Question

pps W ww NCOIS Property Connect MLS - Conn... U U Citrix XenApp - Log... 5 greystar.devsuite.n. EXAMPLE 23.3 Images Formed by a Convex Mirror GOAL Calculate properties of a convex mirror. PROBLEM An object 3.00 cm high is placed 20.0 cm from a convex mirror with a focal length of magnitude 8.00 cm. Find (a) the position of the image, (b) the magnification of the mirror, and (c) the height of the image. STRATEGY This problem again requires only substitution into the mirror and magnification equations. Multiplying the object height by the magnification gives the image height. SOLUTION (A) Find the position of the image. Because the mirror is convex, its focal length is negative. Substitute into the mirror equation 20.0 cm 9 -8.00 cm Solve for a. 9 = -5.71 cm (B) Find the magnification of the mirror. Substitute into the magnification equation 1. (2020.com ) - 0.2 equation. (C) Find the height of the image Multiply the object height by magnification = (3.00 cm)(0.286) = 0.858 cm

Answer 1

Answer:

) - Practice its Question: The image is always Virtual Coptrén given that LE= 20.1om f = - 8.30cm Enge height= 2.52com * * - // - - - - 20.1cm (a) » VEOLTOZ = -5-87 95.9 cm virtual image is behind mirror (b) magnification = - ---509 cm) 70.2937 (C): height of image - (2.52 cm) (0.293) = 10.738 cm) Excerciser u=40 15cm f = -8.30 com So →=-= - - Zuze - r - - 0.36 | @) >> V-361 = (-2.77 cm ib virtual & behind the misser given magnification - --V=-62077cm) o 4.150m 7.6677 2 C) As magnification is positive it's cepšight, height of image = (2.52 cm) (0.667) = 1.68 cm) The image 9s upright & virteal

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