We now use all of the skills we have developed to derive what happens
in the case of *N* identical, very narrow slits, all at equal
separations *d* from one another (Figure 6). The only
new technique in this section is that we use the formula for the sum
of a geometric series.

**Figure 6:** Interference experiment with *N* finite slits

Following our derivation for the two-slit case, we begin with

where we already have used the fact that all slits let through the
same intensity (all *I*'s are equal) and waves arrive at the slits and
thus emerge from them all with the same phase (all 's are the
same).

Once again, we can factor out the constant phases, and
the intensity we would expect from each slit individually, leaving
only a sum of exponentials of path differences times *k*,

The next key step is to realize that the path differences are just multiples of each other. If we define, again, , then , , , . (See Figure 6.) Thus, we must compute the sum

but this is just a geometric series with geometric ratio . The formula you know for real numbers, *G*=1/(1-*r*) still works,
and so the sum is

To compute the magnitude of this, we use the fact that the magnitude of the ratio is the ratio of the magnitudes. Then, we note that both numerator and denominator have the same form, , where for the former case and for the latter . For both cases, we have

where, again, at the last step we use a convenient trig identity. Using this, our final result is

which simplifies to our final result,

where, as before, .

Thu Sep 13 15:26:14 EDT 2001