(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 $$E_P(t) = E_0 cos(kR - \omega t) + E_0 cos(kR - \omega t + \phi)$$ $$+ E_0 cos(kR - \omega t + 2\phi) + . . .$$ $$+ E_0 cos(kR - \omega t + (N - 1)\phi)$$ 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 $e^{iz} = cos z + i \space sin \space z$, where $i = \sqrt{-1}$. In this expression, cos $z$ is the $real$ $part$ of the complex number $e^{iz}$, 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 $$\sum _{n=0} ^{N-1} E_0 e^{i(kR-\omega t+n\phi)}$$ (c) Using the properties of the exponential function that $e^Ae^B = e^{(A+B)}$ and $(e^A)^n = e^{nA}$, show that the sum in part (b) can be written as $$E_0 ( {e^{iN\phi} - 1 \over e^{i\phi} - 1} )e^{i(kR-\omega t)}$$ $$= E_0 ({e^{iN\phi/2} - e^{-iN\phi/2} \over e^{i\phi/2} - e^{-i\phi/2}} )e^{i[kR-\omega t+(N-1)\phi/2]}$$ Then, using the relationship $e^{iz}$ = cos $z$ + $i$ sin $z$, show that the (real) electric field at point $P$ is $$E_P(t) = [E_0 {sin(N\phi/2) \over sin(\phi/2)} ] cos [kR - \omega t + (N - 1)\phi/2]$$ 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 $$I = I_0 [{ sin(N\phi/2) \over sin(\phi/2) }] ^2$$ 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. ($Hint$: Is I0 defined in the same way in both expressions?)