Worksheet 4, Problem 2 - Resolving the Telescopes

Question: CCAT is a 25-meter telescope that will detect light with wavelength 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 the infrared J-band? 

In this problem, we are comparing two telescopes to see how powerful they are. Just by pure numbers, we are given that the CCAT is 25-meters in diameter, versus the MMT, which is only 6.5 meters in diameter. Now, if you haven't seen my previous post about a telescope's angular resolution, please read it! Like seriously!

But of course, that is a very long post, so if you don't want to read it, here's the short version:

TL;DR - The larger the diameter of the telescope, the more angular resolution it has, which basically means the larger the telescope, the more powerful it is.

The other thing this question talks about is the J-band. In the simplest terms, the J-band is light at 1.25 micrometers ($1.25 \cdot 10^{-4}$ centimeters) in wavelength.

Okay, so the formula for angular resolution, which we derived in the previous post (seriously, read it!) of a telescope is:

\begin{align}
\theta_{\text{min}} = \frac{\lambda}{D}
\end{align}

where $\theta_{\text{min}}$ is the minimum angle that the telescope can resolve (how small/far away things a telescope can see), $\lambda$ is the wavelength of light passing through, and $D$ is the diameter of the telescope.


Okay, so for the CCAT, we know that the telescope is 25 meters (2500 centimeters) in diameter, and can detect wavelength up to 850 microns ($8.5 \cdot 10^{-2}$ centimeters). To calculate it's angular resolution when looking at something 850 microns in size, you get:

\begin{align}
\theta_{\text{min}_{CCAT}} = \frac{\lambda}{D} = \frac{8.5 \cdot 10^{-2} \text{ cm}}{2500 \text{ cm}} = 3.4 \cdot 10^{-5} \text{ radians}
\end{align}

That is pretty good resolution for seeing something that small!

However, since the problem states that the telescope is observing the infrared J-band, which is at 1.25 micrometers ($1.25 \cdot 10^{-4}$ centimeters) in wavelength, the angular resolution of the CCAT when observing the J-band is:

\begin{align}
\theta_{\text{min}_{CCAT}} = \frac{\lambda}{D} = \frac{1.25 \cdot 10^{-4} \text{ cm}}{2500 \text{ cm}} = 5.0 \cdot 10^{-8} \text{ radians}
\end{align}

As for the MMT, which is 6.5-meters (650 centimeter), can gather light at the J-band. It's angular resolution is as follows:

\begin{align}
\theta_{\text{min}_{MMT}} = \frac{\lambda}{D} = \frac{1.25 \cdot 10^{-4} \text{ cm}}{650 \text{ cm}} = 1.9 \cdot 10^{-7} \text{ radians}
\end{align}

The ratio of minimum angular resolution of the CCT to the MMT when observing the J-Band is:

\begin{align}
\frac{\theta_{\text{min}_{CCAT}} }{\theta_{\text{min}_{MMT}}} = \frac{5.0 \cdot 10^{-8} \text{ radians}}{1.9 \cdot 10^{-7} \text{ radians}} = 0.26.
\end{align}

Now, having a higher resolution means having a smaller $\theta_{\text{min}}$, so when observing wavelengths at 1.25 microns, the CCAT is only slightly better at 26% better.

When looking at the ratio of the CCAT to the MMT with respect to the CCAT observing at 850 microns, vs MMT's 1.25 microns, things get more interesting:

\begin{align}
\frac{\theta_{\text{min}_{CCAT}} }{\theta_{\text{min}_{MMT}}} = \frac{3.4 \cdot 10^{-5} \text{ radians}}{1.9 \cdot 10^{-7} \text{ radians}} \approx 180
\end{align}

This shows that the CCAT discerning light at 850 microns would be 180 times worse than the MMT looking at 1.25 microns.

So I guess, bigger isn't always better!