I am trying to derive the intensity variation function for a single slit diffraction. Diagram

Sorry for the poor diagram...

So I decided to take the amplitudes of the waves originating from the slit on the left (wherein the variable that denotes distance within the slit is $l$) and integrate the amplitudes over the entire slit width, taking some point at a distance $x$ on the screen to achieve the resultant amplitude of the waves that strike the screen. With this function, I decided I would use the standard expression for intensity (i.e. $I=\kappa A^2)$

The amplitude for a wave originating from a point on the slit should be: $$ y=a\sin{kr}$$ where $r$ is the distance between the point of origin on the slit and point of contact on the screen (and $k$ is the angular wave-number). So: $$ r^2=D^2+(x+l)^2$$ and on approximating: $$ r\approx D+\frac{1}{2D}(x+l)^2$$ So I took the amplitude function (for the screen) as $A(x)$ and: $$ A(x)=a\int_{-l/2}^{l/2}\sin{kD+\frac{k}{2D}(x+l)^2} dl$$ substituting $k(x+l)/2D=u$ (ignoring limits for now): $$ A(x)=a\sqrt{\frac{2D}{k}}(\sin{kD}\int_{l_1}^{l_2}\cos{u^2}du+\cos{kD}\int_{l_1}^{l_2}\sin{u^2}du)$$ I looked these integrals up so I know that they are Fresnel Integrals, but more importantly that they are transcendental functions.

So my questions are:

  1. Are my assumptions flawed?
  2. Is there a flaw somewhere in the procedure?
  3. If what I've done is correct, how shall I proceed?
  • $\begingroup$ There is a flaw in your procedure. You appear to be integrating over x (which would mean that A(x) doesn't depend on x), but what you want is to express how r varies depending on where the light source is, along the slit. This is what StarDrop's answer is doing. $\endgroup$
    – Dr Chuck
    Commented Oct 27, 2015 at 17:06
  • $\begingroup$ @DrChuck : So sorry... I missed the $dl$ (I have edited it in the question). As you can see, I was integrating over $l$, not $x$, to find the amplitude at a point $x$ on the screen. So, $x$ is taken to be constant for the procedure. $\endgroup$ Commented Oct 27, 2015 at 20:40
  • $\begingroup$ The solution is different whether you are interested in the case where $D$ is finite (Fresnel diffraction) or "infinite" (the assumption that $\ell << D$ leads to the Fraunhofer diffraction case, which is the sinc function). Which of these are you trying to get to? Because the Fresnel case is indeed not a closed form - if that is what you are after, you might want to read up on the Cornu spiral... $\endgroup$
    – Floris
    Commented Jun 16, 2016 at 23:36
  • $\begingroup$ @Floris : thank you. I've also been told that the flaw in my approach is that I've used a plane wave equation whereas I should be using a spherical wave equation $\endgroup$ Commented Jun 16, 2016 at 23:39
  • $\begingroup$ The spherical assumption would add a $1/R^2$ term which changes very little over the range of values of your integrand (over the slit), and slightly more over the range of values of $x$ (the position along the screen). Both of these terms vanish in the Fraunhofer case. You didn't answer my question - are you actually looking for a solution to the Fresnel (finite D) case? $\endgroup$
    – Floris
    Commented Jun 16, 2016 at 23:45

2 Answers 2


Try using this method.

enter image description here

To study diffraction of light, laser light is passed through a narrow single slit and the diffraction pattern is formed on a distant screen. An imaginary reference line is drawn perpendicularly from the center of the slit out to the screen (see Figure 3), which is a distance L away. The intensity variation of the diffraction pattern can then be measured accurately as a function of the distance y from the reference line. In the theoretical description of the diffraction pattern, however, it is more convenient to quantify the light intensity as a function of the sine of the angle θ defined accordingly by

$sin θ = y/\sqrt{y2+L2}$

The theory of diffraction predicts that the spatial pattern of light intensity on the viewing screen by a light wave passing through a single rectangular-shaped slit is given by

enter image description here

(4) where I0 is the light intensity at θ = 0◦ and the quantities in parentheses are in radians.


  • 1
    $\begingroup$ Thanks for the input, but please note that I wished to arrive at the expression you've mentioned (or something similar) through a different approach, namely the one I've detailed in the question. $\endgroup$ Commented Oct 27, 2015 at 15:08

It is not possible to write a closed form equation for the Fresnel diffraction pattern. Usually one will use the Cornu spiral to evaluate problems like this.

The Cornu spiral is a graphical tool that maps the phase / amplitude contribution of a infinitesimal element of the aperture.

enter image description here

(image by R. Nave, from http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/cornu.html#c2 )

To get the phase and amplitude at a particular point on the screen, you need to determine the position of the left and right hand edge of the screen in terms of normalized parameters $v$, which represents the phase difference from the point on the aperture to the point on the screen. In your case, you have a plane wave incident, and the parameter $v$ is

$$v_± = \frac{\sqrt{D^2 + (x±\frac{\ell}{2})^2}-D}{\lambda}$$

You then draw the line from $v_-$ to $v_+$ to get a line that represents both amplitude and phase of the wave at a given point on the screen.

A derivation of the shape of the curve (which is a representation of the Fresnel integrals, as you correctly found) can be found here.


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