Diffraction by a Circular Aperture — Question 9

Question 9

An elliptical pupil has direction-dependent first-zero resolution

For a uniformly illuminated elliptical pupil, assume the Airy argument is $u=(\pi/\lambda)\sqrt{D_x^2\theta_x^2+D_y^2\theta_y^2}$ at small angles. The first zero is $u\simeq1.22\pi$. Let $D_x=0.100\,\mathrm m$, $D_y=0.0500\,\mathrm m$, and $\lambda=550\,\mathrm{nm}$.

1. Derive the first-zero angular radius in a direction making angle $\varphi$ with the x axis.

2. Calculate that radius along x, along y, and at $\varphi=45^\circ$.

Question 9 - Solution

An elliptical pupil has direction-dependent first-zero resolution

(a) Method and calculation

$$\theta_x=\theta\cos\varphi,\quad\theta_y=\theta\sin\varphi.$$$$\boxed{\theta_1(\varphi)=\frac{1.22\lambda}{\sqrt{D_x^2\cos^2\varphi+D_y^2\sin^2\varphi}}}.$$

(b) Interpretation and further calculation

$$\boxed{\theta_{1x}=6.710\,\mu\mathrm{rad},\quad\theta_{1y}=13.420\,\mu\mathrm{rad}}.$$$$\boxed{\theta_1(45^\circ)=8.48755\,\mu\mathrm{rad}}.$$

Physical check

The broader pupil dimension gives the smaller diffraction width in its corresponding angular direction.