Question

A monochromatic beam of electrons accelerated by a potential difference 'V' normally falls on the plane containing two narrow slits separated by a distance', The interference pattern is observed on a screen parallel to the plane of slits and at a distance of 'D. from the slits. Fringe width is found out to be 'a'. When the electron beam is accelerated through a potential difference 4V, the fringe width is found out to be 'b'. The ratio $\frac{a}{b}$ is (given $d<<<D)$

• A. 1
• B. 2
• C. 0.5
• D. 4

Answer 1

The correct option is A 1

$K.E$ of electron $=eV=\frac{P^2}{2m_e}$

$\lambda =\frac{h}{P}=\frac{h}{\sqrt{2eV{m}_e}}$

$\lambda \propto \frac{1}{\sqrt{V}}$

Fringe width is proportional to the wavelength

$\frac{w_1}{w_2}=\frac{a}{b}=\sqrt{\frac{V_2}{V_1}}=2$