Past exam of the physics 2 paper of the University of Oxford / 2014 / 7

Let $I_s$ be the intensity that either narrow slit would produce alone, and let $y$ measure screen displacement from the perpendicular bisector. The optical path difference is approximately $\delta =b\sin \theta \simeq by/D$. Equal coherent field amplitudes then giveThe maxima are at $y_m=m\lambda D/b$; the minima are halfway between them. HereThe expression uses the paraxial approximation and neglects the slowly varying single-slit diffraction envelope because the slit widths are unspecified.

The slide adds an optical path length $\Delta =(n-1)t=50\mu \mathrm{m}$ in the covered path. If the covered slit is at positive transverse position, its geometric path to $y>0$ is shorter, giving total path difference $\Delta -by/D$. The zero-difference fringe therefore moves toward that slit:This is about $94$ fringe periods. The leading grating-angle correction, $b\sin \theta =\Delta$ and $y=D\tan \theta$, gives $50.06$ mm, consistent with the paraxial result. In a strictly monochromatic periodic pattern, identifying the continuously tracked central order is essential: the final intensity pattern determines displacement only modulo one fringe spacing.

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.