number of operators of order p>2 generate a group under which each one of these operators has no more than p conjugates then the entire group contains no operator whose order exceeds p. It should however be observed that while this condition is sufficient it is by no means necessary to insure the fact that the resulting group contains no operator whose order exceeds p. It seems well to add that the greater part of this paper aims to present known results in a connected form. The theorem relating to the number of subgroups of order p is, however, supposed to be new and evidently admits of further generalization. The method of proving Frobenius' extension of Sylow's theorem is also supposed to be new. THE THEORY OF OPTICAL SQUARES. By PROFESSOR ARNOLD EMCH, The University of Colorado. 1. Optical squares* as they are used in surveying owe their usefulness to the fact that the incident rays from an object include constant angles with the corresponding final reflected rays. The proof for this constancy is a simple application of the geometric proposition that the sum of the angles of any triangle is equal to two right *For a description of optical squares (angular mirrors) see G. F. Barker's Physics, Advanced Course, pp. 410-414, or any good text-book on surveying. †Traite des proprietes projectives des figures (1822), p. 345, or 2nd ed., tome I, p. See also Cremona: Elements of projective geometry, p. 184. 281. to a plane which is parallel to the rays of light and whose intersections with the mirrors are the lines a1, a2, ɑz, ..........., ɑn. For the sake of generality we make the hypothesis that these mirrors may be indefinitely extended and that reflections may occur along their whole extent and always in such a direction that the ray reflected from the mirror a, will strike the mirror ai+1• In other words, we assume that a ray of light may be indefinitely produced beyond its physical boundaries. This hypothesis is geometrically possible and includes the real physical cases. 2 2 2 3 1 Designate in Fig. 1 the reflecting lines or mirrors by a,, aş, Az, ..........., An. Through A, draw any ray v, striking a, at B1. The reflected ray v2 passes through the point A, which is symmetrical to A, with respect to a,. The ray v2 strikes a at B, and its reflected ray v, passes through A, which is symmetrical to A, with respect to a, as an axis, and so forth. The ray vn+1 reflected from the last line an at B, passes through An+1 which is symmetrical to A, with respect to an as an axis. Let the rays v1 and v2+1 intersect at V. We have now a closed polygon whose sides v1, V2, V3, ..., Un+1 pass through the fixed points points A., A., A ̧, ............., An+1 and whose vertices B1, B2, B3, ..., B2 except V, lie on the fixed lines a1, az, ɑzı 39 ----, an. Hence, when v, turns about A1, V will describe a conic and the problem is reduced to the one explained under (3). 5. This is as far as Poncelet's and Cremona's discussions of these problems go. It will be seen, however, that an investigation of the angular relations between the rays of light forming the polygon v, U2, ..., Un+1 brings to light some interesting facts. 3 Leta, be the angle of incidence of v, on a,, a, the angle of incidence of v2 on 2, 3 of v ̧ on a,, and so forth; 1, 2, 93, the angles which a, and a,, a 2 and and a 4, include. The angles 41, 42, 43 according to the physical requirements of the problem must be chosen in such a manner that always If there are reflecting lines, then the number of angles a is also n, and the number of angles is n-1. To find the deviation of the first ray v, from the final reflected ray vn+1, drop a perpendicular from any point to the lines v1 and Un+1. Erect perpendiculars to a, a, ɑ3, ........., an at B1, B2, B3, ----, Bn respectively. Then, the first and last perpendicular deviate from each other by the angle a1 + (n−2)x +an = ( & 1 + $ 2 + & 3+.......+Pn−1), consequently also the rays v1 and vn+1 by the same angle In this case the angle between the original incident ray and the final reflected ray depends on the angles and the original angle of incidence a1. Their point of intersection V describes a conic and there are two positions of V, real or imaginary, for which the incident and reflected ray make a given angle. From this value of it is seen that in case of an even number of reflecting sides, V describes a circle and the angle is constant. Cremona's optical problem admits either of no solution or of an infinite number of solutions and the angle does not depend on the angles ; of even indices. We have therefore proved the following proposition: If rays of light emanate from a fixed source which, in succession, are reflected ́on n straight lines (mirrors) then the final reflected rays cut the corresponding original rays in points of a conic which is not or is a circle according as n is odd or even. In the first case (n odd) there are two places on the conic where the original and final rays make a given angle. In the second case (n even) there are no such places on the circle, or else an infinite number. In this case (n even) the angle & only depends on the angles between succeeding reflecting lines, whose orders in this succession are odd. 6. APPLICATIONS. Let n=2, then -24,. To make 6, we must choose -45°. This is illustrated in Fig. 2, and is practically applied in Adams' Angle Mirror or Optical Square. 1 To make 0, 0, +3 must be made equal to . Under this condition 3 , and may vary separately. The condition is also satisfied by taking 4,42 ==112° 30′, and this is illustrated in Fig. 3. DEPARTMENTS. SOLUTIONS OF PROBLEMS. ALGEBRA. Solutions of Problem 188 have been received from J. E. Sanders, Hackney, Ohio, and Charles A. Carpenter, Student in Adelphi Academy, Brooklyn, N. Y. 189. Proposed by S. F. NORRIS, Professor of Astronomy and Mathematics, Baltimore City College, Baltimore, Md. *Invented by George Adams in London in the middle of the 18th century and first described by his son in 1791. |