Spring 2004
Dr Mike Fanelli
Solutions to Assigned Problems:
Chapter 5
PROBLEM 5-3: A 2-meter telescope collects a specific amount of light in 1 hour.
Given the same conditions, how much time is required for a 6-meter telescope to
perform the same task ? A 12-meter telescope ?
ANSWER: The light gathering power of telescopes depends on the area of the
collecting surface. Since the area of a circle is = R2, light gathering power goes
as the radius or diameter squared.
(a) A 6 meter mirror is 3 times the diameter of a 2-m, and collects (33) = 9
times as much light. Therefore it can conduct the same observation in 1/9th the
time, about 6.5 minutes.
(b) A 12-m telescope is 6 times the diameter of a 2-m, collects (66) = 36 times
more light in the same period of time. It can conduct the same observation that it
takes a 2-m 1 hour to complete in 1.7 minutes (1/36 the amount of time).
PROBLEM 5-4: A certain space-based telescope can achieve an angular
resolution of 0.05 arcseconds for red light (wavelength about 700 nanometers).
What would this telescope’s resolution be (1) in the infrared part of the spectrum
at 3.5 microns, and, (b) in the ultraviolet at 140 nanometers ?
ANSWER: As described in your notes and the text, the resolution of a telescope
in arcseconds can be expressed using this useful formula:
R = 0.25  wavelength  mirror diameter,
where the wavelength is expressed in units of microns and the mirror diameter
is expressed in meters. You could first solve this expression for the given size of
the mirror. But since the mirror size is not changing, the relative resolution of this
telescope at different wavelengths will be simply the ratio of the wavelengths to
the value given in the question.
(a) 3.5 microns is (3.5 / 0.7) times larger than 0.7 microns (700 nanometers
converted to microns). 3.5 / 0.7 = 5. Since the wavelength is larger, the
resolution will be larger, 5 times larger, which is 0.25 arc-seconds.
(b) 140 nanometers is (140 / 700) = 0.2, or 5 times smaller than 700
nanometers. The resolution for this case will be smaller, 0.05 / 5 = 0.01
arcseconds.
PROBLEM 5-8: A CCD camera replaces a photographic device at the
telescope. Given that a photograph is about 5% efficient at recording the light
that falls on it, and the CCD is about 90% efficient, how much time is saved by
the new CCD camera ?
ANSWER: To determine this ratio, divide 90% / 5%. This ratio = 18. The CCD
camera is 18 times faster in recording the light which falls on it. If the photograph
takes an hour, then the CCD can make the same observation in 60 / 18 = 3.3
minutes. The electronic device is simply more efficient at recording photons that
fall on it, compared to film.
PROBLEM 5-9: Given the Moon’s distance from Earth, 380,000 km, the problem
asks: what sized features could we just detect on the Moon for the best
resolutions obtainable with the Palomar telescope (1), the Hubble Telescope
(0.05 ), and a radio telescope interferometer (0.001) ? The values in
parentheses are the resolutions in arcseconds.
ANSWER: This problem involves application of the very useful small angle
formula, introduced in chapter 1. You are given a distance (the lunar distance)
and three angular sizes, representing the capabilities of three different
telescopes. Plug into the formula to determine the physical size corresponding
to each of these angular sizes at the Moon’s distance.
The small angle formula is expressed:
Physical diameter
=
(angular diameter) x 2 x (distance)
______________________________
360 degrees
In order for the units to work out properly, the angular resolution must be
expressed in the same units as the denominator, that is in degrees. There are
3600 arcseconds in a degree. The result will give an answer in kilometers, which
is what we’d expect.
(a)
=
(1 arcsecond  3600) x 2 x (380,000 km)
____________________________________
360 degrees
= 1.8 km
(b) Plugging into the same formula, for an angular resolution of 0.05, gives
0.092 km or 92 meters.
(c) For a resolution of 0.001, the resolution is 0.0018 km or 1.8 meters.