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# Telescope Specifications Help (page 2)

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## 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.

Fig. 19-12 . A telescope’s absolute field of view q is measured in angular degrees, minutes, and/or seconds of arc.

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

#### Problem 1

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.

#### Solution 1

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.

#### Problem 2

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?

#### Solution 2

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.

#### Problem 3

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?

#### Solution 3

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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