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Therefore, we finally get (n1/s)+(n1/s)=(n2-m1)(1/R1-1/R2) (27.10) Now we note again that if one of the points is at infinity, the other will be at a point which we will call the focal lengths. The focal length is given by l=(n-1)(1/R1-1/R2), (27.11) where n= n2/nl Now, if we take the opposite case, where s goes to infinity, we see that s'is at the focal length,'. This time the focal lengths are equal.(This is another special case of the general rule that the ratio of the two focal lengths is the ratio of the indices of refraction in the two media in which the rays focus. In this particular ptical system, the initial and final indices are the same, so the two focal lengths are equal Forgetting for a moment about the actual formula for the focal length, if we bought a lens that somebody designed with certain radii of curvature and a certain index, we could measure the focal length, say, by seeing where a point at infinity focuses. Once we had the focal length, it would be better to write our equation in terms of the focal length directly, and the formula then is (1/s)+(1/)=1/ (27.12) Now let us see how the formula works and what it implies in different circu stances. First, it implies that if s or s'is infinite the other one is f. That means that parallel light focuses at a distance and this in effect defines f. Another interesting thing it says is that both points move in the same direction. If one moves to the right, the other does also. Another thing it says is that s and s are equal if they are both equal to 2/. In other words, if we want a symmetrical situation, we find that they will both focus at a distance 2f. 27-4M So far we have discussed the focusing action only for points on the axis. Now let us discuss also the imaging of objects not exactly on the axis, but a little bit off, so that we can understand the properties of magnification, When we set up a lens so as to focus light from a small filament onto a"point"on a screen, we notice that on the screen we get a"picture"of the same filament, except of a larger or smaller size than the true filament. This must mean that the light comes to a focus from each point of the filament. In order to understand this a little better, let us analyze the thin lens system shown schematically in Fig. 27-7. We know the follow ing facts 1 )Any ray that comes in parallel on one side proceeds toward a certain par- Fig. 27-7. The geometry of imaging ticular point called the focus on the other side, at a distance from the lens. by a thin ens. 2)Any ray that arrives at the lens from the fo out This is all we need to establish formula(27. 12)by geometry, as follows: Suppose we have an object at some distance x from the focus; let the height of the object be y. Then we know that one of the rays, namely Pe, will be bent so as to pass through the focus r on the other side. now if the lens will focus point P at all, can find out where if we find out where just one other ray goes, because the new focus will be where the two intersect again. We need only use our ingenuity to find the exact direction of one other ray. But we remember that a parallel ray goes through the focus and vice versa: a ray which goes through the focus will come out parallel! So we draw ray PT through U. (It is true that the actual rays which are doing the focusing may be much more limited than the two we have drawn, but they are harder to figure, so we make believe that we can make this ray. Since it would come out parallel, we draw TS parallel to XW. The intersection S is the point we need. This will determine the correct place and the correct height
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