6.2 Single Slit

**Figure 6.2:** Diffraction of light by a single slit of width . The horizontal scale is greatly compressed, and the vertical scale is greatly exaggerated.
$\includegraphics{1slit.eps}$

A single slit also exhibits interference effects due to the range of path lengths traversed by waves emitted over the entire area of the slit.
Instead of the discrete sum we used for the double slit, we will need an integral to take the superposition. The relevant differential is

$\begin{displaymath} d\psi = \frac{c}{r} e^{i(kr - \omega t)} dA \end{displaymath}$ (6.7)

where is a differential area element, and is a constant with the units of [ $\psi$ ] per area.
We will consider a rectangular long slit of length and width with so that interference is only observable along the axis parallel to the width. Then, we have

$\begin{displaymath} \psi = Lc \int_{-\frac{a}{2}}^{\frac{a}{2}} \frac{1}{r - x \sin \theta} e^{i(k(r - x \sin \theta) - \omega t)} dx \end{displaymath}$
Again, making the approximation that the path length difference $x\sin\theta$ only makes a significant contribution in the phase of the exponential, we have

$\begin{displaymath} \psi = \frac{Lc}{r} e^{i(kr - \omega t)} \int_{-\frac{a}{2}}^{\frac{a}{2}} e^{-ikx \sin \theta} dx \end{displaymath}$

which gives

$\begin{displaymath} \psi = \frac{Lc}{r} e^{i(kr - \omega t)} \frac{1}{-ik\sin\th... ...ac{ka}{2} \sin \theta} - e^{i\frac{ka}{2} \sin \theta} \right) \end{displaymath}$
Finally, recognizing the sine function in the complex exponentials, we find

$\begin{displaymath} \psi(\theta) = \frac{Lc}{r} \frac{\sin\left( \frac{ka}{2} \... ...theta \right)} {\frac{ka}{2} \sin \theta} e^{i(kr - \omega t)} \end{displaymath}$ (6.8)
Squaring and averaging over many periods, we find

$\begin{displaymath} \bar{I}(\theta) = \bar{I}_o \left( \frac{r_o}{r(\theta)} \ri... ...in \theta \right)} {\left( \frac{ka}{2} \sin \theta \right)^2} \end{displaymath}$ (6.9)

which for small angles becomes

$\begin{displaymath} \bar{I}(\theta) = \bar{I}_o \frac{\sin^2 \left( \frac{ka}{2} \sin \theta \right)} {\left( \frac{ka}{2} \sin \theta \right)^2} \end{displaymath}$ (6.10)
Interference minima fall at the roots of the intensity function,^6.3

$\begin{displaymath} \frac{ka}{2} \sin \theta_m = m \pi \hspace{1cm} (m=\pm1,\pm2,\pm3,...) \end{displaymath}$

or, in terms of wavelength,

$\begin{displaymath} a \sin \theta_m = m \lambda \hspace{1cm} (m=\pm1,\pm2,\pm3,...) \end{displaymath}$ (6.11)
You will locate the bright fringes for homework.

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