Ever seen the pictures of the dozens of satellite dishes all in a row, here’s why

We are bounded in a nutshell of Infinite Space: Worksheet # 4, Problem #3: Ever seen the pictures of the dozens of satellite dishes all in a row, here’s why

Problem 3: Interferometers. Wait, interfere-what?

Interferometers are giant telescopes comprised of multiple smaller telescopes. Imagine two 1-meter telescopes in the CHARA array, separated by 330 meters (you should look up CHARA later). Use the double-slit experiment to think about how these two telescopes might be able to measure the angular diameter of a star with R = 5 Rd at a distance of 150 light years. What is the maximum wavelength λ at which this star can be resolved? Think about plane-parallel waves arriving from a star that is barely resolved. Instead of a point-source, the star will be a very narrow top-hat on the sky.

In case you haven’t, this is the Very Large Array (VLA), in New Mexico, which is one of the examples of how the properties of light are used to create extraordinary measurements and images from space. Knowing light has a double property of being a particle (known as a photon) and a wave, we can take advantage of the wave property in order combine many sources of the light using constructive interference, a property all waves have (either electromagnetic or vibrational).

Constructive interference is when you take two or more waves that have the same phase (time at which their valleys and peaks pass a certain point), and combine them so their amplitudes are added on to each other to make a more powerful signal. When you have several telescopes, or Satellites, you can receive a signal from a distant star at several points, and by catching each wave at just the right point, you can combine several signals to get a clearer image. (A full explanation of the basic theory for how this works can be found here: http://hyperphysics.phy-astr.gsu.edu/hbase/phyopt/slits.html).

Now, as to the problem at hand, we have a situation like this:

Where the distance from the star to the two satellites is so enormous that the distances from each dish is virtually identical, which also stems from a small angle approximation from the incredible distance. From some other facts from the question, we know the distance to the star is 150 light years, which can be turned into meters with: \[ c_{Speed~of~Light} \times t_{Seconds ~in ~a ~year} \times \delta_{Distance ~to~Star} [ly]  =\delta_{Distance ~to~Star} [m] \] \[ 3\times 10^8 \frac{m}{s} \times 3.15 \times 10^7 \frac{s}{year} \times 150 ly = 1.42 \times 10^{18} m .\]

Next, we find the size of the star in meters. We know this star’s diameter is \(5R_\odot\) ad we must find the diameter. So: \[R_{\odot} = 6.96 \times 10^8 m\] \[  D_\star = 2 \times 5 \times R_{\odot} \] \[ D_\star = 7.0 \times 10^9 m ,\]

Imagining the star’s diameter to be the opposite side, and the distance to it the hypotenuse of a triangle with an extremely small angle, we can use the following approximation to find the angle: \[\theta = frac{D_\star}{ \delta_{Distance ~to~star}} \] \[\theta = \frac{7.0 \times 10^ 9 m}{1.42 \times 10^{18} m}\] \[ \theta = 4.93 \times 10^{-9} .\]

And now we can use the equation established by the Double Slit experiment for Constructive Interference: \[ \theta = \frac{\lambda}{D} ,\] where D is the distance between the two signal recipients , and \(\lambda\) is the wavelength of the signal, what we are trying to find. So, using this equation, we have: \[ \theta \cdot D = \lambda\] \[ \lambda = 4.93 \times 10^{-9} \cdot 200 m \] \[ \lambda = 9.86 \times 10^{-7} = 986 nm ,\] the maximum wavelength which can be resolved.