Worksheet 4, Problem 1: Angular Resolution

To understand the basics of astronomical instrumentation, it is useful to go back to the classic Young's double slit experiment. Draw a double slit setup as a 1-D diagram. The two slits are separated by a distance \(D\), and each slit is \(w\) wide, where \(w\ll D\) such that its transmission function is basically a delta function. There is a phosphorescent screen placed a distance \(L\) away from the slits, where \(L\gg D\). We'll be thinking of light as plane-parallel waves incident on the slit-plane, with a propagation direction perpendicular to the slit plane. Further, the light is monochromatic with a wavelength \(\lambda\).

You may be familiar with Thomas Young's double slit experiment which demonstrated the wave nature of light. In this experiment, Young directed light through two small slits and placed a screen a little distance behind the slits that would detect the incident light from the slits. What he found was an interference pattern of bright and dark fringes similar to what you might see in intersecting water waves/ripples. This problem explores some of the physics behind the double slit experiment and applies it to observational astronomy and telescopes.

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).

Figure 1. Double Slit Experiment

First, using the fact that \(L\gg D\), we can assume that \(\theta\) is very small and that \(r_1\) and \(r_2\) are parallel. This makes the small triangle in the diagram above look like the following:

Figure 2. Approximations of Angles in Diagram for Double Slit Experiment

The light passing through the bottom slit has to travel further than the upper slit in order to reach the point \(y\) on the screen shown in Figure 1. This difference in distance will cause an offset in the wavelengths of the light passing through the two slits, which, in turn, will cause interference in places where the peaks and troughs don't align. In order to understand the pattern of interference observed on the screen, we must find the extent of the offset. Since \(\theta\) is small, we can approximate that the length of \(r_2\) is greater than that of \(r_1\) by \(d\), and from Figure 2, we can see that \(d=D\sin{\theta}\). 

Now, to find the effect of the offset, \(d\), on the interference of the light, we can derive equations for the conditions of constructive (bright fringes) and destructive (dark fringes) interference. For constructive interference, the peaks must perfectly align, so \(d\) has to be an integer multiple of the wavelength.

\(d=n\lambda\)

\(D\sin{\theta}=n\lambda\)          

Where \(n=0,1,2,3\ldots\).

And for destructive interference, the peaks of one wave, should line up exactly with the troughs of the other, so they should be offset by half a wavelength.

\(d=(n+\dfrac{1}{2})\lambda\)

\(D\sin{\theta}=(n+\dfrac{1}{2})\lambda\)      

For \(n=0,1,2,3\ldots\).

Since \(\theta\) is small, we can make the approximation that \(R\approx L\). Using this approximation, for constructive interference, we get:

\(D\sin{\theta}=D\dfrac{y}{L}=n\lambda\)

Which gives us the condition:

\(y_{con}=n\dfrac{\lambda L}{D}\)

And for destructive interference:

\(y_{des}=(n+\dfrac{1}{2})\dfrac{\lambda L}{D}\)

From this, we can already see a periodic nature for the constructive and destructive interference of the waves, which implies that the brightness pattern may be a trigonometric function. Since we have a bright fringe at \(y=0\) for \(n=0\), we would have a cosine function.

The pattern that would be seen on the screen is shown in the following figure, where the right pane depicts the bright and dark spots on the screen from the interference pattern.

Figure 3.
Image Credit: http://ipodphysics.com/prop-of-light-youngs-double-slit.php

Also, note that from the conditions we derived for constructive and destructive interference, we can see that the further away the screen is from the slits (for larger \(L\)) and the larger the wavelength of the light, the more spread apart the bright fringes are. However, the further the slits are apart (for smaller \(D\)), the closer the bright fringes are together.

Another way of understanding the cosine-form brightness pattern on the screen is through Fourier transforms. The two slits can be modeled as delta functions and taking their Fourier Transform, we get a cosine function, as shown in the image below.

Figure 4. The Fourier transform of two delta functions is representative of the interference pattern by two slits
Image Credit: Gray's Stellar Photospheres

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? 

Interference patterns are created by 'adding' the light waves together (a wave can be represented as a sinusoidal function and the overlap of waves which causes interference can be though of as the sum of each light wave's sine function). Due to this fact, we can treat the second set of slits as a separate occurrence, and then add the brightness patterns of both sets of slits together to find what the brightness pattern would be for all four slits. As we found in part (a), the smaller the distance between the two slits, the more spread apart the brightness pattern is, meaning that the second set of slits just inside of the first set will have a smaller separation between slits and therefore a wider resulting cosine function as its brightness pattern. When the two interference patterns are added together, we get something that resembles this:

Figure 5. Superposition of Cosines with Different Wavelengths

So the superposition of the two interference patterns caused the central maximum brightness (at \(n=0\)) to increase in intensity and the other bright fringes decrease in intensity.

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

If we continue the process discussed in part (b) by adding more cosine functions that are even more spread each time, eventually, the resulting brightness pattern will be focused more towards the center, with a very bright central maximum \((n=0)\) and almost imperceptible secondary maxima. The resulting brightness pattern will look like a sinc function as depicted below:

Figure 6. The sinc function is representative of the brightness pattern that would be seen on the screen.

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 set?

We can use Fourier transforms to find the resulting brightness pattern on the screen. As the number of slits goes towards infinity with their separation nearing zero, the resulting system of slits approaches a "top hat," or a larger opening with width \(D\). 

Figure 7. Limit as number of slits approaches infinity and separation approaches zero

The Fourier transform of this top hat is the sinc function, as shown in the image below:

Figure 8. Fourier Transform of Top Hat
(Note: for consistency, we will refer to \(W\) in the image as \(D\))
Image Credit: Gray's Stellar Photospheres

Notice that the distance from the central peak to the first dark fringe, or null, is given as \(\dfrac{1}{D}\).

e) For the top hat function's Fourier transform, 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 you result as a proportionality in terms of only the wavelength of light \(\lambda\) and the diameter of the top hat \(D\).

As explained in part (d), the slit approaches a top hat shape, which is essentially a slit with width \(d\). The following image shows what the brightness pattern would look like and includes the equation of the condition for minima, or nulls.

Figure 9. Single Slit Diffraction
(Note: for consistency, we call the width of the slit \(d\) instead of \(a\), and the distance to the screen \(L\) instead of \(D\))
Image Credit: http://www.gitam.edu/eresource/Engg_Phys/semester_1/optics/diffrac_files/sinslit.gif

Incidentally, the equation for nulls in a single slit setup is the same as the equation for the double slit setup. The distance from the central maximum to the nulls is given by:

\(y=n\dfrac{\lambda L}{D}\)

Additionally, we know from part (a) that larger wavelengths result in a more spread apart brightness pattern, and from the Fourier transform in part (d) we saw that the distance to the first null is inversely proportional to the distance \(D\) between the two outer slits.

Using the conclusions from the single slit setup and from our observations in parts (a) and (d), we can see that the distance to the first null is directly proportional to wavelength and inversely proportional to the width of the slit.

\(y\propto \dfrac{\lambda}{D}\)

f) Take a step back and think about what this activity is trying to teach you and how it relates to a telescope primary mirror. 

Applying the this problem to telescopes, a telescope's primary mirror is essentially the screen in the double slit experiment, and the telescope's aperture is like the slit with width \(D\). This means that the resulting image will have a brightness pattern which can be determined using a similar method as we did for the top hat slit with a width. The difference in this case is that the 'slit' is in two dimensions, so the Fourier transform will be a jinc function, shown below:

Figure 10. The jinc function is representative of the brightness pattern that would be seen from the telescope.
Image Credit: (Left) http://upload.wikimedia.org/wikipedia/en/e/e6/Airy-3d.svg (Right) http://upload.wikimedia.org/wikipedia/commons/1/14/Airy-pattern.svg

The telescope's ability to resolve objects further away (its angular resolution), with smaller angular size, is determined by the central maximum peak (which is the central bright dot in the right image in Figure 10. The brighter (and therefore narrower) the peak, the more focused the image in the telescope and the greater the telescopes ability to resolve. This implies that the angular resolution \(\theta\) is directly proportional to the distance to the null \(y\) in the brightness pattern. As we found in part (e), \(y\) is directly proportional to wavelength and indirectly proportional to width, or diameter \(D\) in this case. So we have:

\(\theta \propto \dfrac{\lambda}{D}\)

This gives us the minimum possible angular size that the telescope can resolve at a given wavelength. 

I would like to thank Charles Law, Daniel Chen, and Andy Mayo for their help in solving this problem.