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# lecture9 - Lecture 9 Notes: 07 / 13 Multiple-lens systems...

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Lecture 9 Notes: 07 / 13 Multiple-lens systems If we have one lens behind another, we can simply treat the image formed by the first lens as an object for the second lens. For example, suppose we have two convergent lenses, with focal lengths f 1 and f 2 , separated by a distance L . The object is located at a distance p 1 in front of the first lens. We want to locate the image and find the magnification. This is a typical ray diagram, to show our distances and sign conventions: For this particular diagram, all the quantities are positive. If one of the lenses was divergent, its focal length would be negative; if one or both of the images was on the same side as its corresponding object, the value of q for that image would be negative. Let us calculate q 2 for given p 1 , L , f 1 and f 2 . The equations for the two lenses are We can simply use the first equation to calculate q 1 and plug into the second. For example, suppose that the object is 50 cm away, the lenses are 120 cm apart, the first lens has a focal length of 30 cm and the second lens has a focal length of 100 cm . Then, our equation will give us This means that the final image will be virtual, and will be 82 centimeters in front of the second lens (in the direction of the object). This is unlike the ray diagram above, where the final image is real and behind the second lens.

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We can calculate the magnification of this two-lens contraption. Since the magnification of the first lens is M1 = -h 1 / h , where h is the height of the object and h 1 is the height of the first image, and M2 = -h 2 / h 1 , since the first image is now the object and h 2 is the height of the second image, M 1 M 2 = h 2 / h = M is the total magnification from the object to the final image (the sign cancels because a sequence of two inverted real images give an upright real image, so the magnification for a real image is positive.) Since M 1 = -q 1 / p 1 and M 2 = - q 2 / p 2 = q 2 / ( L - q 1 ), the magnification is For our example, this gives The final image is therefore inverted, and 2.7 times larger than the original. Example: We might ask how the magnification depends on the distance L ; perhaps our apparatus is adjustable, and we want find a length that gives a certain magnification. M 1 = -q 1 / p 1 does not depend on the distance L , but M 2 does. We have The overall magnification is thus equal to What should we make the length of our apparatus if we want a magnification of, say, -5? Solve this equation for L : So if we separated the lenses by 145 cm and left the object 50 cm from the front lens, the magnification would now be -5.
Suppose we have an object at infinity, and we look at it using lenses of focal lengths f 1 and f 2 separated by a distance L . Where does the image form? The first lens will form an image at the focal point, since parallel rays coming from the

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## This note was uploaded on 09/19/2011 for the course PHYS 1C taught by Professor Smith during the Spring '07 term at UCSD.

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lecture9 - Lecture 9 Notes: 07 / 13 Multiple-lens systems...

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