1
$\begingroup$

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?
$\endgroup$
  • $\begingroup$ 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 Oct 27 '15 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. $\endgroup$ – Prish Chakraborty Oct 27 '15 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 Jun 16 '16 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 $\endgroup$ – Prish Chakraborty Jun 16 '16 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 Jun 16 '16 at 23:45
0
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

http://www.physics.nus.edu.sg/~ephysics/documents/PC2232-Diffraction-revised.pdf

$\endgroup$
  • 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. $\endgroup$ – Prish Chakraborty Oct 27 '15 at 15:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.