= Solution
The band has mean wavelength $\bar\lambda\simeq530\,\mathrm{nm}$ 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 $D\bar\lambda/b$ gives
$$
|m|\sim\frac{\bar\lambda}{\Delta\lambda}\simeq26.5.
$$
Equivalently, the <coherence length> scale is $\ell_c\sim\bar\lambda^2/\Delta\lambda\simeq14\,\mu\mathrm m$. Thus the usual bandwidth estimate is
$$
\boxed{\text{about }26\text{ orders on either side, or order }50\text{ fringes in total}.}
$$
This is a washout scale, not a sharp definition of “clearly”. For a spectrum uniform in inverse wavelength $\sigma=1/\lambda$, the <interference visibility> is $|\operatorname{sinc}(\pi\Delta\sigma\,\delta)|$, with $\operatorname{sinc}u=\sin u/u$; 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.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2014-double-slit-bandwidth.png]
{title=Double-slit fringes and bandwidth-induced loss of contrast}
{description=The band example assumes equal spectral intensity per unit inverse wavelength from 520 to 540 nm. Dashed envelopes show the visibility decline and its first zero near 26.5 central-wavelength orders.}
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