Question

CHAPTER 24 GEOMETRIC OPTICS As an object moves from just inside the focal point of a sonverzing lens to just gutside it, he image goes from andto and A. large; inverted, large; upright. B. large; upright; large; inverted. C. small: inverted; small; upright. D. small: upright; small; inverted. 7. 8. When the reflection of an object is seen in a concave mirror, the image is A. Virtual and erected. B. Real and inverted. C. Virtual and inverted. D. Real and erected. 9. What should be the Power of a.converging lens located at 12.5 em from an ant 1 image with height of 5.35x10 m B. +3.5 D C.-7D D +7D E. None of above 10. In order to form an image by means of a convex spherical mirror (curvature radius of 2em) that is three times smaller as the object, how far away from the mirror should the object be placed? A. 1 cm B. 2 cm C. 3 cm D. None of above

Answer 1

Large, upright large, inverted Due to Reflection the image is virtual & inverted M = h_i/h_0 = -d_i/d_0 = 5.35 times 10^-4/1 times 5^-3 m = -d_i/d_0 = 0.535 d_i = -0.535 times 12.5 d = -6.6875 foci length d_i d_0/d_i + d_0 = -6.6875 times 12.5/-6.6875 + 12.5 f = 15.735 cm Power (P) = 1/-f = 1/-15.735 times 10^-2 P = -7D foci length (f) = -2R f = -2 times 2 = -4cm -h_i/h_0 = 1/3 = > h_i = 1/3 h_0 -d_i/d_0 = 1/3 = > -d_0/3 therefore d_0 = d_i f/d_i - f = -d_0/3 times (-4)/-d_0/3 + 4 = > d_0 (-d_0 + 12) = 4 d_0 d_0 = 3 cm at 3 cm