## 1.

## Introduction

It is well known that different lithographic features have different optimal settings of the numerical aperture (NA) and illumination condition.^{1, 2} Typically the depth of focus or the process latitudes are optimized. To achieve the optimal settings, modern steppers and scanners have a projection lens with a variable numerical aperture, where the NA value can be reduced to about 70% of its maximum value. As a consequence, a smaller fraction of the lens pupil and a corresponding smaller fraction of the phase aberrations is actually involved in the imaging process.

In the literature (see Ref. 3 for a survey) the problem of computing the aberration Zernike coefficients of scaled pupils has been discussed at various places. The recent result of Dai^{3} gives the Zernike coefficients of the scaled pupil in terms of those of the unscaled pupil in analytic form [see Eq. 18].

In this letter we give an alternative expression for Dai’s result. Our main result in Eq. 4 agrees mathematically with Dai’s Eq. 18 but has the advantage that it is very simple and direct in terms of Zernike polynomials. Thus it leads to the simple and explicit results in Eqs. 7, 9 from which the sensitivity of the aberration coefficients and Strehl ratios can be assessed when NA is close to its maximum value. The short and elegant proof of our main result in Eq. 4 is given in Appendix A.

## 2.

## Zernike Coefficients of Scaled Pupils

We consider a pupil function

on a unit disk $0\u2a7d\rho \u2a7d1$ with real phase $\Phi $ , and we assume that $\Phi $ is expanded as a Zernike series according to:## 2

$$\Phi (\rho ,\theta )=\sum _{n,m}{\alpha}_{n}^{m}{R}_{n}^{m}\left(\rho \right)\mathrm{cos}\left(m\theta \right),$$^{4}of azimuthal order $m$ and degree $n$ . For simplicity we only consider cosin terms. Scaling to a smaller pupil with relative size $\u03f5=\mathrm{NA}\u2215{\mathrm{NA}}_{\mathrm{max}}\u2a7d1$ means that we have to expand the scaled phase $\Phi (\u03f5\rho ,\theta )$ into a Zernike expansion:

## 3

$$\Phi (\u03f5\rho ,\theta )=\sum _{n,m}{\alpha}_{n}^{m}\left(\u03f5\right){R}_{n}^{m}\left(\rho \right)\mathrm{cos}\left(m\theta \right),\phantom{\rule{1em}{0ex}}0\u2a7d\rho \u2a7d1,\phantom{\rule{1em}{0ex}}0\u2a7d\theta \u2a7d2\pi .$$## 4

$${\alpha}_{n}^{m}\left(\u03f5\right)=\sum _{{n}^{\prime}}{\alpha}_{{n}^{\prime}}^{m}[{R}_{{n}^{\prime}}^{n}\left(\u03f5\right)-{R}_{{n}^{\prime}}^{n+2}\left(\u03f5\right)],\phantom{\rule{1em}{0ex}}n=m,m+2,\dots ,$$## 3.

## Examples and Sensitivity Analysis

Figure 1 shows an example where we scale low-order coma ${\alpha}_{3}^{1}\left(\u03f5\right)$ with ${\alpha}_{3}^{1}=2\pi \left(0.016\right)$ and ${\alpha}_{5}^{1}=2\pi \left(0.016\right)$ . For this special case, Eq. 4 and Dai’s Eq. 18 become

## 5

$${\alpha}_{3}^{1}\left(\u03f5\right)={\alpha}_{3}^{1}{R}_{3}^{3}\left(\u03f5\right)+{\alpha}_{5}^{1}[{R}_{5}^{3}\left(\u03f5\right)-{R}_{5}^{5}\left(\u03f5\right)]$$## 6

$${\alpha}_{3}^{1}\left(\u03f5\right)={\u03f5}^{3}[{\alpha}_{3}^{1}+4{\alpha}_{5}^{1}({\u03f5}^{2}-1)],$$A consequence of Eq. 4 is that

## 7

$$\text{\hspace{0.17em}}{\phantom{\mid}\frac{\partial {\alpha}_{n}^{m}}{\partial \u03f5}\mid}_{\u03f5=1}=n{\alpha}_{n}^{m}+2(n+1)[{\alpha}_{n+2}^{m}+{\alpha}_{n+4}^{m}+\dots ].$$The Strehl ratio^{4} is approximated as:

## 9

$$\text{\hspace{0.17em}}{\phantom{\mid}\frac{\partial S}{\partial \u03f5}\mid}_{\u03f5=1}=2\sum _{n,m}\frac{{\left({\alpha}_{n}^{m}\right)}^{2}}{{\gamma}_{m}(n+1)}-2\sum _{m}\frac{1}{{\gamma}_{m}}{\left(\sum _{n}{\alpha}_{n}^{m}\right)}^{2}=2[\overline{{\mid \Phi \mid}_{\mathit{\text{disk}}}^{2}}-\overline{{\mid \Phi \mid}_{\mathit{\text{rim}}}^{2}}],$$^{5}Intuitively it is expected that the Strehl ratio increases when the NA is decreased. However, in general this is not true. In the special case shown in Fig. 2, the Strehl ratio decreases when the NA is decreased from its maximum value. This result can be seen as follows: as in our example ${\sum}_{n}{\alpha}_{n}^{m}=0$ for each $m$ value, the phase aberration $\Phi (1,\theta )$ at the rim of the pupil equals 0. From Eq. 9 it then follows that the slope at $\u03f5=1$ is positive.

## Appendices

#### Appendix A: Proof of the Main Result, Eq. 4

By decoupling per azimuthal order
$m=0,1,\dots $
, and normalization and orthogonality^{4} of
${R}_{n}^{m}\left(\rho \right),n=m,m+2,\dots $
, the Zernike coefficients
${\alpha}_{n}^{m}\left(\u03f5\right)$
of
$\Phi (\rho \u03f5,\theta )$
and the Zernike coefficients
${\alpha}_{n}^{m}$
of
$\Phi (\rho ,\theta )$
are related by

## 10

$${\alpha}_{n}^{m}\left(\u03f5\right)=2(n+1)\sum _{{n}^{\prime}}{\alpha}_{{n}^{\prime}}^{m}{M}_{n{n}^{\prime}}^{m}\left(\u03f5\right),\phantom{\rule{1em}{0ex}}n=m,m+2,\dots .$$## 11

$${M}_{n{n}^{\prime}}^{m}\left(\u03f5\right)={\int}_{0}^{1}{R}_{n}^{m}\left(\rho \right){R}_{{n}^{\prime}}^{m}\left(\rho \u03f5\right)\rho \phantom{\rule{0.3em}{0ex}}\mathrm{d}\rho ,\phantom{\rule{1em}{0ex}}n,{n}^{\prime}=m,m+2,\dots .$$## 12

$${M}_{n{n}^{\prime}}^{m}\left(\u03f5\right)=\frac{1}{2(n+1)}[{R}_{{n}^{\prime}}^{n}\left(\u03f5\right)-{R}_{{n}^{\prime}}^{n+2}\left(\u03f5\right)];$$^{6}

## 13

$${R}_{l}^{k}\left(\rho \right)={(-1)}^{(l-k)\u22152}{\int}_{0}^{\infty}{J}_{l+1}\left(r\right){J}_{k}\left(\rho r\right)\phantom{\rule{0.3em}{0ex}}\mathrm{d}r,\phantom{\rule{1em}{0ex}}0\u2a7d\rho <1,$$## 14

$${M}_{n{n}^{\prime}}^{m}\left(\u03f5\right)={(-1)}^{({n}^{\prime}-m)\u22152}{\int}_{0}^{\infty}{J}_{{n}^{\prime}+1}\left(r\right)\left[{\int}_{0}^{1}{R}_{n}^{m}\left(\rho \right){J}_{m}\left(\rho \u03f5r\right)\rho \phantom{\rule{0.3em}{0ex}}\mathrm{d}\rho \right]\phantom{\rule{0.3em}{0ex}}\mathrm{d}r.$$## 15

$${\int}_{0}^{1}{R}_{n}^{m}\left(\rho \right){J}_{m}\left(\rho v\right)\rho \phantom{\rule{0.3em}{0ex}}\mathrm{d}\rho ={(-1)}^{(n-m)\u22152}\left[\frac{{J}_{n+1}\left(v\right)}{v}\right]$$^{4}and we get

## 16

$${M}_{n{n}^{\prime}}^{m}\left(\u03f5\right)={(-1)}^{({n}^{\prime}+n-2m)\u22152}{\int}_{0}^{\infty}\frac{{J}_{{n}^{\prime}+1}\left(r\right){J}_{n+1}\left(\u03f5r\right)}{\u03f5r}\phantom{\rule{0.3em}{0ex}}\mathrm{d}r.$$^{7}

## 17

$$\frac{{J}_{n+1}\left(\u03f5r\right)}{\u03f5r}=\frac{{J}_{n}\left(\u03f5r\right)+{J}_{n+2}\left(\u03f5r\right)}{2(n+1)},$$#### Appendix B: Dai’s Formula

We reproduce Dai’s formula^{3} for the scaling of Zernike coefficients: