Telescope Specifications Help (page 2)
Several parameters are significant when determining the effectiveness of a telescope for various applications. Here are the most important ones.
The magnification , also called power and symbolized ×, is the extent to which a telescope makes objects look closer. (Actually, telescopes increase the observed sizes of distant objects, but they do not look closer in terms of perspective.) The magnification is a measure of the factor by which the apparent angular diameter of an object is increased. A 20× telescope makes the Moon, whose disk subtends about 0.5° of arc as observed with the unaided eye, appear 10° of arc in diameter. A 180× telescope makes a crater on the Moon with an angular diameter of only 1 minute of arc ( of a degree) appear 3° across.
Magnification is calculated in terms of the focal lengths of the objective and the eyepiece. If f o is the effective focal length of the of the objective and f e is the focal length of the eyepiece (in the same units as f o ), then the magnification factor m is given by this formula:
m = f o / f e
For a given eyepiece, as the effective focal length of the objective increases, the magnification of the telescope also increases. For a given objective, as the effective focal length of the eyepiece increases, the magnification of the telescope decreases.
The resolution , also called resolving power , is the ability of a telescope to separate two objects that are not in exactly the same place in the sky. It is measured in an angular sense, usually in seconds of arc (units of 1/3600 of a degree). The smaller the number, the better is the resolving power.
The best way to measure a telescope’s resolving power is to scan the sky for known pairs of stars that are appear close to each other in the angular sense. Astronomical data charts can determine which pairs of stars to use for this purpose. Another method is to examine the Moon and use a detailed map of the lunar surface to ascertain how much detail the telescope can render.
Resolving power increases with magnification, but only up to a certain point. The greatest image resolution a telescope can provide is directly proportional to the diameter of the objective lens or mirror, up to a certain maximum dictated by atmospheric turbulence. In addition, the resolving power depends on the acuity of the observer’s eyesight (if direct viewing is contemplated) or the coarseness of the grain of the photographic or detecting surface (if an analog or digital camera is used).
The light-gathering area of a telescope is a quantitative measure of its ability to collect light for viewing. It can be defined in centimeters squared (cm2) or meters squared (m2), that is, in terms of the effective surface area of the objective lens or mirror as measured in a plane perpendicular to its axis. Sometimes it is expressed in inches squared (in2).
For a refracting telescope, given an objective radius of r , the light-gathering area A can be calculated according to this formula:
A = π r 2
where π is approximately equal to 3.14159. If r is expressed in centimeters, then A is in centimeters squared; if r is in meters, then A is in meters squared.
For a reflecting telescope, given an objective radius of r , the light-gathering area A can be calculated according to this formula:
A = π r 2 − B
where B is the area obstructed by the secondary mirror assembly. If r is expressed in centimeters and B is expressed in centimeters squared, then A is in centimeters squared; if r is in meters and B is in meters squared, then A is in meters squared.
Absolute Field Of View
When you look through the eyepiece of a telescope, you see a circular patch of sky. Actually, you can see anything within a cone-shaped region whose apex is at the telescope (Fig. 19-12). The absolute field of view is the angular diameter q of this cone; q can be specified in degrees, minutes, and/or seconds of arc. Sometimes the angular radius is specified instead of the angular diameter.
The absolute field of view depends on several factors. The magnification of the telescope is important. When all other factors are held constant, the absolute field of view is inversely proportional to the magnification. If you double the magnification, you cut the field in half; if you reduce the magnification to one-quarter of its previous value, you increase the field by a factor of 4.
The viewing angle—that is, the apparent field of view—provided by the eyepiece is important. Some types of eyepieces have a wide field, such as 60° or even 90°. Others have narrower apparent fields, in some cases less than 30°.
Another factor that affects the absolute field of view is the ratio of the objective diameter to its focal length. In general, the larger this ratio, the wider is the maximum absolute field of view that can be obtained with the telescope. Long, narrow telescopes have the smallest maximum absolute fields of view; short, fat ones have the widest maximum fields.
Telescope Specifications Practice Problems
How much more light can a refracting telescope with a 15.0-cm-diameter objective gather compared with a refracting telescope having an objective whose diameter is 6.00 cm? Express the answer as a percentage.
Light-gathering area is proportional to the square of the objective’s radius. Therefore, the ratio of the larger telescope’s light-gathering area to the smaller telescope’s light-gathering area is proportional to the square of the ratio of their objectives’ diameters. Let’s call the ratio k . Then in this case
k = 15.0/6.00 = 2.50
k 2 = 2.50 2 = 6.25
The larger telescope gathers 6.25 times, or 625 percent, as much light as the smaller one.
Suppose that a telescope has a magnification factor of 100× with an eyepiece of 20.0 mm focal length. What is the focal length of the objective?
Use the formula given in the section entitled, “Magnification.” The value of f o in this case is the unknown; f e = 20.0 mm, and m = 100. Therefore
m = f o / f e
100 = f o /20.0
f o = 100 × 20.0 = 1,000 mm
Technically, we are justified in expressing the answer to only three significant digits. We can legitimately say that f o = 2.00 m.
Suppose that the absolute field of view provided by the telescope in Problem 19-6 is 20 arc minutes. If the 20-mm eyepiece is replaced with a 10-mm eyepiece that provides the same viewing angle as the 20-mm eyepiece, what happens to the absolute field of view provided by the telescope?
The 10-mm eyepiece provides twice the magnification of the 20-mm eyepiece. Therefore, the absolute field of view of the telescope using the 10-mm eyepiece is half as wide or 10 arc minutes.
Practice problems of these concepts can be found at: Optics Practice Test
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