24_InstSolManual_PDF_Part3

24_InstSolManual_PDF_Part3 - Geometric Optics 24-3 Solve:...

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Solve: (a) so and The image is 7.5 cm in front of the mirror. The image is 4.0 mm tall and is inverted. (b) The image is 10.0 cm in front of the mirror, at the location of the object. The image is 8.00 mm tall and is inverted. (c) The image is 5.00 cm behind the mirror. The image is 16.0 mm tall and is erect. (d) The image is 5.00 cm in front of the mirror. The image is 0.040 mm tall and is inverted. Reflect: From we see that the image is real if and virtual if Real images are in front of the mirror and are inverted. Virtual images are behind the mirror and are erect. 24.6. Set Up: To find the location of the image use For a convex mirror To find the height of the image use Solve: (a) so The image is 3.75 cm behind the mirror. It is 2.00 mm tall and is upright. (b) The image is 3.33 cm behind the mirror. It is 2.67 mm tall and is upright. (c) The image is 1.67 cm behind the mirror. It is 5.33 mm tall and is upright. (d)
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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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