(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
$$
\begin{aligned}
E_{0} &\left(\frac{e^{i N \phi}-1}{e^{i \phi}-1}\right) e^{i(k R-\omega t)}
\\\
&=E_{0}\left(\frac{e^{i N \phi / 2}-e^{-i N \phi / 2}}{e^{i \phi / 2}-e^{-i
\phi / 2}}\right) e^{j[k R-\omega t+(N-1) \phi / 2]}
\end{aligned}
$$
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 \(4
.\) (Hint: Is \(I_{0}\) defined in the same way in both expressions?)
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