In an interference experiment, interference was observed using...

Fringe width in an interference pattern is given by $\beta = \dfrac{\lambda D}{d}$

The de-Broglie wavelength of an electron accelerated through a potential difference $V$ is $\lambda = \dfrac{h}{\sqrt{2meV}}$

Substituting the value of $\lambda$, we get $\beta = \dfrac{h D}{d \sqrt{2meV}}$

From the above relation, $\beta \propto \dfrac{1}{\sqrt{V}}$

To increase the fringe width $\beta$, the electron gun voltage $V$ must be decreased. Increasing the slit separation $d$ or moving the screen closer (decreasing $D$) will decrease the fringe width.

Answer: electron gun voltage be decreased.