Illumination; the uniform lighting of a plane
In this section we turn to a completely different physical problem—we want to illustrate the great variety of possibilities. This time we will do something that leads to the same kind of integral that we found in electrostatics. (If we have a mathematical problem which gives us a certain integral, then we know something about the properties of that integral if it is the same integral that we had to do for another problem.) We take our example from illumination engineering. Suppose there is a light source at the distance $z$ above a plane surface. What is the illumination of the surface? That is, what is the radiant energy per unit time arriving at a unit area of the surface? (See Fig. 12–9.) We suppose that the source is spherically symmetric, so that light is radiated equally in all directions. Then the amount of radiant energy which passes through a unit area at right angles to a light flow varies inversely as the square of the distance. It is evident that the intensity of the light in the direction normal to the flow is given by the same kind of formula as for the electric field from a point source. If the light rays meet the surface at an angle $\theta$ to the normal, then $I_n$, the energy arriving per unit area of the surface, is only $\cos\theta$ as great, because the same energy goes onto an area larger by $1/\cos\theta$. If we call the strength of our light source $S$, then $I_n$, the illumination of a surface, is \begin{equation} \label{Eq:II:12:39} I_n=\frac{S}{r^2}\,\FLPe_r\cdot\FLPn, \end{equation}  where $\FLPe_r$ is the unit vector from the source and $\FLPn$ is the unit normal to the surface. The illumination $I_n$ corresponds to the normal component of the electric field from a point charge of strength $4\pi\epsO S$. Knowing that, we see that for any distribution of light sources, we can find the answer by solving the corresponding electrostatic problem. We calculate the vertical component of electric field on the plane due to a distribution of charge in the same way as for that of the light sources.1 Consider the following example. We wish for some special experimental situation to arrange that the top surface of a table will have a very uniform illumination. We have available long tubular fluorescent lights which radiate uniformly along their lengths. We can illuminate the table by placing the fluorescent tubes in a regular array on the ceiling, which is at the height $z$ above the table. What is the widest spacing $b$ from tube to tube that we should use if we want the surface illumination to be uniform to, say, within one part in a thousand?  Answer: (1) Find the electric field from a grid of wires with the spacing $b$, each charged uniformly; (2) compute the vertical component of the electric field; (3) find out what $b$ must be so that the ripples of the field are not more than one part in a thousand. In Chapter 7 we saw that the electric field of a grid of charged wires could be represented as a sum of terms, each one of which gave a sinusoidal variation of the field with a period of $b/n$, where $n$ is an integer. The amplitude of any one of these terms is given by Eq. (7.44): \begin{equation*} F_n=A_ne^{-2\pi nz/b}. \end{equation*}  We need consider only $n=1$, so long as we only want the field at points not too close to the grid. For a complete solution, we would still need to determine the coefficients $A_n$, which we have not yet done (although it is a straightforward calculation). Since we need only $A_1$, we can estimate that its magnitude is roughly the same as that of the average field. The exponential factor would then give us directly the relative amplitude of the variations. If we want this factor to be $10^{-3}$, we find that $b$ must be $0.91z$. If we make the spacing of the fluorescent tubes $3/4$ of the distance to the ceiling, the exponential factor is then $1/4000$, and we have a safety factor of $4$, so we are fairly sure that we will have the illumination constant to one part in a thousand. (An exact calculation shows that $A_1$ is really twice the average field, so that $b\approx0.83z$.) It is somewhat surprising that for such a uniform illumination the allowed separation of the tubes comes out so large.