Step 1: Recall the fringe width.
In Young's double-slit experiment the fringe width is the gap between two bright or two dark bands. \[ \beta = \frac{\lambda D}{d} \] Here $\lambda$ is the wavelength, $D$ is the slit-to-screen distance, and $d$ is the slit separation.
Step 2: Spot the wavelength term.
The wavelength $\lambda$ is on top, so the fringe width grows with the wavelength. \[ \beta \propto \lambda \]
Step 3: Check the screen distance.
The distance $D$ is also on top, so larger $D$ gives wider fringes. But this is a separate factor and not the one asked here.
Step 4: Check the slit separation.
The separation $d$ is on the bottom, so a larger $d$ makes the fringes narrower. So fringe width is inversely related to $d$, not directly.
Step 5: Compare with the choices.
Among the given options, only the plain wavelength $\lambda$ correctly shows a direct proportion. The options $1/\lambda$, $d$, and $1/D$ do not match.
Step 6: State the answer.
The fringe width is proportional to the wavelength. \[ \boxed{\beta \propto \lambda} \]