CHAPTER VII. THE THEORY OF OPTICAL INSTRUMENTS. 69. Preliminary discussion. There is a limit to the power of every instrument, due to the finite size of the wave-length of light. According to the laws of geometrical optics, the image of a star formed in a parabolic mirror should be a mathematical point, and if this were the case the sole consideration to be attended to in the construction of optical instruments would be the avoidance of aberrations. According to the wave theory of light, however, the image of a point source is never a point, however perfect the instrument may be in other respects, and the longer the wave-length the more does the light spread out sideways from the geometrical image. It is therefore useless to try to avoid aberrations beyond a certain point, and it becomes a matter of primary importance to define the natural limit of the power of an instrument, so as to be able to form a clear idea as to how far the optician may usefully spend labour in the refinement of his surfaces. Let a wave divergent from a point source A (Fig. 89) be limited by wave surfaces become portions of spheres concave towards a point P. It will be necessary to calculate the amplitude in the light in the neighbourhood of P, and a preliminary proposition will help to simplify the problem. Trace the rays AS, AS', limiting the beam, according to the laws of geometrical optics, and let TU, T'U' be portions of ese rays. Place a screen at KK' with an aperture just sufficient to enclose these rays all round, or in other words, let the edge of the aperture KK' coincide with the geometrical shadow of the opaque portions of the screen SS'. The proposition to be proved is, that the introduction of this screen does not alter the distribution of light in the neighbourhood of P, and that the screen SS' may now be removed, leaving all the amplitudes near P as they were. The truth of the proposition depends on the fact that all portions of the wave surface passing through KK' contribute equally to the amplitude of P, as P being a point of convergence of the rays, its optical distance to any point of KK' is the same. The screen KK' obliterates only the waves which have spread out laterally before they have reached the plane of the screen. The portions so obliterated form a very small fraction of the light forming the image at P which is due to the combined action of the complete wave. The same is true for the resultant amplitude at Q so long as the aperture KK' only contains a small number of Fresnel zones drawn from Q as centre. In order that students should not be misled to apply this proposition erroneously, we may take an example where it does not hold. T SS1 (Fig. 90) is a screen limiting a parallel beam of light, RS being the edge of the geometrical shadow. SS, cannot in this case be replaced by a screen TT1 placed so as to touch the same limiting rays, because tracing Fresnel zones from Q backwards, the locus of the division between two zones is a parabola (Art. 51). Such a parabola QS will trace the limiting zone for the screen SS1, while if this were replaced by TT, the limiting curve would be a different parabola QT. If the angular space TQS includes an odd number of zones, the change of position of the screen from SS, to TT, would cause a difference in amplitude equal to that of a complete zone, so that a maximum of light might be changed into a minimum or vice versa. P Fig. 90. 70. Image formed by a Lens. It is convenient to imagine the beam to be now limited by a diaphragm just inside the lens which concentrates the light at F. The wave-fronts are then circles with Fas centre. If D=2R is the diameter of the lens, p the distance of any point P from F, and f= OF, A Fig. 91. and p being very sinall compared with ƒ, AP-BP = 2 Rp .(1). If we were only to consider rays in the plane of the paper, then destroyed by interference if light at P would be f (2), = (n + 1 ) λ, and if we imagine the figure to revolve round the axis OF of the lens, the luminous appearance in the plane through F, at right angles to the axis, would be a luminous disc fading outwards until the intensity becomes zero when p=fX/D. This disc would be surrounded by dark and bright rings, the brightest parts of the rings corresponding to the distance p = (n + })ƒλ/D. Owing to the rays which do not lie in the plane of the paper the destruction of light takes place at a distance somewhat greater from F than that given by the above approximate calculation. Sir George Airy* was the first to solve the problem of the distribution of light in the diffraction image of a point source. His solution depends on the summation of a series. Lommel gave the solution in terms of Bessel functions. The main effect is obtained more simply by the above elementary considerations. The diffraction image is a disc surrounded by bright rings, which are separated by circles at which the intensity vanishes. If we write p=m fλ (3), the values of m for the circles of zero intensity are given in the following Table. They differ very nearly by one unit, but instead of being integers, as the approximate theory would indicate, approach a number which exceeds the nearest integer by about one quarter. The third column in the above table gives the amount of light lying outside each ring. The first number 161 indicates that 839 of the total light goes to form the central disc while the difference between the first and second number gives the fraction of the total light which forms the first ring. These differences are put down in Table IX. which is mainly intended to give the values of m for the circles of maximum illumination and the corresponding intensities. The third column contains the intensity at the maximum in terms of the central intensity, while the fourth column gives the fraction of the total light which goes to form the central image and each successive ring. Fig. 92. Fig. 93. Fig. 92 gives in diagrammatic form the relative sizes of the central disc and the first three rings. Fig. 93 shows the images of two sources of light placed at such a distance apart that the centre of the bright disc of one falls on the first dark ring of the other. 71. Resolving Power of Telescopes. It has long been known to all astronomers working with high powers, that the image of a star in a telescope has the appearance roughly represented in Fig. 92, and it is a matter of experience that a close double star may be recognized as such when the relative position of the stars is not closer than that represented by Fig. 93. This allows us to calculate the angular distance between the closest double star which the telescope can recognize as such. The radius of the first dark ring being p and the focal length of the telescope being ƒ, the angle & subtended at the centre of the object glass by two stars which occupy such a position that the centre of the diffraction image of one falls on the first dark ring of the other is plf, which by (3) gives А Fig. 94. 0 = 1.22 X/D (4). This is equal to the angular distance between the stars when they are on the point of resolution. No subsequent refraction of light through lenses can increase this angle. The images may be enlarged but the rings and discs are always enlarged in the same ratio. This is an important fact which may be more formally proved in this way: If the rays crossing at any point of the diffraction image Q (Fig. 94) are brought by a lens or system of lenses to cross again at a point Q', the optical distance from Q to Q' along all paths must be the same, and hence the retardation of phase between any two rays at Q is accurately reproduced again at Q'. If there is neutralization at Q, there must also be neutralization at Q'. As is the geometrical image of Q, the diffraction pattern in the plane of 'must be the geometrical image of the diffraction pattern in the plane of Q. Our result may therefore be applied to eye observations through a telescope, the plane of Q' representing the plane of the retina. It appears from the above that the power of a telescope to resolve double stars is proportional to the diameter of the lens. This is a result of the wave-theory of light, for if the rays were propagated by the laws of geometrical optics, the size of the object glass would not enter into the question, while the angular separation due to greater focal length could be increased at will by using a magnifying arrangement. We also see that the smaller the wave-length, the more nearly are the laws of geometrical optics correct. To resolve stars at an angular distance of 1 second of arc (4·84 × 10−6 in angular measure), we should for λ = 5 × 10-5 require a linear aperture of D= 1.22 × 5 × 10-5 4.84 × 10-6 The Yerkes telescope with an aperture of 100 cms. should be able therefore to resolve two stars at a distance of one-eighth of a second of This calculation is based on the supposition that the whole of the light which passes through the telescope enters the eye. By a well arc. |