= Solution
For the centred screen, $|y|\le0.30$ m. Retaining the screen-angle geometry gives
$$
\Delta_{\max}=\frac{d(0.30)}{\sqrt{0.50^2+0.30^2}}=5.145\,\mu\mathrm m,\qquad
\bar\lambda=\frac1{\bar\sigma}=512.20\,\mathrm{nm}.
$$
There are about $\bar\sigma\Delta_{\max}=10.045$ carrier orders on each side. With equal spectral intensities, direct extrema of the intensity formula give
$$
\boxed{21\text{ bright fringes: the central maximum and ten on each side}.}
$$
Their contrast declines toward the edges: the first equal-intensity visibility zero occurs at $\Delta=1/(2\Delta\sigma)=5.25\,\mu\mathrm m$, just beyond the screen's $5.145\,\mu\mathrm m$ limit. Edge visibility is only about $0.031$, so fewer fringes may be discernible at a practical contrast threshold.
The screen reaches $|y|/D=0.6$, where a paraxial conversion of its full width is inaccurate. Also, wavelength-independent detector sensitivity does not itself establish equal source intensities. The boxed count specifies the usual equal-intensity idealization; without spectral weights and a visibility threshold, a unique experimental count is not fixed.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-two-colour-fringes.png]
{title=Two-colour double-slit fringes across the stated 60 cm screen}
{description=Equal intensities at 500 and 525 nm are added incoherently between colours. The actual screen-angle geometry gives 21 local maxima, with low contrast near the edges.}
Back to article page
