In a double slit experiment, interference is obtained from electron waves produced in an electron gun supplied with voltage V. If `lambda` is wavelength of the beam, D is the distance of screen, d is the spacing between coherent source, h is Planck's constant, e is charge on electron and m is mass of electron, then fringe width is given as

Options

A. (a) `(hD)/(sqrt(2meVd))`
B. (b) `(2hD)/(sqrt(meVD))`
C. (c) `(hd)/(sqrt(2meVD))`
D. (d) `(2hd)/(sqrt(meVD))`

Answer

A

Explanation

`beta=(Dlambda)/(d)` and `lambda=(h)/(sqrt(2meV)`.