24_InstSolManual_PDF_Part24

# 24_InstSolManual_PDF_Part24 - 24-24 Chapter 24 This image...

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This image is 51.3 cm to the right of the lens. so the image is real. so the image is inverted. Image formed by the light that first reflects off the mirror: First consider the image formed by the mirror. The candle is 20.0 cm to the right of the mirror, so The image formed by the mirror is at the location of the candle, so and The second image is 51.3 cm to the right of the lens. so the final image is real. so the final image is erect. Reflect: The two images are at the same place. They are the same size. One is erect and one is inverted. 24.66. Set Up: Refer to Figure 24.15a in Section 24.2. R and are negative, so and For small angles, and and Eliminating gives Substituting in the expressions for the angles gives Dividing by h and replacing by and by gives Rearranging gives which is Equation (24.4). Now refer to Figure 24.15b in Section 24.2. Triangles PVQ and are similar, so so which is Equation 24.7. 24.67. Set Up:
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## This note was uploaded on 03/06/2009 for the course PHYS 114 taught by Professor Shoberg during the Spring '07 term at Pittsburg State Uiversity.

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