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Question

A monochromatic beam of electrons accelerated by a potential difference V falls normally on the plane containing two narrow slits separated by a distance d. The interference pattern is observed on a screen parallel to the plane of the slits and at a distance of D from the slits. Fringe width is found to be w1. When electron beam is accelerated by the potential difference 4V, the fringe width becomes w2. Find the ratio w1w2. (Given d << D)

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Solution

Kinetic energy of electrons =eV=P22me
Here P is linear momentum of electron
λ=hP=h2eVme
λ1V

Fringe width w=λDdwλ.
As fringe width is proportional to wave length.
w1w2=λ1λ2=V2V1=4VV=2

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