Chapter 24 - 24.1. Model: Each lens is a thin lens. The...

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24.1. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize: The figure shows the two lenses and a ray-tracing diagram. The ray-tracing shows that the lens combination will produce a real, inverted image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 50 cm from the second lens and the height of the final image is 4.5 cm. (b) s 1 = 15 cm is the object distance of the first lens. Its image, which is a virtual image, is found from the thin- lens equation: 11 1 111 1 1 5 40 cm 15 cm 120 cm sf s =−= = 1 24 cm s ′=− The magnification of the first lens is ( ) 1 1 1 24 cm 1.6 15 cm s m s =− = The image of the first lens is now the object for the second lens. The object distance is s 2 = 24 cm + 10 cm = 34 cm. A second application of the thin-lens equation yields: 22 2 111 1 1 20 cm 34 cm s 2 680 cm 48.6 cm 14 s == The magnification of the second lens is 2 2 2 48.6 cm 1.429 34 cm s m s The combined magnification is ( )( ) 12 1.6 1.429 2.286 mm m == − = . The height of the final image is (2.286)(2.0 cm) = 4.57 cm. These calculated values are in agreement with those found in part (a).
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24.2. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize: The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a virtual, inverted image in front of the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 20 cm in front of the second lens and the height of the final image is 2.0 cm. (b) s 1 = 60 cm is the object distance of the first lens. Its image, which is a real image, is found from the thin-lens equation: 11 111 1 1 1 40 cm 60 cm 120 cm s sfs =−= = 1 120 cm s ′ = The magnification of the first lens is 1 1 1 120 cm 2 60 cm s m s = −= = The image of the first lens is now the object for the second lens. The object distance is s 2 = 160 cm 120 cm = 40 cm. A second application of the thin-lens equation yields: 22 2 1 1 1 40 cm 40 cm ss f =− + = + +− 2 20 cm s ′=− The magnification of the second lens is 2 2 2 20 cm 0.5 40 cm s m s = = + The overall magnification is ( )( ) 12 20 . 5 1 . 0 mm m == = . The height of the final image is ( + 1.0)(2.0 cm) = 2.0 cm. The image is inverted because m has a negative sign. These calculated values are in agreement with those found in part (a).
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24.3. Model: Each lens is a thin lens. The image of the first lens is the object for the second lens. Visualize: The figure shows the two lenses and a ray-tracing diagram. The ray tracing shows that the lens combination will produce a real, upright image behind the second lens. Solve: (a) From the ray-tracing diagram, we find that the image is 10 cm behind the second lens and the height of the final image is 2 cm.
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This note was uploaded on 01/30/2011 for the course PHYS 131 taught by Professor E.salik during the Winter '10 term at Cal Poly Pomona.

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Chapter 24 - 24.1. Model: Each lens is a thin lens. The...

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