Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / Section D / D12 / Solution

Across a uniformly illuminated slit, the phase difference between its edges is $2\pi a\sin \theta /\lambda$. Integrating the complex amplitude across the slit gives the single-slit diffraction factorThus $\beta =\pi a/\lambda$ in the small-angle amplitude expression. The first noncentral zero has $q=\pi$, so ${\theta }_0\simeq \lambda /a$. At the stated approximate subsidiary maximum,The exact first side maximum solves $\tan q=q$, giving $q\simeq 4.493$ and intensity ratio $0.0472$; the requested approximation need not be mistaken for the exact maximum.

A virtual object is a point toward which incident rays are already converging and would meet if the lens were removed. With real-is-positive object/image distances it has $u<0$. The thin-lens equation gives $1/v=1/f-1/u$; as $u\to -\infty$, $v\to f$, so a distant virtual object images in the rear focal plane.

Without the lens, $y\simeq L\theta$ puts the central maximum at zero and the first dark fringes at $y=\pm L{\lambda }_0/a=\pm 0.100\mathrm{m}$. Subsequent zeros are integer multiples, and weaker subsidiary maxima lie between them.

For the finite-distance pattern compression, take the old screen as the prospective image plane at $L=6\mathrm{m}$, with $u=-L$, and the new screen at $d=0.10\mathrm{m}$. ThenThe plane surface has infinite radius of curvature. The transverse scale is $d/L=1/60$, yieldingThe finite-distance scaling can also be checked without treating an intensity pattern as a collection of geometrical point objects: the slit-to-screen Fresnel integral after a thin lens has quadratic coefficient $1/d-1/f=1/L$ and linear term $-y/d$. It therefore reproduces the old diffraction pattern at transverse coordinate $Y=Ly/d$, up to an overall amplitude and phase. If the old far-field plane is treated as infinitely distant, the equally useful leading approximation is $f=d$, $R=6.0\mathrm{cm}$; the finite-distance correction is only $1.7\%$.

The colour discussion below assumes a smooth white incident spectrum and neglects lens dispersion when comparing the ideal diffraction patterns.