= Solution
For a <positive photoresist>, let $D_0$ be the largest exposure preserving the initial film thickness and $D_c$ the clearing dose. The <photoresist contrast> is $\gamma=1/\log_{10}(D_c/D_0)$. An aerial image $I=\bar I[1+M\cos(2\pi\nu x)]$ can clear its maxima while retaining its minima only if
$$
\frac{1+M}{1-M}\geq\frac{D_c}{D_0}=10^{1/\gamma},\qquad
M\geq M_c=\frac{10^{1/\gamma}-1}{10^{1/\gamma}+1}.
$$
Thus $M_c=0.6991$ and $0.2006$ for the lower- and higher-contrast materials. Reading those levels on the supplied <modulation transfer function> gives \b[maximum normalized <spatial frequencies> approximately $\nu/\nu_0=0.40$ and $0.60$], respectively. The stated partial coherence is already represented by that curve; do not multiply the read <frequencies> by another coherence factor.
The unambiguous pattern dimensions are full pitch $P=1/(x\nu_0)$ and equal line/space width $w=P/2$, where $x$ is the read normalized <frequency>. To convert the paper's Rayleigh-frequency label into <lengths>, explicitly use the circular-pupil <Rayleigh criterion> $d_R=0.61\lambda/NA$ and $\nu_0=1/d_R$. With the ArF vacuum wavelength $193\ \mathrm{nm}$ and the stated <numerical aperture>, $d_R=235.5\ \mathrm{nm}$, giving
$$
\begin{array}{c|cc}
\gamma&P\ (\mathrm{\mu m})&w\ (\mathrm{\mu m})\\\hline
1.33&0.589&0.294\\
5.66&0.392&0.196
\end{array}
$$
\b[The higher-contrast material resolves a line/space feature about two-thirds as wide.] The pitch and half-pitch distinction matters: the point-separation <Rayleigh criterion> is not itself an equal-line/space width. If a course defines $\nu_0$ as an optical cutoff rather than $1/d_R$, use that specified <frequency> in $w=1/(2x\nu_0)$; the normalized graph readings and their ratio are unchanged. No absolute clearing dose is provided.
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