= Solution
For <prism sweep under different rotation conventions>, let the apex angle be $A=60^\circ$. Snell's law at both faces gives
$$
r_1=\arcsin(\sin i/1.5),\qquad r_2=A-r_1,\qquad e=\arcsin(1.5\sin r_2),\qquad D=i+e-A.
$$
Here $e$ is measured from the second face's outward normal, whereas $D$ is the deviation from the incident direction. The two settings give
$$
\begin{array}{c|cc}
i&e&D\\\hline
50^\circ&47.2093^\circ&37.2093^\circ\\
60^\circ&38.8768^\circ&38.8768^\circ
\end{array}
$$
For a fixed laser and rotating prism, the outgoing beam therefore sweeps
$$
\boxed{|\Delta D|=1.6675^\circ\approx1.67^\circ.}
$$
If instead the prism stays fixed and the input beam is steered, its exit-face normal stays fixed and the answer is \b[$|\Delta e|=8.3325^\circ$]. The source does not explicitly say which component moves, so the laboratory-angle convention matters. Both settings transmit through the second face; total internal reflection does not occur.
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