# Interference of N Waves

To motivate this section, we will note our previously statement that the fringe pattern for two interfering waves follows approximately from the $cos(\delta)$ term, where $\delta = \frac{2\pi}{\lambda_0}*OPD$. Since $\delta$ is dependent on $\lambda_0$, one may think that you could measure the wavelength of the light source that created the pattern by measuring the spacing between the fringes. However, this is not very accurate as there would typically be a lot of noise on the measurement that would prevent any high resolution. As such, we should then interfere many waves to perform this; thus, N wave interference!

INTERFERENCE FROM N IDENTICAL SLITS

We can start by making the same approximation that we made in the double slit experiment, that the screen is effectively at infinity compared with the total extent of the slits so we only need to consider parallel rays from the slits. Alternatively, we can also place a lens after the slits in order to get rid of this assumption. The situation can then be described by the figure below, where the “grating” denotation gives the hint that a grating is an example of N-slit interference. Also notice that in the figure below we have the second situation we had described above, where the screen is at infinity because of the lens.

Then, the total field at the point “$P_1$” is just given by the sum of each of the fields that are converging at this point, i.e.

\begin{align} E & = E_0(r_1)e^{i\omega-kr_1} + E_0(r_2)e^{i(\omega{t}-kr_2)}+ … + E_0(r_N)e^{i(\omega{t}-kr_n)} \\ & \approx E_0(D)e^{i\omega{t}}[e^{-ikr_1}+e^{-ikr_2}+ … +e^{-ikr_N}] \end{align}

where $r_i$ is the distance from the slit $i$ to the point $P_1$. We have also made the approximation in the second relation that because the difference in path length from each slit to the point $P_1$ is negligible compared with the total distance “$D$” from the grating to the screen, we can replace each $E(r_i)$ with $E(D)$. To pull out the first term and to make each of the other terms a function of $r_1$, we can rewrite the above relation as:

$$E = E_0(D)e^{i(\omega{t}-kr_1)}[1 + e^{-ik(r_2-r_1)}+ … + e^{-ik(r_N-r_1)}] \label{rewrite}$$

Now, to simplify this further, we can note that each of the slits are evenly spaced, such that:

$$r_2-r_1 = r_3 – r_2 = r_4 – r_3 = …. = r_N – r_{N-1}$$

which means that we can rewrite each of the exponentials in Equation \eqref{rewrite} as multiples of $r_2 – r_1$. For example:

$$r_3 – r_1 = 2(r_2-r_1)$$

$$r_4 – r_1 = 3(r_2-r_1)$$

$$r_N – r_1 = (N-1)(r_2-r_1)$$

Now, we know from Young’s double slit experiment that the optical path length difference between $r_2$ and $r_1$ is given as:

$$r_2 – r_1 = dsin(\theta) = \frac{\delta}{k}$$

where $\delta$ is the phase difference arising from the path length difference between two slits. We can then write:

$$E = E_0e^{i(\omega{t}-kr_1)}[1 + e^{-i\delta} + e^{-2i\delta} + … + e^{-(N-1)i\delta}]$$

which is equivalent to:

$$E = E_0e^{i(\omega{t}-kr_1)}[1 + e^{-i\delta} + (e^{-i\delta})^2 + … + (e^{-i\delta})^{N-1}]$$

If you have any experience with geometric series, then you may now recognize why we tried to isolate a “$1$” in the front of the series: we can now write the terms inside of the brackets as a fraction! First, the series itself is given as:

$$1 + x + x^2 + … + x^{N-1}$$

which we can see is simply:

$$\sum 1 + x + x^2 + … + x^{N-1} = \frac{x^{N}-1}{x-1}$$

So now, we can rewrite the field as:

\begin{align} E & = E_0e^{i(\omega{t}-kr_1)}[\frac{e^{-iN\delta}-1}{e^{-i\delta}-1}] \\ & = E_0e^{i(\omega{t}-kr_1)}[\frac{e^{-iN\delta/2}(e^{-iN\delta/2}-e^{iN\delta/2})}{e^{-i\delta/2}(e^{-i\delta/2}-e^{i\delta/2})}] \\ & = E_0e^{i(\omega{t}-kr_1-(N-1)\delta/2)}[\frac{sin\frac{N\delta}{2}}{sin\frac{\delta}{2}}] \end{align}

In the last relation, we can rewrite the term in the exponential if we consider that the distance to P from the center of the grating D is given as:

$$D = r_1 + \frac{1}{2}(N-1)dsin(\theta)$$

so that we therefore have:

$$E = E_0(D)e^{i(\omega{t}-kD)}[\frac{sin(N\delta/2)}{sin(\delta/2)}]$$

Now, the intensity is, as usual, given by the modulus of the field:

$$I = I_0[\frac{sin^2(N\pi\frac{dsin\theta}{\lambda})}{sin^2(\pi\frac{dsin\theta}{\lambda})}]$$

Now, we can note that the intensity is maximized whenever the denominator equals to zero, which corresponds to the condition:

$$\pi\frac{dsin\theta}{\lambda} = m\pi,~~~~m = 0, \pm{1},\pm{2},…$$

or, after canceling $\pi$ terms:

$$dsin(\theta) = m\lambda$$

We can see that this condition is governed by the $\theta$ term, so now we can consider what happens to the total I when it is maximized. To do so, we will consider the limit where $theta = 0$, which means that $m=0$. In this case, we can make the small-angle approximation such that $sin(\theta) \approx \theta$, so that the modulating term can be rewritten as:

$$\frac{sin^2(N\pi\frac{dsin\theta}{\lambda})}{sin^2(\pi\frac{dsin\theta}{\lambda})} \approx \frac{sin^2(N\pi\frac{d\theta}{\lambda}}{sin^2{\pi\frac{d\theta}{\lambda}} \approx \frac{(N\pi\frac{d\theta}{\lambda})^2}{(\pi\frac{d\theta}{\lambda})^2} = N^2$$

so therefore

$$I = N^2I_0$$

which, if we were to substitute in the condition $\delta/2 = m\pi$ instead, we would also have the same condition. All other values of $\theta$ would cause the denominator to be not equal to zero, which means that $I$ would be lowered. So the principle maxima occur when $dsin(\theta)=m\lambda$.

FERMAT’S PRINCIPLE REVISITED

FEYNMAN’S BOOK!!