Worksheet 4, Problem 2: Two Telescopes

#2. CCAT is a 25-meter telescope that will detect light with wavelengths up to 850 microns. How does the angular resolution of this huge telescope compare to the angular resolution of the much smaller MMT 6.5-meter telescope observing in the infrared J-band? Ask a TF in class or the internet about the meaning of “J-band.” 

Angular resolution describes how much detail of an image you can resolve. As we learned in class, the angular resolution of telescopes is determined by \(\theta_{min}\). Telescopes with a smaller \(\theta_{min}\) will have a greater resolution. From what Prof Johnson told us in lecture, and from what we worked out Problem 1, \(\theta_{min}\:=\:\frac{\lambda}{D}\) with \(\lambda\) being the wavelength of light and D being the diameter of the telescope mirror.

After looking up J-band, we found out that a J-band is  a  band of light centered at 1.2 microns = 1.2 x 10^-4 cm.  Also, 850 microns is equivalent to 8.5 x 10^-2 cm. Thus, all we had to do was solve for the \(\theta_{min}\) of each telescope by plugging in the values we already knew.

For the CCAT: \[\theta_{min}\:=\:\frac{\lambda}{D}\] \[\theta_{min}\:=\:\frac{8.5\:\times\:10^{-2}\:cm}{2.5\:\times\:10^{3}\:cm}\] \[\theta_{min}\:=\:3.4\:\times\:10^{-5}\]

For the MMT: \[\theta_{min}\:=\:\frac{\lambda}{D}\] \[\theta_{min}\:=\:\frac{1.2\:\times\:10^{-4}\:cm}{6.5\:\times\:10^{2}\:cm}\] \[\theta_{min}\:=\:1.85:\times\:10^{-7}\]


Thus the MMT, surprisingly, has a better resolution because it has a smaller theta min even though it has a much smaller diameter.