xx2d)/(6000 xx 10^(-9) D) "or" x=(12 xx 600 xx 10^(-9))/(2d)`
Also `Y_(2)=(400 xx 10^(-9)D)/(2d)`
`:. " "n_(2)=(x)/(y_(2))=(12 xx 600 xx 10 ^(-9)D xx 2d )/(2D xx 400 xx 10^(-9))D=18`