Fraunhofer Diffraction

Fresnel diffraction gives a picture of how a beam may be diffracted, and therefore how it may look on a screen, in the near field. Fraunhofer diffraction then describes how light would look on a screen many Rayleigh ranges away. We can derive the formula for Fraunhofer diffraction from the Fresnel diffraction integral, however, so don’t fret that you have to wrap your brain around a whole new physical picture!

Table of Contents:

  1. Derivation
  2. Example: Rectangular Aperture
  3. Example: Circular Aperture


To derive the Fraunhofer diffraction integral, we will start with the Fresnel integral just derivated:

\begin{equation} E(x,y,z) = i\frac{e^{-ikz}}{\lambda{z}}\iint^{\infty}_{-\infty}E(x’,y’;0)e^{-ik[(x-x’)^2+(y-y’)^2]/2z}dx’dy’ \end{equation}

If we expand the phase term:

\begin{equation} k\frac{(x-x’)^2+(y-y’)^2}{2z} = k[\frac{x^2+y^2}{2z}-\frac{xx’+yy’}{z}+\frac{x’^2+y’^2}{2z}] \end{equation}

Since x’ and y’ give the coordinates for the aperture, if the position of the observation screen at a distance z away from the aperture is much larger than the size of the aperture, then the last term becomes very small such that $1 >> \frac{k(x’^2+y’^2)}{2z}$, or

\begin{equation} \frac{z}{k} >> \frac{x’^2+y’^2}{2} \end{equation}

This is the simplifying assumption of the Fraunhofer far field picture: the observation screen is at such a large distance from the aperture that the size of the aperture effectively vanishes. We can state this assumption more precisely by letting $W = max(x’,y’)$ be the maximum value of x’ and y’ in the aperture. We then require:

\begin{equation} z >> \frac{kW_{max}^2}{2} = \frac{\piW^2_{max}}{\lambda} \end{equation}

We can note here that the form of this condition looks exactly like $z >> z_R = \pi\omega_0^2/\lambda$ we had from Gaussian beams. Thus the pattern is in the far field if the field is “many Rayleigh ranges” away from the aperture.

For a larger aperture, it can be hard to satisfy this condition. In order to make measurement of the far field diffraction easier, we can use a lens since it effectively “puts the screen at infinity”! (This is because the role of a lens is to collimate each of the rays i.e. smooth all of the wavefronts such that they would have a wavefront curvature $\rightarrrow \infty$ like they would have if the screen was at infinity.)

By rewriting the phase, the diffraction formula is now:

\begin{equation} E(x,y;z) = \frac{ie^{-ikz}}{\lambda{z}}e^{-ik(x^2+y^2)/2z}\times\iint_AE(x’,y’;0)e^{ik(xx’+yy’)/z}dx’dy’ \end{equation}

We can see from this that the effect of the Fraunhofer diffraction is to give a linear phase variation in the integral, as opposed to quadratic in the Fresnel diffraction case. We can then look at the linear phase terms $kxx’/z$ and $kyy’/z$. Starting with the x-term, we will have $tan\theta_x \approx sin\theta_x \approx \theta_x = x/z$, and a similar relation for $kyy’/z$, just in terms of “y” of course. The phase term for the x-component is then just $k\theta_x{z}$, which we can then approximate as $k\theta_x \approx ksin\theta_x = k_x$, which is just the spatial frequency in the x-direction. Again, of course, we would have the same for the y-direction: $kyy’/z=k\theta_yy’=k_yy’$. The Fraunhofer integral is thus:

\begin{equation} E(x,y,z) = i\frac{e^{-ikz}}{\lambda{z}}e^{-ik(x^2+y^2)/2z}\times\iint_AE(x’,y’;0)e^{i(k_xx’+k_yy’)}dx’dy’ \end{equation}

We’ve done it! The main thing to take away from the Fraunhofer diffraction integral is that the diffraction pattern is proportional to the Fourier decomposition of the spatial frequencies required to “construct” the aperture field E(x’,y’;0). It is also important to note that every spatial frequency maps to a particular position (x,y) at a distance z in the observation plane.

So, the prescription to find the field at an observation screen is to first find the field at the aperture, and then solve the integrals to find what the diffraction pattern in the far field will be.


Consider a rectangular aperture with an incident uniform plane wave. Let the box length be $2x_0$ and the box width be $2y_0$, and let the box be centered at the origin. The field at z = 0 is would then be equal to $E_0$ if the coordinate in the aperture $|x’| \leq x_0$, and $|y’| \leq y_0$. A problem of this form is called a “rect” function and can be written explicitly as:

\begin{equation} E(x,y;z) = iE_0\frac{e^{-ikz}}{\lambda{z}}e^{-ik(x^2+y^2)2z}\int_{-x_0}^{x_0}e^{ik_xx’}dx’\int_{-y_0}^{y_0}e^{ik_yy’}dy’ \end{equation}

The integration of this would then yield two sinc functions. The intensity for this case is given as:

\begin{equation} I = I_0(sinc^2\frac{kx_0}{z}x)(sinc^2\frac{ky_0}{z}y) \end{equation}

Note that sinc(0) = 1, so the intensity is equal to $I_0$ at the center of the screen. The width of the pattern is the distance between the zeros, $\lambda{z}/x_0$, which means that a smaller slit will lead to a larger pattern and vice versa!


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