Triple slit question help...
The first step is to recognize that the differential length d_l of each vector pointing from each slit to the observation point P is given as slit separation times the sine of the angle formed by the triangle formed by the slit, the observation point, and a point directly across from the slit.
Now if we assume that the observation point is closest to the topmost slit, we can say that l_2 = l_1 + d_l and likewise for l_3.
So the path difference for slits 2 and 3 are d_l and 2*d_l respectively.
Knowing that the phase difference is defined as the wave number times path difference, we obtain expressions for phi_2 and phi _3 (no phase difference on the topmost slit it's the reference!)
Then, begins the algebraic slog... Using superposition theorem, we generate an expression for the net electric field at the observation point.
A big equation with sine terms whose arguments are sums? Sum of sines formula!
Then it's time to factor out the time dependent sin(wt) and cos(wt) terms.
Then comes the part I don't *really* understand...
1 + cos(phi) + cos(2*phi) is defined as Rcos(v), so we combine it all into one resulting amplitude and phase shift??? Same goes for sin(phi) + sin(2*phi) = Rsin(v)???
And then the big kicker: the whole equation is supposed to be just R*sin(wt + v) ?! How does that work?? (I guess it's time to review trig....)
At this point I'm lost in the algebra. We square R ( I'm beginning to suspect that R here is the net electric field amplitude) and since intensity is proportional to the square of the amplitude, and the incident intensities are I_0, we get I/I_0 = 1 + 4cos(phi) + 2cos(2*phi)
Did anyone else try this problem themselves??? What is this black magic going on in the algebra here?? It's the same answer as in the back of the book but that trick with the R*R really confused me
I thought we were supposed to use the |E|^2 = E times E* the complex number method to generate the interference terms ???