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Single-Slit Experiment

One should discuss the nature of the Fraunhofer diffraction pattern produced by a single slit3. Some important features of this problem can be deduced by examining waves coming from various portions of the slit, as shown in Figure 1a. According to Huygens´ principle, each portion of the slit acts as a source of waves. Hence, light from one portion of the slit can interfere with light from another portion, and the resultant intensity on the screen will depend on the angle $\theta$.

Figure 1: Diffraction by single narrow slit and its intensity distribution [3]
\includegraphics[width=\textwidth]{pic/jik2}

To analyze the diffraction pattern, it is convenient to divide the slit in two halves, as in Figure 1a. All the waves that originate from the slit are in phase. Consider waves 1 and 3, which originate from the bottom and center of the slit, respectively. Wave 1 travels farther than wave 3 by an amount equal to the path difference $(a/2)\sin{\theta}$, where $a$ is the width of the slit. Similarly, the path difference between waves 2 and 4 is also $(a/2)\sin{\theta}$. If this path difference is exactly one half of a wavelength $(\lambda/2)$, the two waves cancel each other and destructive interference results. This is true, in fact, for any two waves that originate at points separated by half the slit width because the phase difference between two such points is $180^{\circ}$. Therefore, waves from the upper half of the slit interfere destructively with waves from the lower half of the slit when
\begin{displaymath}
\frac{a}{2}\sin{\theta}=\frac{\lambda}{2}
\end{displaymath} (7)

or when
\begin{displaymath}
\sin{\theta}=\frac{\lambda}{a}
\end{displaymath} (8)

If we divide the slit into four parts rather than two and use similar reasoning, we find that the screen is also dark when
\begin{displaymath}
\sin{\theta}=\frac{2\lambda}{a}
\end{displaymath} (9)

Likewise, we can divide the slit into six parts and show that darkness occurs on the screen when
\begin{displaymath}
\sin{\theta}=\frac{3\lambda}{a}
\end{displaymath} (10)

which can be simply understood from [3] Therefore, the general condition for destructive interference is
\begin{displaymath}
a \sin{\theta}=m \lambda \quad \quad m=\pm1,\pm2,\pm3,\ldots
\end{displaymath} (11)

Equation 11 gives the values of $\theta$ for which the diffraction pattern has zero intensity, that is, where a dark fringe is formed. However, it gives no information about the variation in intensity along the screen. The general features of the intensity distribution along the screen are shown in Figure 1b. A broad central bright fringe is observed, flanked by much weaker, alternating bright fringes. The various dark fringes (point of zero intensity) occur at the values of $\theta$ that satisfy Equation 11. The position of the points of constructive interference lies approximately halfway between the dark fringes. Note that the central bright fringe is twice as wide as the weaker bright fringes. In this section was referred to the alternating dark and bright bands on the screen as a diffraction pattern, while in the case of Young´s double-slit experiment (see fig. 2) a similar pattern is referred to

Figure 2: Young´s double-slit experiment [3]
\includegraphics[width=0.9\textwidth]{pic/young_ex}

an interference pattern. Also, there should be noted that the derivation of the equations associated with interference and diffraction effects are similar in the way that they consider the addition of waves that are either in or out of phase. The effects of interference and diffraction can be distinguished as follows. When the waves to be added come from two or more openings (sources), as in Young´s experiment, the resulting pattern is called an interference pattern. When various portions of a single wave interfere, as in this section, the result is a diffraction pattern.
next up previous
Next: Grating Up: Gratings Previous: Diffraction phenomena
root 2002-05-23