Diffraction by a Circular Aperture — Question 7

Question 7

A central obstruction changes both throughput and normalized peak curvature

An annular pupil has outer diameter D and inner-to-outer diameter ratio $\epsilon$. Its normalized amplitude is $F(u)=2[J_1(u)-\epsilon J_1(\epsilon u)]/[(1-\epsilon^2)u]$. Use $J_1(u)=u/2-u^3/16+O(u^5)$ and equal incident irradiance for obstructed and unobstructed pupils.

1. Find the transmitted-power and on-axis-intensity ratios for $\epsilon=0.30$.

2. Expand the normalized amplitude and intensity near u=0 and compare central curvature with an unobstructed aperture.

Question 7 - Solution

A central obstruction changes both throughput and normalized peak curvature

(a) Method and calculation

$$\boxed{P_{\mathrm{ann}}/P_{\mathrm{full}}=1-\epsilon^2=0.91,\quad I_{c,\mathrm{ann}}/I_{c,\mathrm{full}}=(1-\epsilon^2)^2=0.8281}.$$$$J_1(u)-\epsilon J_1(\epsilon u)=(1-\epsilon^2)u/2-(1-\epsilon^4)u^3/16+\cdots.$$

(b) Interpretation and further calculation

$$\boxed{F(u)=1-(1+\epsilon^2)u^2/8+O(u^4)}.$$$$\boxed{|F(u)|^2=1-(1+\epsilon^2)u^2/4+O(u^4)};\quad\text{curvature factor}=1.09.$$

Physical check

A steeper normalized central curvature does not mean greater throughput or a universally better image; the obscuration redistributes light into the full diffraction pattern.