13.4.14∗.The point light source S is located at a distance a < 4f from the flat screen. How will the illumination of the screen change at point A if a lens with a focal length of $f$ is placed between the source and the screen, at a distance of $x$
For problem $13.4.14$ from the source? At what $x$ is the maximum illumination at point A?
Solution
first of all the thin lens equation: $-\frac{1}{f} + \frac{1}{x} = \frac{1}{F}$ . The distance from center to image is $f = \frac{Fx}{F-x}$ , F - focal length.The flux conservation , $\int \vec{E} \cdot \vec{dS} = P= const.$ , yields the following relation $E4\pi x^{2}=E_{0}4\pi f^{2}$, from which it is obtained the intensity of the flux in point at which the rays are emitted . Now second time applying the flux conservation(Gauss's law) yields: $Ea^{2} = E_{1}(a-x-f)^{2}$ , which gives : $E_{0}(\frac{Fx}{F-x})^{2} \frac{a^{2}}{x^{2}}\frac{(F-x)^{2}}{ ((a-x)(F-x)-Fx)^2 }$,simplifying yields :
