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971_PartUniversity Physics Solution

# 971_PartUniversity Physics Solution - Geometric Optics...

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Geometric Optics 34-9 (b) The image is inverted since both the image and object are real ( 0, 0). s s ′ > > (c) 1 1 1 1 1 0.0732 m, 0.0741m 5.93 m f f s s = + = + = and the lens is converging. E VALUATE : The object is close to the lens and the image is much farther from the lens. This is typical for slide projectors. 34.29. I DENTIFY : Apply 1 2 1 1 1 ( 1) n f R R = . S ET U P : For a distant object the image is at the focal point of the lens. Therefore, 1.87 cm f = . For the double- convex lens, 1 R R = + and 2 R R = − , where 2.50 cm R = . E XECUTE : 1 1 1 2( 1) ( 1) n n f R R R = = . 2.50 cm 1 1 1.67 2 2(1.87 cm) R n f = + = + = . E VALUATE : 0 f > and the lens is converging. A double-convex lens is always converging. 34.30. I DENTIFY and S ET U P : Apply 1 2 1 1 1 ( 1) n f R R = E XECUTE : We have a converging lens if the focal length is positive, which requires 1 2 1 2 1 1 1 1 1 ( 1) 0 0. n f R R R R = > > This can occur in one of three ways: (i) 1 R and 2 R both positive and 1 2 R R < . 1 2 (ii) 0, 0 R R (double convex and planoconvex). (iii) 1 R and 2 R both negative and 1 2 R R > (meniscus). The three lenses in Figure 35.32a in the textbook fall into these categories. We have a diverging lens if the focal length is negative, which requires 1 2 1 2 1 1 1 1 1 ( 1) 0 0. n f R R R R = < < This can occur in one of three ways: (i) 1 R and 2 R both positive and 1 2 R R > (meniscus). (ii) 1 R and 2 R both negative and 2 1 R R > . (iii) 1 2 0, 0 R R (planoconcave and double concave). The three lenses in Figure 34.32b in the textbook fall into these categories. E VALUATE : The converging lenses in Figure 34.32a are all thicker at the center than at the edges. The diverging lenses in Figure 34.32b are all thinner at the center than at the edges. 34.31. I DENTIFY and S ET U P : The equations 1 1 1 s s f + = and s m s = − apply to both thin lenses and spherical mirrors. E XECUTE : (a) The derivation of the equations in Exercise 34.11 is identical and one gets: 1 1 1 1 1 1 , and also . s f sf s f s m s s f s f s fs s f s f s + = = = = = − = (b) Again, one gets exactly the same equations for a converging lens rather than a concave mirror because the equations are identical. The difference lies in the interpretation of the results. For a lens, the outgoing side is not that on which the object lies, unlike for a mirror. So for an object on the left side of the lens, a positive image distance means that the image is on the right of the lens, and a negative image distance means that the image is on the left side of the lens. (c) Again, for Exercise 34.12, the change from a convex mirror to a diverging lens changes nothing in the exercises, except for the interpretation of the location of the images, as explained in part (b) above.

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