interferometer

**Figure 1:** *The experimental setup, consisting of the 0.83 meter antennas, 4 GHz C-band feed horns and low-noise block converters (from 4 to 1 GHz), coaxial cables, T-junction, total power box, and voltmeter.*
$\includegraphics[scale=1.0]{interferometer-small.ps}$

Introduction: In the previous observational exercise, you used a single 0.83-meter dish to measure the equivalent blackbody temperature of the Sun at a frequency of 4 GHz. In this exercise, we will combine two such telescope beams in phase, thereby creating an interferometer and using it to measure the angular size of the Sun.

Interferometer Basics: As you have learned in basic physics courses, light passing through adjacent slits can interfere constructively or destructively, as a result of the path difference being an integral number of wavelengths or half-wavelengths, respectively. An excellent example of an instrument used to measure distances to a small fraction of a wavelength is a Michelson Interferometer, a device in which two light beam path distances can be carefully altered with respect to each other. The resulting pattern of constructive and destructive interference fringes can be used to measure stresses in physical objects, vibrations, and other microscopic motion.

Light from multiple telescopes can also be combined interferometrically. The most significant gain is angular resolution; a two-telescope interferometer with spacing D has a comparable angular resolution to a single telescope with a primary ``mirror'' of aperture D. Clearly, it is easier and more cost-effective to build two telescopes spaced 1 kilometer apart, than to build a 1 kilometer-wide telescope. In order to obtain an interferometric signal, the telescope beams must be combined coherently, in phase. This means that the optical/electrical path length from the various telescopes must be an integral number of wavelengths. At long-wavelengths, as in the radio portion of the electromagnetic spectrum, this is relatively straight-forward, although it is extremely difficult at infrared and visible wavelengths. Interferometry also requires that all optical and electrical elements en route must preserve the light wave phase information. If the phase of the light waves is randomized or muddled with noise or destabilized, coherent combination of the telescope beams will not occur, and no fringes will be seen and interferometry will be impossible. In such a case, the beam pattern will look merely like that of a single telescope, although the amount of signal being collected increases with the number of telescopes.

How do we analyze an interferometric signal? The fringe pattern we see depends on the wavelength of light being observed, the antenna spacing, and the angular size of the object being observed. Let's explore how these parameters affect the interference fringe pattern.

First, the spacing between fringes is associated with the interferometer angular resolution - much in the same way that the half-power width of a single telescope defines its angular resolution. The angular spacing of fringes on the sky, which we will term $\Delta \theta$ , is given as for a single telescope: $\Delta \theta = 1.22 \cdot \lambda/D$ , where $\lambda$ is the wavelength of the received radiation, and D is the spacing between the telescopes. The peak-to-peak fringe amplitude when looking directly at the source, i.e. the peak interferometric signal is called the Visibility Function, or V_o. The visibility function for two telescopes can be written as:

$\begin{displaymath}V_o = \frac{sin \left[ 2\pi S (\alpha/2) \right]}{\left[ 2\pi S (\alpha/2) \right]}\end{displaymath}$

where $\alpha$ is the object's angular size (in radians), and S is the separation (in units of wavelength; i.e. $D/\lambda$ ).

**Figure 2:** Interferogram of an object unresolved by the interferometer. The visibility function, or normalized contrast between the maxima and minima, is identically 1. The ``envelope'' traced by the maxima looks like the beam pattern of a single telescope (with twice the amplitude).
$\includegraphics[scale=0.9]{interferogram.ps}$

Looking at the above expression, it is clear to see that if an observed object is unresolved by the interferometer (i.e. it has a smaller angular extent than the interferometer's angular resolution), then $\alpha$ becomes very small and V_o becomes 1. Thusly, the normalized contrast between fringe peak and minimum is identically 1; there is perfect constructive interference at the maxima, and perfect destructive interference at the minima. Figure 2 shows an example of what such an interferogram would look like.

For an object resolved by the interferometer, $\alpha$ is no longer insignificant and the contrast between minima and maxima drops; there is imperfect constructive and destructive interference:

**Figure 3:** *Interferogram of an object resolved by the interferometer. The visibility function, or normalized contrast between the maxima and minima, is less than 1.*
$\includegraphics[scale=1.0]{interferogram2.ps}$

Important question! Measure the visibility function from Figure 3. Now consider a source that has a very large extent compared to the interferometric angular resolution; i.e. in the limit of large $\alpha$ . What does the interferogram look like? Sketch it! What is the visibility function in this case? Is ``more interferometric angular resolution'' (i.e. very wide telescope spacing) always better? Defend your answer.

This lab will be performed using the $\sim$ 0.83-meter radio telescope located on the roof adjacent to the 4'' James refractor. Follow the steps outlined below and write-up the lab using the same guidelines used in earlier lab reports. Show ALL work and be sure to answer all questions in this handout!

In this experiment, you will align each telescope to look at the Sun, peaking the total power of each telescope separately. Then you will combine the signals from the two telescopes at a T-junction, which sends the combined signal to an amplifier, whose output is fed to a total power box (see previous lab handouts for a description of the total power box). You will monitor and record the total power as the Sun drifts out of the beam of both telescopes. The resulting fringe pattern will allow you to measure the visibility function, V_o, and thereby $\alpha$ , the angular size of the Sun. The antenna spacing is 4 meters, and the central frequency of observation is 4 GHz.

Additional questions to address for your writeup: Because we didn't detect a useful fringe pattern, please analyze the attached interferogram of a nearly-identical setup instead. The antenna spacing is 4 meters, and the effective frequency of observation is 4 GHz. From this interferogram, measure the angular size of the Sun (i.e. solve for $\alpha$ in the visibility function equation). Describe (and draw) the experimental setup clearly, and discuss any possible reasons we might not have detected fringes (we talked about this extensively in lab). For example, the electrical length of coax from each telescope is 314 cm and 201.5 cm. Do these lengths of coax allow the signals to arrive in phase at the T-junction? Was the Sun detected by both antennae, etc.? How did your derived value of $\alpha$ compare with the optical angular extent of the Sun?