= Solution
Retain the question's common crossing condition $y'(0)=1$, and replace the slope-ratio condition of part (iii) by the specified initial slope. Then part (i) gives
$$
(1+\alpha)^{-1/3}=\frac12,
$$
so $\alpha=7$. Substitution in part (ii) gives the ray
$$
\boxed{y(x)=\begin{cases}
\displaystyle6-\frac32(1-x)^{2/3},&-7\leq x<0,\\[5pt]
\displaystyle\frac52+2\sqrt{1+x},&0\leq x\leq7.
\end{cases}}
$$
It starts at $(-7,0)$ with slope $1/2$, passes through $(0,9/2)$ with slope one, and reaches the receiver with slope $1/\sqrt8$. The final height is
$$
\boxed{y(7)=\frac52+4\sqrt2\approx8.15685}.
$$
The left-hand curvature is $y''=\frac13(1-x)^{-4/3}>0$; the right-hand curvature is $y''=-\frac12(1+x)^{-3/2}<0$. Thus the graph is increasing throughout, with a change from upward to downward curvature at the interface.
\Image[/past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2018/paper-1-q9-ray.png]
{title=Optical ray with initial slope one half and unit slope at the interface}
If the word “instead” were also taken to remove the common crossing condition, the initial slope alone would leave $\alpha$ undetermined: it would give $p=\mu(-\alpha)/\sqrt5$ for any selected source position. The numerical answer above uses the crossing condition stated before all four subparts.
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