Chapter 10: Problem 5
Show that the fraction of the total energy in the circular-aperture diffraction pattern out to the radius specified by \(u_{\operatorname{msx}}\) is $$ \frac{1}{2} \int_{0}^{u_{\max }}\left[\frac{2 J_{1}(u)}{u}\right]^{2} u d u=1-J_{0}^{2}\left(u_{\max }\right)-J_{1}^{2}\left(u_{\max }\right) . $$ Hence, verify the final column of Table \(10.2\).
Short Answer
Step by step solution
Key Concepts
These are the key concepts you need to understand to accurately answer the question.