= Solution
A \b[<negative photoresist> is removed at low dose and retained at high dose]. Its remaining-thickness curve rises from zero above a gel or retention threshold and approaches one above a larger full-retention threshold. A useful schematic model is
$$
f_-(D)=\begin{cases}
0&D\leq D_a,\\
\dfrac{\ln(D/D_a)}{\ln(D_b/D_a)}&D_a<D<D_b,\\
1&D\geq D_b.
\end{cases}
$$
The developed profile is $t(x)/t_0=f_-[\tau I(x)]$. A central intensity maximum therefore leaves a central ridge, with cleared regions towards the darker sides. \b[Halving exposure narrows and can lower the retained ridge; doubling it broadens the ridge and can raise its shoulders.] A saturated central plateau cannot become thicker than the original film in this simple model. If the half-dose peak is below $D_a$, no ridge survives; if it lies between $D_a$ and $D_b$, the centre survives only partially.
The negative-resist thresholds are not supplied, so there is no unique numerical profile or linewidth. The original schematic below declares its own illustrative thresholds and intensity function; it explains the dose mapping without treating a positive-resist curve as measured negative-resist data.
\Image[/past-exam-of-the-mphil-in-micro-and-nanotechnology-enterprise-course-of-the-university-of-cambridge/2022/paper-1-resist-response.png]
{title=Negative-resist dose mapping and the effect of exposure time}
{description=Original illustrative model with declared retention thresholds. The intensity is a smooth test function rather than a copy of the examination graph; the numerical edge readings in the preceding parts come from the original PDF.}
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