Answer
Please see the work below.
Work Step by Step
We know that bright fringes in a two-slit experiment are given as $dsin\theta=m\lambda$
and the location of bright fringes in a two-slit experiment is given as $dsin\theta=(m+\frac{1}{2})\lambda$
If the wavelength is greater than the separation of slits, then
$\frac{d}{\lambda}\lt 1$
and $\frac{d}{\lambda}=\frac{m}{sin\theta}$
Thus $\frac{m}{sin\theta}\lt 1$
$\implies m\lt sin\theta$
As the range of the sine function is $(-1, 1)$
$\implies -1\lt m\lt 1$
$m=0$ indicates that there is only a central bright fringe which is shown on screen.
