Problems

#03

The amount of light collected depends on the area of the telescope aperture, which in turn depends on the square of the aperture diameter. So this is a proportions problem. If a 2 m telescope collects a certain amount of light in 1 hour, a 6 m can do the same in less time. How much less? Answer: the ratio of their diameters squared, or $2^2/6^2 \times 1 = (1/9)= 0.11$ hrs, or 6.7 min. A 12 m takes even less time, $2^2/12^2 \times 1 = (1/36) = 0.28$ hrs, or 1.7 min.

#04

Resolution depends on the ratio of the wavelength of observation divided by the telescope aperture. The formula from the text is

$$\alpha = 0.25'' \times \frac{\lambda}{D}$$

where $\alpha$ is the angular resolution in arcsec, $\lambda$ is the wavelength in microns, and $D$ is the telescope diameter in meters. The value given is $\alpha=0.05''$ at $\lambda=700$ nm. the problems are then comparisons to this reference for the same telescope (hence, the $D$ is not changing).
(a) 3.5 microns is 3500 nm, which is 5 times larger than 700 nm. The angular resolution must therefore be worse (i.e., bigger) by 5 times, or $5\times 0.05 = 0.25''$.
(b) Now the wavelength is shorter: 140 nm is 0.2 of 700 nm. So the resolution is better (i.e., smaller) by 0.2, or $0.2 \times 0.05 =0.01''$.