Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

To understand the phenomenon of diffraction as an interference effects of several dipole oscillators (like in case of several symmetrical, not sawtooth, scratches in a diffraction grating), we consider a linear array of $N$ particles, each of which acts as a source of EM wave, and their interference produces the diffraction pattern. This can also be used in case of width-zero slits, $N$ in number, arranged linearly.
For a single slit diffraction pattern, we use an array of infinite sources (with separation $d\rightarrow 0$)along the width of the slit, a justification lies either in the Huygen's principle (sources of secondary wavelets) or as explained by assuming a plug at the slit in Feynman's lectures on physics.

How do we extend this methodology to study the diffraction pattern at the edge of an opaque object, producing a shadow? Do we consider infinite sources extending from the edge till infinity? Or we consider sources only upto some defined distance?

enter image description here enter image description here

share|improve this question
2  
I've added some background on the Cornu spiral, which may interest you. –  WetSavannaAnimal aka Rod Vance Dec 21 '13 at 2:17

1 Answer 1

up vote 3 down vote accepted

One way to study this case is through the numerical analysis of diffraction, as described in my other answer to you.

You can also do this pretty much as you describe through Huygens's principle or as Feynman describes in his popular QED book. If you set up an equation to describe what you've said, you'll see that the amplitude at a point with transverse co-ordinate $X$ on a screen at an axial distance $d$ from the plane with the knife edge is:

$$\psi(X) \approx\int\limits_0^\infty\exp\left(i\,k\,\sqrt{(X-x)^2+d^2}\right)\,{\rm d}\,x\qquad(1)$$

where the line of sources runs from $x=0$ to $w$ (the width of the bright region), where we can take $w\to+\infty$ if we like. We have neglected the dependence of the magnitude of the contribution from each source on the distance $\sqrt{(X-x)^2+d^2}$. This is because we now invoke an idea from the method of stationary phase, whereby only contributions from the integrand in the neighbourhood of the point $x=X$ where the integrand's phase is stationary will be important. Thus for $x\approx 0$ we can assume $|X-x|\ll d$ and so:

$$\psi(X) \approx\int\limits_0^w\exp\left(i\,k\,\frac{(X-x)^2}{2\,d}\right)\,{\rm d}\,x\qquad(2)$$

an integral which can be done in closed form:

$$\begin{array}{lcl}\psi(X) &\approx& \sqrt{\frac{2\,d}{k}}\int\limits_{\sqrt{\frac{k}{2\,d}}(X-w)}^{\sqrt{\frac{k}{2\,d}} X} e^{i\,u^2}\,{\rm d}\,u \\ &=& \sqrt{\frac{d}{2\,k}} e^{i\frac{\pi}{4}} \sqrt{\pi} \left({\rm Erf}\left(e^{3\,i\frac{\pi}{4}}\sqrt{\frac{k}{2\,d}}(x-w)\right)-{\rm Erf}\left(e^{3\,i\frac{\pi}{4}}\sqrt{\frac{k}{2\,d}}\, x\right)\right) \\ &=& \sqrt{\frac{d}{2\,k}} \left(C\left(\sqrt{\frac{k}{2\,d}} X\right) + i\,S\left(\sqrt{\frac{k}{2\,d}}X\right) -\right.\\ & & \qquad\left.\left(C\left(\sqrt{\frac{k}{2\,d}}(X-w)\right) + i\,S\left(\sqrt{\frac{k}{2\,d}}(X-w)\right)\right)\right)\end{array}\qquad(3)$$

where:

$$\begin{array}{lcl} C(s) &=& \int\limits_0^s\, \cos(u^2)\,{\rm d}\,u\\ S(s) &=& \int\limits_0^s\, \sin(u^2)\,{\rm d}\,u\\ \end{array}\qquad(4)$$

where $C(s)$ and $S(s)$ are called the Fresnel integrals.

If I plot the squared magnitude of this function (related to the Fresnel integrals) in normalised units when $k=d=1$ and $L\to\infty$ (noting $C(\infty)=S(\infty) = -1/2$) for $X\in[-10,20]$ I get the following plot:

Knife Edge Diffraction

which I believe is exactly your plot with a shrunken horizontal axis (yours is likely mine with the transformation $x_S = 2\,\pi\,x_R$ where $x_S$ is Satwik's $x$-co-ordinate and $x_R$ Rod's).

Footnote: One of the loveliest curves from eighteenth and nineteenth century mathematics is the Cornu Spiral, which is a special case of the Euler Spiral. $\psi(X)$ in (3) traces a path in the complex plane parameterised by $X$, which turns out to be the arclength $s$ of the spiral path in $\mathbb{C}$ such that:

$$\begin{array}{lcl}x &=& {\rm Re}(\psi(s)) \propto C(s) + \frac{1}{2}\\ y &=& {\rm Im}(\psi(s)) \propto S(s) + \frac{1}{2}\end{array}\qquad(4)$$

and I plot the normalised and shifted path $z = C(s) + i\,S(s)$ I get the lovely spiral below. The curly bits spiral all the way in to $\pm(1+i)/2$ as $s\to\infty$. The shifting and then taking magnitude squared explains why the intensity plot above is not symmetric about $X=0$, oscillating as $X\to\infty$ and dwindling monotonically as $X\to-\infty$.

Cornu Spiral

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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