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Astronomy Principles and Practice Fourth Edition.pdf

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Refractors 251<br />

Figure 16.13. A ray diagram illustrating the effect of chromatic aberration resulting from the use of a single<br />

positive lens.<br />

which can be obtained. The image in this plane of sharpest focus is not a point image. It is known as<br />

the circle of least confusion <strong>and</strong> its size can be determined from the physical properties of the lens.<br />

The spread of the image along the optic axis is known as longitudinal chromatic aberration <strong>and</strong> the<br />

spread of the image in the plane of sharpest focus is known as lateral chromatic aberration.<br />

It will be remembered that it was first supposed that chromatic aberration could not be removed<br />

from lens systems but, eventually, Doll<strong>and</strong> proposed a method for achromatism. It is possible by<br />

using in combination with a positive lens, a negative lens of different material—<strong>and</strong> hence refractive<br />

power—to cancel the dispersion without complete cancellation of the refractive power. This is easily<br />

demonstrated as follows.<br />

If two lenses of focal lengths F 1 <strong>and</strong> F 2 are fixed together by optical cement, the focal length of<br />

the combination, F,isgivenby<br />

1<br />

F = 1 + 1 .<br />

F 1 F 2<br />

Hence,<br />

F = F 1 F 2<br />

. (16.18)<br />

F 1 + F 2<br />

By expressing the individual focal lengths in terms of the shape of each lens <strong>and</strong> their refracting<br />

powers, equation (16.18) can be rewritten as<br />

K 1 K 2<br />

F =<br />

K 2 (n 1 − 1) + K 1 (n 2 − 1) . (16.19)<br />

Now the aim of the combination is to provide a system whose focal length is identical for the blue<br />

<strong>and</strong> red ends of the spectrum. If the refractive indices of the materials of the lenses are denoted by n 1B ,<br />

n 2B , for blue light <strong>and</strong> n 1R , n 2R , for red light, the achromatic condition is achieved when<br />

K 1 K 2<br />

K 2 (n 1B − 1) + K 1 (n 2B − 1) = K 1 K 2<br />

K 2 (n 1R − 1) + K 1 (n 2R − 1)<br />

i.e.<br />

K 1<br />

=− n 1B − n 1R<br />

. (16.20)<br />

K 2 n 2B − n 2R<br />

Since n 1B > n 1R <strong>and</strong> n 2B > n 2R by the general properties of optical materials, K 1 /K 2 must be<br />

negative. Hence, achromatization can only be achieved by combining a positive <strong>and</strong> a negative lens.

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