= 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> factor
$$
\frac{I(\theta)}{I(0)}=\left[\frac{\sin q}{q}\right]^2,\qquad q=\frac{\pi a\sin\theta}{\lambda}\simeq\frac{\pi a\theta}{\lambda}.
$$
Thus $\beta=\pi a/\lambda$ 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,
$$
\boxed{\frac{I(1.5\lambda/a)}{I(0)}\simeq\frac4{9\pi^2}=0.0450.}
$$
The exact first side maximum solves $\tan q=q$, giving $q\simeq4.493$ 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 $u\to-\infty$, \b[$v\to f$], 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 $y=\pm L\lambda_0/a=\pm0.100\,\mathrm m$. 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 $d=0.10\,\mathrm m$. Then
$$
\frac1f=\frac1d-\frac1L,\qquad
f=\frac{Ld}{L-d}=0.101695\,\mathrm m,\qquad
\boxed{R=(n-1)f=0.0610\,\mathrm m.}
$$
The plane surface has infinite <radius of curvature>. The transverse scale is $d/L=1/60$, yielding
$$
\boxed{y_{0,\rm new}=\frac dL y_{0,\rm old}=\frac{d\lambda_0}{a}=1.67\,\mathrm{mm}.}
$$
The 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 $1.7\%$.
\Image[/past-exam-of-the-physics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2013/paper-1-ia2013-slit-patterns.png]
{title=Original single-slit intensity plots at the original and compressed screens, and wavelength-dependent angular patterns for three representative visible colours}
The colour discussion below assumes a smooth white incident spectrum and neglects lens dispersion when comparing the ideal <diffraction> patterns.
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