Answer
increase
Work Step by Step
This is a diffraction grating, so we can look at equation (36-31) to see how the wavelength affects $\Delta \lambda$.;
$$R=\frac{\lambda_{avg}}{\Delta \lambda}$$
If the resolving power, R, remains constant, then as the wavelength, $\lambda$, increases, the difference in wavelength, $\Delta \lambda$, must also increase.
