Optics Problem 2

Main Author: David Vartanyan
Co-authors: Nathaniel Baskin, Mee Chatarin, Eric Mukherjee

Abstract: I will attempt to explain the basics of Young's double-slit experiment parallel with Fourier transforms so as to arrive at an understanding of astronomical instrumentation.

Introduction: Light behaves as both a particle and a wave. Young's double-slit experiment showed that the light patterns we get when we bombard photons through two small slits is not a simple summation of the pattens we get when we illuminate a single slit. Rather, we get a light and dark pattern suggesting that light behaves as a wave near the slit edges with interference patterns.

a. The brightest spot emerging from the slits occurs when the light emerging from both slits coincides constructively on the frame. Constructive interference is defined as when two or more waves meet in phase, with peaks aligned with peaks and troughs aligned with troughs. The amplitudes thus add, and we have a bright spot.









We see that r1 and r2 are equal in length, so the difference in path length that the waves travel is s=d*sin(Theta). Since L>>D, s~D*[Theta]. In order for the waves to be in phase,
D*[Theta]=n*[Lambda], where n is an integer, and [Lambda] is wavelength.

To understand why the brightness pattern of light is a cosine function as a function of angle, remember that light waves are electromagnetic fields, with field strength E(z)=E(0)*cos(kz-wt), where w is the angular frequency, t is time, k is the wavenumber 2Pi/[Lambda], and z is position.

We want the amplitudes of both waves, E(z1) and E(z2) to be in phase. Assume a light source infinitely far away. At concentric circles of increasing radius around it, each point on each circle is an equal radius away from the source and hence is at equal phase. As the light approaches the double slits we can assume a plane wave of equal phase arriving at the slits. The light from both slits approaches point P at same time t with same frequency w and wavenumber k. Now we want to make sure that the distance they travel, 'z', is an integer number of wavelengths so they will be in phase at P. Define kz-wt as phase Q.
Thus, the electric field strength of one wave E(z1) has to equal that of the second E(z2) for constructive interference.

E(z1)=E(0)*cos(Q1+deltaQ)=E(0)*cos(Q2)=E(z2); deltaQ=kd[Theta] since the only variable in phase is z=d*[Theta].
Consequently:
E(z1)=cos(Q+kd[Theta]) . By trigonemetric expansion,
so E=cos(Q)*cos(k*d*[Theta])-sin(Q)*sin(k*d*Theta)
Setting the phase equal to 0 at point P, we have
E=E(0)*cos(2Pi*d*[Theta]/[Lambda])

b. Jackie was fundamental in helping us understand the Fourier transform of waves. Assume we have a perfectly pitched voice (worse assumptions have been made in science). Plotting against frequency, we have a delta function with varying amplitude for a constant frequency. Now, switch to its Fourier pair and plot the sound wave against time or space. We now have a sinusoidally varying sound wave as a function of time/space. Consequently, we see that the Fourier transform of delta function pairs are sinusoidal waves. Now assume we have an additional pair of slits carved inward near the first pair.
We can consider this as 2 pairs of delta function pairs, each at a slightly different frequency (because of interference). Our interference pattern now depends on two cosine waves at different frequencies; as a result, our brightness/darkness fringes are spread farther out since peaks and troughs will align perfectly less frequently, giving us a bright center bright spot but dimmer peripheral spots.

c. Extending this idea for n slits with decreasing separation we end up linearly superimposing multiple cosine waves, essentially giving us a bright center fringe with increasingly damped peripheral oscillations. Each peak becomes narrower and more intense, and increasing the number of slits improves resolution.

 

d. The logical limit of this continuous set of slits is a top hat transmission function where we essentially slice a slit across d.  The Fourier transform of this is:

 

where as in d) we see a larger increasing resolution for each peak.

e. Professor Johnson helped us realize that the transmission function is diffraction limited. Rather than see a perfect geometrically focused point we see a blown up circle at each bright spot because light is bent around the edges of the slits. We cannot infinitely resolve an image because we are limited by the size of our telescope lens.

From a), E~cos(2Pi*d[Theta]/[Lamda], and d*[Theta]~n*[Lambda]. Since the slits are a distance d apart, and the nulls a distance d*[Theta] apart, our angular resolution is limited to [Theta]=[Lamda]/d.

Conclusion:

f. Assuming we can create an ideal telescope (preferably in space), the limiting factor on our resolution would be the diffraction limit. The larger the telescope lenses (d) or the smaller the wavelength [Lambda] (higher frequency), the better we can resolve two very close stars. Take for example a rotating binary system. Those two points stars emit at different frequency (if at all) because each is Doppler shifted differently. We would need a large telescope to be able to resolve such two points so close and so similar to each other.

An interesting result is that the Fourier transform of infinite delta functions (a constant line) is a delta function, since as we superimpose more delta functions we get increasing attenuation outside the central peak.

Also, single-electron emitters have also created diffraction patterns indicating that an electron can interact with itself and in essence pass through both slits. Just be sure not to look!