lecture18

# lecture18 - 18 Lecture 18 18.1 Lens equation Going back to...

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18. Lecture 18 18.1 Lens equation Going back to the convergent lens, we can once again find an equation that determines the position and size of the image given the position and size of the object. The sign convention for the image’s position is now that s > 0 if the image is on opposite sides of the lens as the object and s < 0 if the image is on the same side of the object. This makes sense because we look at the lens from behind whereas we looked at the mirror from the front. That is a real image has s > 0 in both cases. With these conventions and looking at figure 101 we can derive, similarly as for the mirror: h s f = h f (18.1) and also h s f = h f (18.2) Since the equations are the same as for the mirror we can immediately derive s = sf s f (18.3) h = hf s f (18.4) or equivalently: 1 s + 1 s = 1 f (18.5) h s = h s (18.6) It is easy to see that once again these equations are valid if s > f or s < f . 18.2 Divergent Lens When rays arrive parallel to the axis into a divergent lens they become divergent and seem to originate from a focal point situated on the same side of the lens from where the rays arrived. As seen in fig. 104 the image is always virtual and up-right. The equation for the lens can be applied to this case if we take f < 0. – 118 –

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± ±² ³´µ ¶·µ ³²·µ ¶·µ Figure 104: Image of an object by a divergent lens. It is always virtual and upright.
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