 Solution

The band has mean wavelength $\bar{\lambda }\simeq 530\mathrm{nm}$ and width $\Delta \lambda =20\mathrm{nm}$. At order $m$, the positions of the extreme-wavelength maxima differ by about $|m|D\Delta \lambda /b$. Comparing this spread with a fringe spacing $D\bar{\lambda }/b$ givesEquivalently, the coherence length scale is ${\ell }_c\sim {\bar{\lambda }}^2/\Delta \lambda \simeq 14\mu \mathrm{m}$. Thus the usual bandwidth estimate isThis is a washout scale, not a sharp definition of “clearly”. For a spectrum uniform in inverse wavelength $\sigma =1/\lambda$, the interference visibility is $|\mathrm{sinc}(\pi \Delta \sigma \delta )|$, with $\mathrm{sinc}u=\sin u/u$; its first zero occurs near order $26.5$. Fringes well inside that limit have stronger contrast. A half-fringe-smear criterion would instead retain about $13$ orders each side. The precise visible number needs the spectral distribution and a contrast threshold, neither supplied. If the slide remains in place and dispersion is neglected, the envelope is centred on the shifted zero-path-difference position and this order-count scale is unchanged.