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Unformatted text preview: PHY6938 Proﬁciency Exam Spring 2002 April 5, 2002 Optics and Thermodynamics 1. A combination of two thin, convex lenses are placed as shown in the ﬁgure. An object is placed 4 cm in front of the ﬁrst lens, which has a focal length of 12 cm. The second lens is 12 cm behind the ﬁrst and has a focal length of 6 cm.
Object Lens 1 Lens 2 4 cm 12 cm (a) Use the thin-lens formula to ﬁnd the image of the ﬁrst lens, and show it through a ray-diagram on the sketch. We can ﬁnd everything we need from the lens formula(e) 1 1 s 1 + = , m=− , ss f s (1) The lenses are convex, this means that the focal points f1 and f2 are positive. Since the object is in front of the ﬁrst lens then s1 is positive too. Substitute the values of f1 and s1 in Eq.1 and check the result for s1 with the ﬁgure. f1 s’ 1 s1 f1 1 1 1 1 1 1 − = − =− , = s1 f1 s1 12 cm 4 cm 6 cm This means that s1 = −6 cm. (2) (b) Is the image real or virtual ? Since s1 = −6 cm, this means that the image is in front of the ﬁrst lens. Therefore the image is virtual. (c) Find the image for the lens combination using the thin-lens formula, and use a ray-diagram to mark the image at the correct location on the sketch. To ﬁnd the image for this combination of lenses we use Eq.1 again. The image of the ﬁrst lens, as calculated in part (a) is located at 6 cm in front of the ﬁrst lens. Therefore the distance between this image and the second lens s2 = 18 cm. f2 s’ 2 s2 f2
1 1 1 1 2 1 = − = − =− , s2 f2 s2 6 cm 18 cm 18 cm s2 = 9 cm (3) (d) Is the resulting image real or virtual ? Since s2 = 9 cm, is positive, the image is behind the second lens. Therefore, the image is real. ...
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This note was uploaded on 03/11/2012 for the course PHY 3900 taught by Professor Staff during the Fall '1 term at FSU.
- Fall '1