PHYS212 : Lectures

6.1 Double Slit, Take I

Figure 6.1: Diffraction of light by two narrow slits separated by a distance $d$. The horizontal scale is greatly compressed, and the vertical scale is greatly exaggerated.

• Waves from a single source is incident on a pair of narrow slits separated by a distance $d$. The transmitted intensity is observed at a point $P$ at an angular position $\theta$ relative to the normal as shown in Fig. 6.1.

• Provided the distance between the slits and point $P$ is less than the coherence length $\Delta x$ (see Eq. 2.78) of the source, interference effects are observed, because the path lengths of traversed by the waves from the two slits reaching point $P$ differ.

• The path length difference is
  

  $$r_2 - r_1 \approx d \sin \theta$$ (6.1)

  
  
  to a good approximation, provided the two rays are nearly parallel. Hence, we require that $d « r_1, r_2$.

• We will treat the slits as a pair of point sources, emitting power isotropically to the right of the slits. The wave function of one such point source is given by
  

  $$\psi = \frac{C}{r} e^{i(kr - \omega t)}$$ (6.2)

  
  
  where $C$ is a constant with the units of [$\psi$] per length. This is a spherical wave propagating along the radial unit vector $\hat{r}$. You will show that these are solutions to the wave equation in spherical coordinates for homework.

• At point $P$, we have the superposition
  
  $$\psi = \frac{C}{r_1} e^{i(kr_1 - \omega t)} + \frac{C}{r_2} e^{i(kr_2 - \omega t)}$$
  
  

• In the denominators, $r_1 \approx r_2 \approx r$ is an excellent approximation. However, if we observe interference effects, the path length difference must be significant relative to the wavelength of the incident light, so we have
  

  $$E = \frac{C}{r} \left( e^{i(kr_1 - \omega t)} + e^{i(kr_2 - \omega t)} \right)$$ (6.3)

  
  

• Taking the real part explicitly and following Eq. 2.2, Eq. 6.3 becomes
  $$\begin{eqnarray*} E &=& \frac{2 C}{r} \cos\left(\frac{k(r_2 - r_1)}{2}\right) ... ...os \left( \frac{kd}{2} \sin \theta \right) \cos(kr - \omega t) \end{eqnarray*}$$
  
  

• For all of the waves we have considered thus far, the intensity is proportional to $\psi^2$.^6.1 Generally, we are concerned with average intensity. Averaging over many periods removes the factor $\cos(kr - \omega t)$, leaving
  
  $$\bar{I}(\theta) = \frac{i_o}{r(\theta)^2} \cos^2 \left( \frac{kd}{2} \sin \theta \right)$$
  
  
  where $i_o$ is related to the power emitted by the source. ^6.2

• We can also relate $i_o$ to the intensity and $r(\theta)$ at $\theta = 0$ via
  
  $$\bar{I}_o = i_o r_o^2$$
  
  
  yielding
  

  $$\bar{I}(\theta) = \bar{I}_o \left( \frac{r_o}{r(\theta)} \right)^2 \cos^2 \left( \frac{kd}{2} \sin \theta \right)$$ (6.4)

  
  

• If we limit ourselves to small $\theta$ (or constant $r$), we make the approximation that $r$ does not vary significantly, yielding
  

  $$\bar{I}(\theta) = \bar{I}_o \cos^2 \left( \frac{kd}{2} \sin \theta \right)$$ (6.5)

  
  

• Dark fringes, or interference minima, are located at the roots of the intensity function, or
  
  $$\frac{kd}{2} \sin \theta_m = \frac{(2m+1)\pi}{2} \hspace{1cm} (m = 0, \pm1, \pm2,...)$$
  
  
  which is usually expressed in terms of the wavelength
  

  $$d \sin \theta_m = \left( m + \frac{1}{2} \right) \lambda \hspace{1cm} (m = 0, \pm1, \pm2,...)$$ (6.6)

  
  

• Note that if we were not concerned with predicting $I(\theta)$, we could have found Eq. 6.6 by requiring that that path length difference $r_2 - r_1$ be an odd number of half wavelengths, as is usually done in introductory physics courses.

Copyright © 2003-2009, Lewis A. Riley

Updated Wed Jan 18 09:51:28 2006