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Unformatted text preview: 29. We place an object far away from the composite lens and ﬁnd the image distance i. Since the image is
at a focal point, i = f , where f equals the eﬀective focal length of the composite. The ﬁnal image is
produced by two lenses, with the image of the ﬁrst lens being the object for the second. For the ﬁrst
lens, (1/p1 ) + (1/i1 ) = (1/f1 ), where f1 is the focal length of this lens and i1 is the image distance
for the image it forms. Since p1 = ∞, i1 = f1 . The thin lens equation, applied to the second lens, is
(1/p2 ) + (1/i2 ) = (1/f2 ), where p2 is the object distance, i2 is the image distance, and f2 is the focal
length. If the thicknesses of the lenses can be ignored, the object distance for the second lens is p2 = −i1 .
The negative sign must be used since the image formed by the ﬁrst lens is beyond the second lens if i1
is positive. This means the object for the second lens is virtual and the object distance is negative. If
i1 is negative, the image formed by the ﬁrst lens is in front of the second lens and p2 is positive. In the
thin lens equation, we replace p2 with −f1 and i2 with f to obtain
−
or 1
1
1
+=
f1
f
f2 1
1
1
f1 + f2
=
+
=
.
f
f1
f2
f1 f2 Thus,
f= f1 f2
.
f1 + f2 ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics

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