Past exam of the physics 2 paper of the University of Oxford / 2013 / 4

Let $J_1,J_2$ be the single-slit intensities at the two wavelengths. At a screen position $y$, the path difference is approximately $\Delta =d\sin \theta$, with $\sin \theta =y/\sqrt{D^2+y^2}$. Fields from the two slits interfere at each wavelength, while a detector averaging over the optical beat period adds the intensities of the two distinct wavelengths:For equal single-slit intensities $J_1=J_2=J$, define $\bar{\sigma }=(1/{\lambda }_1+1/{\lambda }_2)/2$ and $\Delta \sigma =1/{\lambda }_2-1/{\lambda }_1$. ThenThe first cosine gives the interference visibility envelope. Near the centre one may further replace $\Delta$ by $dy/D$. The formula neglects unspecified single-slit envelopes and the slow change of single-slit intensity across the screen.

For the centred screen, $|y|\le 0.30$ m. Retaining the screen-angle geometry givesThere are about $\bar{\sigma }{\Delta }_{\max }=10.045$ carrier orders on each side. With equal spectral intensities, direct extrema of the intensity formula giveTheir 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.