Worksheet 4, Problem 1: Double Slits

In one of the most elegant and fundamental experiments in physics, Thomas Young helped to demystify the behavior of light and photons through his double slit apparatus. In this problem, we explore some of these properties, as well as their relationship to the Fourier Transform, while also interpreting the implications on telescopes.

The configuration of Young's Experiment involves casting a light source through two slits in a screen and on to a phosphorescent backdrop. The pattern shown on the second screen reveals alternating dark and light bands. These occur as a result of constructive and destructive interference between waves of light traveling through the two slits. This phenomenon can is illustrated below.

http://www.askamathematician.com/2014/08/q-before-you-open-the-box-isnt-schrodingers-cat-alive-or-dead-not-alive-and-dead/

To help understand the distribution of this interference pattern, we can consider the path that each light source travels before reaching the final screen.

Simplified interference pattern distribution sketch 

a) Convince yourself that the brightness pattern of light on the screen is a cosine function. (HINT: Think about the conditions for constructive and destructive interference of the light waves emerging from each slit).

To approach this thought exercise, we began by drawing the image shown above of the simplified interference pattern distribution. The pattern created on the far screen is a result of the wavelength of the light source, \( \lambda\), the distance between slits, D, and the angle from the central axis, \(\theta\). At the central axis, equidistant between the two slits, the waves of light travel the same distance to the screen, and thus are in phase. This results in a constructive interference, which creates a light spot. However, at all other points on the screen, the two waves of light from the slits travel different distances to backdrop. When the distance traveled by the light is off by a full wavelength, there is again constructive interference from the waves. When the distance is off by a half wavelength, however, there is destructive interference, which results in a dark spot. This property of constructive and destructive interference of waves can be seen below.

http://method-behind-the-music.com/mechanics/physics/

To further understand this distribution pattern, we can apply some simple geometry to derive an equation that guides this behavior. From our simplified interference pattern sketch, we can see that a right triangle can be drawn whose shortest leg depicts the difference in distance traveled by the light waves. Using trigonometry, we can relate the sides of the triangle with the top angle, \(\theta\). \[Sin(\theta) = \frac{X}{D}\]
X represents the difference in distance traveled by the two waves, and for constructive interference to occur, we know that X must be equal to \( \lambda\). Since in the scale we are looking at, the distance between the two screens, L, is much larger than the wavelength of light, we know that the angle (\theta\) must be extremely small. As a result, we can make the following estimation. \[Sin(\theta) \approx \theta\]
This is known as the small angle approximation. This allows us to then simplify our relational equation to: \[\theta = \frac{\lambda}{D}\]
As we discussed above, when \(\theta = 0\), or \(\theta = \frac{\lambda}{D}\), we get constructive interference. Then, for each half phase of \(\lambda\), or \(\theta = \frac{\lambda +0.5}{D}\), we get destructive interference. Using the knowledge that this distribution will be continuous and periodic across the screen, as well as the fact that constructive interference occurs at \(\theta = 0\), we can represent this distribution as a cosine function.

b) Now imagine a second set of slits placed just inward of the first set. How does the second set of slits modify the brightness pattern on the screen?

Adding a second set of slits to the screen introduces two more waves to the interference area. Since the slits are positioned closer together, the value of D will be smaller. According to \(\theta = \frac{\lambda}{D}\), this will create a wider interference pattern than our first two slits. When we consider what effect this will have on the overall pattern, we can consider how this new interference pattern will interact with our original pattern. At \(\theta = 0\), there will still be constructive interference, which will further add to the peak from the original interference pattern. As we move away from the origin, however, we will start to destructive interference occurring from the wider waves, which will make the central peak narrower. Additionally, the constructive peaks that exist away from the origin will be smaller, since the waves aren't perfectly in sync. Ultimately, the greatest effect of these new slits will be making the central peak brighter and narrower.

c) Imagine a continuous set of slit pairs with ever decreasing separation. What is the resulting brightness pattern?

The more slit pairs that we add with decreasing separation will amplify the effect described in part b, continuing to reduce the outer peaks, while making the central peak brighter and narrower. In a sense, this is increasing our resolution of discerning objects at a distance, since we can width of our central peak is generated by a smaller and smaller \(\theta\).

d) Notice that this continuous set of slits forms a “top hat” transmission function. What is the Fourier transform of a top hat, and how does this compare to your sum from the previous step?

The Fourier transform of a top hat is a Sinc function, which is depicted below.

http://www.diracdelta.co.uk/science/source/s/i/sinc%20function/source.html#.VONqQ1MbDmY

This agrees very well with our pattern from the sum of slits in part d. We can see the very narrow, intense central peak, and the greatly reduced outer peaks in this pattern.

e) For the top hat function’s FT, what is the relationship between the distance between the first nulls and the width of the top hat (HINT: it involves the wavelength of light and the width of the aperture)? Express your result as a proportionality in terms of only the wavelenght of light λ and the diameter of the top hat D.

In the Sinc function, the x-intercepts correspond to fully destructive interference, or dark spots on our interference pattern. These intercepts occur at \(\frac{\lambda}{D}\). This translates to a top hat width of  \(\frac{D}{\lambda}\).

This is an incredibly important equation, because it tells us how to optimize our aperture size and light source wavelength to get the best resolution, and narrowest peak. To do so, we would want to have a very large D value, and very small wavelength, λ.

f) Take a step back and think about what I’m trying to teach you with this activity, and how it relates to a telescope primary mirror.

This is vitally important to the design of telescopes, because it essentially tells us how to make telescopes with a high resolution. If we want to be able to distinguish between entities, we need to have the narrowest light intensity peaks possible. As we have shown, in order to do this, we would need to utilize a very wide telescope.

Acknowledgements: Johnathan Budd & Willie Pirc