Diffraction by a Circular Aperture — Question 5

Question 5

Encircled energy distinguishes a dark-ring radius from a power boundary

The fraction of Airy-pattern power inside radius parameter u is $\mathcal E(u)=1-J_0(u)^2-J_1(u)^2$. At the first dark ring $u_1=3.831705970$, use $J_1(u_1)=0$ and $J_0(u_1)=-0.402759396$. Total transmitted power is $P=12.0\,\mathrm{mW}$.

1. Find the fraction and power inside the first dark ring.

2. Find the power outside that ring and decide whether a detector ending there captures 90 percent of the power.

Question 5 - Solution

Encircled energy distinguishes a dark-ring radius from a power boundary

(a) Method and calculation

$$\boxed{\mathcal E(u_1)=1-(-0.402759396)^2=0.837785}.$$$$\boxed{P_{\mathrm{inside}}=0.837785(12.0)=10.0534\,\mathrm{mW}}.$$

(b) Interpretation and further calculation

$$\boxed{P_{\mathrm{outside}}=(1-\mathcal E)P=1.94658\,\mathrm{mW}}.$$$$0.837785<0.90\Longrightarrow\boxed{\text{a larger collecting radius is required}}.$$

Physical check

Zero intensity on a ring does not imply zero energy outside it; the surrounding rings contain a finite share of the total power.