Chapter 28 - Physical Optics: Interference and Diffraction - Problems and Conceptual Exercises - Page 1007: 69

One

We know that $m=\frac{1}{N\lambda_{max}}$ We plug in the known values to obtain: $m=\frac{1}{(7600cm^{-1})(\frac{100cm}{1m})(700\times 10^{-9}m)}$ $m=1.87$ The value of $m$ is less than 2. Thus, we have only one complete visible spectrum formed on either side of the central maximum.