Intensity Pattern of $N$ Slits. (a) Consider an arrangement of $N$ slits with a distance $d$ between adjacent slits. The slits emit coherently and in phase at wavelength $\lambda$. Show that at a time $t$, the electric field at a distant point $P$ is
$$
\begin{aligned}
E_{P}(t)=& E_{0} \cos (k R-\omega t)+E_{0} \cos (k R-\omega t+\phi) \\
&+E_{0} \cos (k R-\omega t+2 \phi)+\cdots \\
&+E_{0} \cos (k R-\omega t+(N-1) \phi)
\end{aligned}
$$
where $E_{0}$ is the amplitude at $P$ of the electric field due to an individual slit, $\phi=(2 \pi d \sin \theta) / \lambda, \theta$ is the angle of the rays reaching $P$ (as measured from the perpendicular bisector of the slit arrangement), and $R$ is the distance from $P$ to the most distant slit. In this problem, assume that $R$ is much larger than $d$. (b) To carry out the sum in part (a), it is convenient to use the complex-number relationship $e^{i z}=\cos z+i \sin z,$ where $i=\sqrt{-1}$. In this expression, $\cos z$ is the real part of the complex number $e^{i z},$ and $\sin z$ is its imaginary part. Show that the electric field $E_{P}(t)$ is equal to the real part of the complex quantity
$$
\sum_{n=0}^{N-1} E_{0} e^{i(k R-\omega t+n \phi)}
$$
(c) Using the properties of the exponential function that $e^{A} e^{B}=e^{(A+B)}$ and $\left(e^{A}\right)^{n}=e^{n A},$ show that the sum in part (b) can be written as
$$
\begin{array}{l}
E_{0}\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)} \\
\quad=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i \phi / 2}}\right) e^{i[k R-\omega t+(N-1) \phi / 2]}
\end{array}
$$
Then, using the relationship $e^{i z}=\cos z+i \sin z,$ show that the (real) electric field at point $P$ is
$$
E_{P}(t)=\left[E_{0} \frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right] \cos [k R-\omega t+(N-1) \phi / 2]
$$
The quantity in the first square brackets in this expression is the amplitude of the electric field at $P$. (d) Use the result for the electric-field amplitude in part (c) to show that the intensity at an angle $\theta$ is
$$
I=I_{0}\left[\frac{\sin (N \phi / 2)}{\sin (\phi / 2)}\right]^{2}
$$
where $I_{0}$ is the maximum intensity for an individual slit. (e) Check the result in part (d) for the case $N=2$. It will help to recall that $\sin 2 A=2 \sin A \cos A .$ Explain why your result differs from Eq. (35.10) the expression for the intensity in two-source interference, by a factor of 4.