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Astronomy Principles and Practice Fourth Edition.pdf

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Chapter 17<br />

Visual use of telescopes<br />

17.1 Magnifying power<br />

In order to make a visual inspection of the images which the telescope collector provides, some kind<br />

of eyepiece must be used. The effect of the use of an eyepiece is illustrated in a simplified way in<br />

figure 17.1, where the virtual image is at a position on or beyond the near point of the eye. (The actual<br />

final image is formed on the retina of the eye.) Under this circumstance, it can be seen that the image<br />

produced by the collector lies inside the focus of the eyepiece. The figure illustrates that, with this<br />

simple eyepiece, the viewed image is inverted. The magnifying power, m, of the overall optical system<br />

is defined as the ratio of the angle subtended by the virtual image at the eye, α e , <strong>and</strong> the angle, α c ,<br />

subtended by the object at the collector. Thus,<br />

m = α e<br />

α c<br />

. (17.1)<br />

For astronomical observations, it is usual practice to adjust the position of the eyepiece so that<br />

the virtual image appears to be at infinity. It is obvious that in this adjustment, the eye lens must be at<br />

the exact distance of its focal length from the image produced by the collector in its focal plane. This<br />

situation is represented in figure 17.2. The collector aperture acts as the entrance pupil <strong>and</strong> the image<br />

of the collector aperture formed by the eyepiece acts as the exit pupil. The distance from the eye lens<br />

(or the last lens of a compound eyepiece) to the exit pupil is known as the eye relief. As all the rays<br />

from all parts of the field which can be viewed by the telescope pass through the exit pupil, it is at this<br />

position that the eye should be placed. The ray passing through the centre of the collector is known<br />

as the chief ray. It is not deviated at the collector aperture <strong>and</strong> it can be seen from figure 17.2 that it<br />

passes through the centre of the exit pupil.<br />

From figure 17.2, it can be seen that<br />

tan α e = h v<br />

<strong>and</strong><br />

tan α c = h u<br />

where h is the distance of the chief ray from the optic axis when it arrives at the eyepiece, u <strong>and</strong> v<br />

are, respectively, the object <strong>and</strong> image distances corresponding to the entrance <strong>and</strong> exit pupil distances<br />

from the eyepiece. As both α c <strong>and</strong> α e are small angles,<br />

α e ≈ h v<br />

(17.2)<br />

272

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