The theory of optical instruments

This historic book may have numerous typos and missing text. Purchasers can download a free scanned copy of the original book (without typos) from the publisher. Not indexed. Not illustrated. 1907 Excerpt: ...in a medium of index /, its height / being at right angles to the plane of the diagram. Let the instrument form an image / of 0, in a medium of index /»', by a thin sagittal pencil whose plane is at right angles to the plane of the diagram, and whose chief ray OPQI makes an angle O with the axis initially, and O' finally. Let a denote the angle between the extreme rays of the pencil initially, and let a be the final value of this angle: and suppose that df is the angle between the meridian planes which pass through the extreme rays of the pencil, so a = sin O. df, a = sin O', df. Clausius' equation (§ 7) gives at once fwl = fi'o'/', or /sinfl. l = p'smO'. /', so the linear magnification of a small object, when the image is formed by rays which pass through this zone on the refracting swrfaces, is This result is true for all optical instruments, independently of whether they are affected with spherical aberration or not. Suppose now that the instrument is corrected for spherical aberration, so that the images of 0 formed by different zones are situated at the same point of the axis. In order that the images of a small object at 0 may be in all respects identical, they must be of the same size; and therefore the equation wsintf H smO where m is the linear magnification for the image formed by the paraxial rays, must be satisfied by every ray which issues from the axial point 0. This equation is called the sine-condition. As might be expected, the sine-condition also ensures that the images formed by meridian pencils have the same magnification, whatever be the zones through which the pencils pass. For again applying Clausius' equation (§7) /i. COS i/. la = p' COS. l'a, we have in this case (the object and image being taken in the plane of the d...