 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 ifThus $M_c=0.6991$ and $0.2006$ for the lower- and higher-contrast materials. Reading those levels on the supplied modulation transfer function gives 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, $d_R=235.5\mathrm{nm}$, givingThe 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.