Worksheet 4, Problem 2: The bigger, the better?

The problem: 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.”

What we know:

• From class, we know the equation of angular resolution is: \[\theta_{min} \approx \frac{\lambda}{D}\] where the lambda is the wavelength of light and D is the diameter of a telescope mirror.

Source

• From the internet we know "In infrared astronomy, the J band refers to an atmospheric transmission window centred on 1.25 micrometres(in the near-infrared)." (Source) 
• A micron is the same thing as a micrometer, both equivalent to \( 1 x 10^{-4} cm\).

Solve:

For CCAT: \[\theta_{CCAT} \approx \frac{8.5 x 10^{-2} cm }{2.5  x 10^{3} cm} \approx 3.4  x 10^{-5} radians \]
For MMT:\[\theta_{MMT} \approx \frac{1.25 x 10^{-4} cm }{6.5  x 10^{2} cm} \approx 1.9 x 10^{-7}radians \]

Update:

Converting these answers to arcseconds gives:
\[\theta_{CCAT}  \approx 3.4  x 10^{-5} radians \cdot \frac{2 \times 10^5 \: arcsecond}{1 \: radian} \approx  6.8 \: arcseconds\]

\[\theta_{MMT} \approx 1.9 x 10^{-7} radians  \cdot \frac{2 \times 10^5 \: arcsecond}{1 \: radian} \approx  3.8 \times 10^{-2} \: arcseconds\]

This exemplifies the relationship in the equation: when diameter is bigger, the angular resolution is smaller. In this case, the MMT angular resolution is about 100 times smaller than the angular resolution for CCAT.
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Acknowledgements: Thanks to Barra, Eden, and April for struggling through this worksheet with me and to Andrew for helping us at TALC.

Bonus: Here is a CCAT telescope (Source) and an MMT telescope (Source).