Worksheet 4, Problem 2: Telescopes

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.”

In Worksheet 4, Problem 1, we explored how wavelength and aperture width relate to the resolution of a telescope. Now, we get to put this to the test, and see just how significant their impact is. As a reminder, the relationship is as follows: \[\theta = \frac{\lambda}{D}\] \[\theta = angle\,from\,central\,axis\] \[\lambda = wavelength\] \[D = Distance\,between\,slits\]
We can now use this to calculate the angular resolution, \(\theta_{CCAT}\), of the CCAT:
\[\theta_{CCAT} = \frac{\lambda}{D} = \frac{850 \times 10^{-4}cm}{25\times10^2cm} = 3.4 \times 10^{-5}\]
For the MMT, we need to determine the wavelength of the J-Band. This band correspond to 1.5cm - 3cm. As an estimate of this, we can use a wavelength of 2cm.

Now we can calculate the angular resolution, \(\theta_{MMT}\), of the MMT:
\[\theta_{MMT} = \frac{\lambda}{D} = \frac{2cm}{6.5 \times 10^2cm} = 3 \times 10^{-3}\]

To then compare the angular resolution of the two telescopes, we can divide the two, and determine how many times greater the resolution of the CCAT is than that of the MMT.
\[\frac{\theta_{MMT}}{\theta_{CCAT}} = \frac{3 \times 10^{-3}}{3.4 \times 10^{-5}} \approx 100\]
This shows that the CCAT resolution is approximately 100 times that of the MMT.

Acknowledgements: Johnathan Budd & Willie Pirc