Dr Mike Fanelli

Fall 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 observing 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 = pR2, 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 arc seconds for red light (with a wavelength of 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 arc seconds 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 (both are units of size). 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 arc seconds.

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% by 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 ¸ 8) = 3.3 minutes. The electronic device is simply more efficient at recording photons 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 arc second), the Hubble Telescope (0.05 arc second), and a radio telescope interferometer (0.001 arc second) ?

ANSWER: This problem involves an 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 small angle formula to determine the physical size corresponding to each of these angular sizes at the Moon’s distance.

The small angle formula is expressed:

(angular diameter) x 2p x (distance)

Physical diameter = ______

360 degrees

In order for the units to work out properly, the angular resolution used in the numerator must be expressed in the same units as the denominator, that is in degrees. There are 3600 arc seconds in a degree. The result will be expressed in kilometers, since the distance used in the numerator uses that unit.

(1 arc second ¸ 3600) x 2p x (380,000 km)

(a) = ______

360 degrees

= 1.8 km

(b) Plugging into the same formula, for an angular resolution of 0.05², gives

0.092 km or 92 meters.