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 $w_1$. When electron beam is accelerated by the potential difference $4V,$ the fringe width becomes $w_2$. Find the ratio $\frac{w_1}{w_2}$. (Given d << D)

Answer 1

Kinetic energy of electrons $=eV=\frac{P^2}{2m_e}$
Here $P$ is linear momentum of electron
$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2eV{m}_e}}$
$\lambda \propto \frac{1}{\sqrt{V}}$

$\begin{array}{c}\because \text{Fringe width}w=\frac{\lambda D}{d}\\ w\propto \lambda \end{array}.$
As fringe width is proportional to wave length.
$\frac{w_1}{w_2}=\frac{{\lambda }_1}{{\lambda }_2}=\sqrt{\frac{V_2}{V_1}}=\sqrt{\frac{4V}{V}}=2$