Question

A converging lens has a focal length of 13.0 cm. For each of two objects located to the left of the lens, one at a distance of 81 = 18.5 cm and the other at a distance of S2 = 6.50 cm , determine the following. Part A For related problem-solving tips and strategies, you may want to view a Video Tutor Solution of Image formed by a diverging lens. Determine the image position. Express your answer in centimeters separated by a comma to three significant figures. V ALQ R O a ? 81, 8, = Submit Request Answer Part B Determine the magnification. Express your answer in centimeters separated by a comma to three significant figures. IVO A MO ? mi, m2 = Submit Request Answer

Answer 1

For first object- Let si is the image Object distance, S = -18.5.com distance to - Ś Ć I Lens relation] : op, Oy, - vhoisis or, 5 = 18.5 – 13.0 18.5 – 13.0 13.0x18.5 S.5 .: S;' = 13.0 x 18.5 cm per E 43.727 cm second object object distance, S is the image distance S2 = - 6. Socom for Let ot, eta + 6s0 13.00 Oy, 13.0-350 ' os, = 6.50 13.0 13.0 X 6.50 308 650 . S- - (13-0x Goso) * Sz' - - 13.0 cm PART-A S;' = 43. 727 S = – 13. 0 cm Cm object 1 is PART-B Magnification for m, = ls = 43. 727 18.5 = 2.36h

magnification of the Second object, m = ( 13.0 e .. 6.50 = 2 D PART-C The the image of image of the the first second object is real and object is virtual PART-D The image of image of the and the first object is inverted second object is upright,