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In a double slit experiment, interferenc...

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

A

`(hD)/(sqrt(2meVd))`

B

`(2hD)/(sqrt(meVd))`

C

`(hd)/(sqrt(2meV D))`

D

`(2hd)/(sqrt(meV D))`

Text Solution

Verified by Experts

The correct Answer is:
A
`beta = (lambda D)/(d) , lambda = (h)/(sqrt(2mVe))`
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