# Locating Images of Extended Objects by Drawing Rays

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** Locating Images of Extended Objects by Drawing Rays**

Figure 35-l4a shows an object 0 outside focal point FI of a converging lens. We can graphically locate the image of any off-axis point on such an object (such as the tip of the arrow in Fig. 35-14a) by drawing a ray diagram with any two of three special rays through the point These special rays, chosen from all those that pass

through the lens to form the image, are the following. A ray that is initially parallel to the central axis of the lens will pass through focal point F2 (ray I in_2. A ray that initially passes through focal point FI will emerge from the lens parallel to the central axis_3. A ray that is initially directed toward the center of the lens will emerge from the lens with no change in its direction (ray 3 in Fig. 35-14a) because the ray encounters the two sides of the lens where they are almost parallel. The image of the point is located where the rays intersect on the far side of the lens, The image of the object is found by locating the images of two or more of its points. Figure 35-14b shows how the extensions of the three special rays can be used to locate the image of an object placed inside focal point FI of a converging lens, Note that the description of ray 2 requires modification (it is now a ray whose backward extension passes through F1). You need to modify the descriptions of rays I and 2 to use them to locate an image placed (anywhere) in front of a diverging lens. In , for example,we find the intersection of ray 3 and the backward extensions of rays I and 2

SOLUTIOII: The Key Ideas here are these: 1. Because the lens is symmetric, r. (for the surface nearer the

object) and r2 have the same magnitude r. 2. Because the lens is a converging lens. the object faces a convex

surface on the nearer side and so r. = +r. Si ilarly, it faces a

concave surface on the farther side and so r 2 = – r.3. We can relate these radii of curvature to the focal length f via

the lens maker’s equation. Eq, 35-10 (our only equation involving

the radii of curvature of a lens).

4. We can relate f to the object distance p and image distance;