Question

In a YDSE, the separation between slits is `2 mm` where as the distance of screen from the plane of slits is `2.5 m`. Light of wavelengths in the range `200-800 nm` is allowed to fall on the slits. Find the wavelengths in the visible region that will be present on the screen at `1 mm` from central maximum. Also find the wavelength that will be present at that point of screen in the infrared as well as in the ultraviolet region.

Answer 1

Here, `d=2xx10^(-3)m, D=2.5m x=10^(-3)m`
For a wavelenght to be present on the screen, interference must be constructive.
`therefore (xd)/(D)=n lambda`
`lambda=(xd)/(nD)`
Put the values of x,d and D to obtain values of `lambda` for n-1,2,3...