Week 4, Post 3

Worksheet 4, Problem 1

This question reviews Young's double-slit experiment, which demonstrates the wave behavior of light. When light is shined from a distance onto a pair of very thin slits, they form an interference pattern on a screen far away from the slits. If light were a particle, we would expect to see two bright lines on the screen, corresponding to the two slits. However, we actually see multiple bright lines, as illustrated in this diagram from Physics StackExchange:

Intensity of Diffraction on Double Slit

a)

To make things easier to understand, I'll refer to a drawing from my high school physics textbook (Giancoli 6th Ed.), which is much clearer than my attempt in Paint.

Figure 1: Wave behavior of light

We can label angle $\theta$ as the angle between the ray of light coming out of the slit and the horizontal. Figure 1d illustrates the geometry of the light rays. We see that there is a small triangle on the left, also with angle $\theta$ (because of similar triangles). The screen is placed at a distance L away from the slits, where L >> d, the distance between the slits. Because the screen is so far away, we can approximate the two rays $d_1$ and $d_2$ as parallel. The difference in path length between the two rays is therefore $d_2 - d_1$.

$\sin(\theta) = (d_2 - d_1)/d$

$d_2 - d_1 = d \sin(\theta)$

When the difference in path length $d_2 - d_1$ is a whole number multiple of the wavelength $\lambda$, we see constructive interference. When the difference in path length is off by half a wavelength, we see destructive interference.

Figure 1a shows that when $\theta = 0$, we get constructive interference, as there is no path length difference. The sine function is a phase shift of the cosine function, and $\cos(0^{\circ}) = 1$, so I am convinced that the brightness pattern of light is a cosine function.

b)

When we add a second set of slits, we get another cosine function brightness pattern. When we move the slits closer together, the brightness pattern becomes stretched out. As explained in Monday's crash course on Fourier transforms: when something is skinny in one domain, it Fourier transforms to something wide in the other domain.

Since we have two sets of slits now, their resulting interference patterns will be superimposed, adding up the two cosine functions. We can see this illustrated in the following graph:

Blue: $\cos(x)$ (first set of slits)
Red: $\cos(0.8x)$ (second set of slits, just inward)
Green: $\cos(x) + \cos(0.8x)$ (sum of both patterns)

The addition of the second set of slits causes the central peak to be brighter, but the peaks on the sides become attenuated as we go out further from the middle.

c)

Let's add even more cosine plots together and see what we get:

Sum of cosines

$\cos (x)+\cos(0.95x)+\cos(0.9x)+\cos(0.85x)+\cos(0.8x)+\cos(0.75x)$

This pattern resembles the sinc function, which is $\frac{\sin(x)}{x}$:

Sinc function

d)

The Fourier transform of top hat function (which is basically the ideal low pass filter) is the sinc function:

$sinc(x) = \frac{\sin(x)}{x}$

This finding agrees with the result from part c). Let's plot the two graphs together:

Red: sum of cosines
Blue: sinc function

e)

Following what we learned about Fourier transforms, a skinnier top hat would result in a wider sinc function, and vice versa. This means that the width of the top hat is inversely proportional to the distance between the two nulls.

The top hat function is basically a square wave, which is composed of a sum of cosines. To make a narrower top hat, we will need cosines of higher frequency, or lower wavelength. Likewise, higher wavelengths result in a wider top hat. This means that wavelength is directly proportional to width of the top hat.

We can combine these relations to get the following proportionality:

$d \propto \frac{\lambda}{D}$

Where $d$ is the width between the nulls.

f)

A telescope is basically a top hat in two dimensions, rather than just the 1-dimensional line that we've been working with up to this point. Now we have light coming in as a 2D plane to the telescope. From the examples we looked at, a wider top hat results in a narrower peak in the sinc function. Likewise, a wider telescope mirror would focus light into a sharper area, providing better resolving power so we can resolve smaller (farther away) objects.


Acknowledgements: Worked on this problem in class with Awnit, Simon, and August.

Worksheet 4, Problem 2

Here, we are comparing telescopes and their angular resolutions. We don't need to draw a diagram out for this one.

For each telescope, we are given its size and the wavelength it is tuned to observe:

• CCAT: $d_C = 25$m, $\lambda_C = 850$ microns
• MMT: $d_M = 6.5$m, $\lambda_M = 1.2$ microns

We are told that MMT operates in the J-band, which is a sub-band of IR, ranging from 1 to 1.4 microns. For the purposes of determining angular resolution, we'll use the middle of the J-band, which is 1.2 microns.

Using the Rayleigh's criterion, we can determine angular resolution for each telescope, given their respective $d$ and $\lambda$.

CCAT:

$\theta_C = 1.22 \frac{\lambda_C}{d_C} = 1.22 \frac{850 \times 10^{-6} \ m}{25 \ m} = 4.1 \times 10^{-5}$ rad

MMT:

$\theta_M = 1.22 \frac{\lambda_M}{d_M} = 1.22 \frac{1.2 \times 10^{-6} \ m}{6.5 \ m} = 2.3 \times 10^{-7}$ rad

We see that MMT is a smaller telescope, but its angular resolution is much smaller than that of CCAT because the wavelengths it observes are much smaller.

Acknowledgements: Worked on this problem in class with Awnit and Simon.

Largest telescope you can find 

and maximum resolution it is able to achieve

According to WorldAtlas.com, the biggest telescope on Earth is the Gran Telescopio Canarias (GTC). Their website states that it is a 10.4 m telescope with a segmented primary mirror.

The page on the GTC's optics says that its useful wavelength ranges from 365 nm to 25 µm.

The maximum resolution it can achieve comes from the smallest objects it can resolve, so let's use the 365 nm figure.

$\theta = 1.22 \frac{\lambda_{min}}{d_{GTC}} = 1.22 \frac{365 \times 10^{-9}}{10.4 \ m} = 4.28 \times 10^{-8}$ rad

We find that the GTC's maximum resolution is $4.28 \times 10^{-8}$ rad.

General Post

https://www.calcalistech.com/ctech/articles/0,7340,L-3756562,00.html

This article discusses the interesting and unique concerns over intellectual property rights in space.
IP rights are meant to incentivize innovation - patents grant the holder a limited monopoly, allowing them to profit from innovation and recoup the development costs that were spent on that new technology. Patents are especially important in sectors where research and development has high upfront costs, such as biopharmaceuticals.

Therefore, we would expect the field of space technology to have similar protections, since it has high risk and cost of development. However, it turns out that this field may not have as robust protection, because of loopholes that enable infringing parties to sidestep regulation. For example, consider a foreign company that develops a product that infringes on US patents. The device is sent to space on a Russian ship and brought on board the International Space Station (ISS). The device passes through the US modules of the space station before reaching the European modules. Whether or not this case presents an infringement of US patents is up for debate. Because patent protections are territorial, space makes things complicated. There are no international patents, so a US patent can only protect the holder from infringement in the US, but not in China, for example.

The US has been expanding its patent protections beyond its own territories. The Supreme Court found that if components of an infringing product are supplied from the US, the exporter of those parts can be liable for that infringement, even if the original infringement did not take place in the US. Specifically with regards to space, we can refer to 35 U.S. Code Section 105(a), protecting “a space object or component thereof under the jurisdiction or control of the U.S.” However, following precedent from maritime law, it stands to reason that a company can circumvent these regulations by registering with another foreign country. Private companies in the field of space technology are still new, so it remains to be seen if these unclear regulations have yet had an adverse impact on innovation.