In order to cut off reflected light from the inside of the tube, a stop is placed midway between the front and back lens ; but not so as to intercept any of the legitimate rays of light Such is the construction of the orthoscopic lens. It is essentially a VIEW-LEN s, and is not intended for portraiture. We have therefore to discuss its merits as a view-lens when compared with the ordinary form, and in doing so must direct our attention more particularly to the following points :— 1st, —Flatness of field, and the included angle of view. 2nd,—Freedom from distortion.
3rd,—Equality of illumination.
4th,—Perfection of focus, and freedom from spherical aberration. 5th,—Coincidence of the visual and actinic foci.
6th,—General convenience, freedom from diffused light, copying powers when the focus is elongated, power of rendering aerial per spective, and other good qualities.
We shall discuss these matters in the order in which they stand.
lst—Flatness of field, and the included angle of view.
In determining the flatness of field of any lens, we have to compare the course of the most oblique with that of a direct pencil ; and the simplest plan is to suppose the pencils cylindrical, or that the lens is pointed at extremely distant objects ; should it be found to answer well in this case it will be equally good for all ordinary purposes.
In the above figure the lenses are represented by straight lines strong and black, the front lens passing through A and the back lens through C ; A Cy being their common axis. A stop is placed behind the back lens, and in contact with it. In order to fix the ideas, and render what we have to say more intelligible, we shall suppose the lens to be a No. 1, having a combined focus of rather more than 11 inches, and covering a picture 10 x 8, with a half-inch stop.
It will be seen, from the above figure, that the oblique pencil passes wentrically through the front lens and centrically through the back lens. If, then, q be the focus of the direct pencil from a distant point Q, and, the focus of an oblique pencil from a distant point P, we have to compare the length Cp with Cy, in order to discover the flatness of field, and how far it deviates from a §phere whose centre is C ; and we would observe that unless it does deviate, and that pretty considerably, from such a sphere, the lens would be next to worthless for the purpose intended. We have to show, then, that
the focal length, Cp, of the oblique pencil, is greater than the focal length, Cy, of the direct pencil, and to calculate the difference be twe,en them.
Let us consider first the case of the direct pencil incident at A. After refraction through the front lens it converges towards m, the the principal focus of that lens ; the distance Am being 8 inches (in round numbers). This converging pencil is then refracted by the posterior lens, the positive focal length of which is 18 inches (in round numbers). The effect of this is to diminish the convergency of the rays and bring them to a focus at q, which is further than m from C. The distance AC being one inch, Cm is 7 inches, and Cy is then found in the follovving way :— Multiply 7 by 18, and divide the product by their difference ;— that is, divide 126 by 11. This gives Cq=---11 A inches.
Next, let us consider the oblique pencil which proceeds from a distant point P, is incident excentrically on the front lens at B, and passes centrically through the back lens at C.
Through A, the centre of the front lens, draw a dotted line AO, parallel to BP, and with A as centre, and Am as radius, strike an arc of a circle cutting A 0 at O. Then, A 0 equals 8 inches ; and the oblique pencil at P will, after passing through the front lens, converge towards the point 0 (as shown by the dotted lines).
Now we come to the pith of the matter. What happens at the second lens ? We have at the second lens an oblique pencil, incident centrically, and converging towards O. Join therefore CO, and produce it to p.
Also, with C as centre and Cy as radius, strike a circle cutting Cp at n. Cn is therefore equal to Cy.
Now, adopting the same formula as in the former case in order to find Cp, we must multiply CO by 18 and divide the product by their difference. What then is the length of COP In the reply to this query will be seen the great ingenuity of M. Petzval's arrangement ; for it appears that CO is greater than Cm.
The proof of this is easy enough. Any two sides of a triangle are, together, greater than the third, therefore 0 C and C A are to gether greater than AO, and therefore than Am. Take away the common part AC, and CO is proved to be greater than Cm.