= Solution
For a <Lorentz boost> along $x$, the <Lorentz transformation> is
$$
t'=\gamma\left(t-\frac{vx}{c^2}\right),\qquad x'=\gamma(x-vt),\qquad y'=y,\quad z'=z,\qquad\gamma=(1-v^2/c^2)^{-1/2}.
$$
To calculate a photograph, compare photons that arrive together, which were generally emitted at different times. Put the lamina in the $xy$ plane, moving along $+x$, and take the distant camera to lie along negative $y$. Let $(X,Y)$ be a material point's coordinates in the lamina's rest frame. At a laboratory time $t$, its position satisfies $x=X/\gamma+vt$, $y=Y$.
The distant-camera approximation makes the variable part of the light travel time $Y/c$. Equal arrival time therefore requires $t+Y/c=\text{constant}$. Choose that constant to be zero. The point's photographed transverse coordinate becomes
$$
x_{\mathrm{image}}=\frac X\gamma-\beta Y.
$$
A stationary lamina rotated in its plane by $\theta$ would instead have the projected coordinate $X\cos\theta-Y\sin\theta$. The two projections agree when
$$
\boxed{\cos\theta=\frac1\gamma,\qquad\sin\theta=\beta,\qquad\theta=\arcsin\frac vc.}
$$
This is <Terrell rotation>. The shape defined by simultaneous laboratory positions still has <length contraction>; the photographic rotation comes from combining that contraction with unequal emission times. Reversing the viewing or motion direction reverses the apparent rotation's sign.
\Image[/past-exam-of-the-physics-1-paper-of-the-university-of-oxford/2016-terrell-edge-on-projection.png]
{title=Equal-arrival-time projection of an edge-on lamina agrees with a rotated square, rather than the simultaneous contracted square}
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