Two narrow parallel slits illuminated by a coherent monochromatic light produces an interference pattern on a screen placed at a distance $\mathrm{D}$ from the slits. The separation between the dark lines of the interference pattern can be increased by

The fringe width $\beta$ in a Young's double slit experiment is given by the formula $\beta = \dfrac{\lambda D}{d}$, where $\lambda$ is the wavelength of the monochromatic light, $D$ is the distance between the slits and the screen, and $d$ is the separation between the two slits.

The separation between the dark lines (fringe width) is directly proportional to the wavelength $\lambda$ and the distance $D$, and inversely proportional to the slit separation $d$.

To increase the fringe width $\beta$, one must either increase $\lambda$, increase $D$, or decrease $d$.

Since frequency $f$ is related to wavelength by $\lambda = \dfrac{c}{f}$, using light of a higher frequency corresponds to a shorter wavelength, which would decrease the fringe width. Increasing the distance between the slits $d$ would also decrease the fringe width. Decreasing the distance $D$ would decrease the fringe width. Therefore, using light of a longer wavelength is the correct method to increase the fringe width.

Answer: using monochromatic light of a longer wavelength