Question

(10%) Problem 10: An object of height 2.8 cm is placed 28 cm in front of a diverging lens of focal length 16 cm. Behind the diverging lens, and 13 cm from it, there is a converging lens of the same focal length. 25% Part (a) Find the location of the image created by the diverging lens in centimeters relative to the location of that lens. Use the standard sign convention such that your answer will be negative if the image is on the same side as the object. s'= - 10.18 s'=-10.18 Correct! 25% Part (b) What is the magnification of the image created by the diverging lens alone? Include its sign to indicate its orientation with respect to the object. m1 = 2.23 X Attempts Remain ~ 25% Part (c) Find the location of the final image, in centimeters beyond the converging lens (such that a negative answer would indicate the final image is on the same side of the converging lens as the object) s"= 51.65 Correct! Par (o) What is the magnification of the final image include its sign to indicate to orientation with respect to the object * 25% Part (d) What is the magnification of the final image? Include its sign to indicate its orientation with respect to the object. Grade Summary Mtotal = 2.23 Deductions 3%

Answer 1

Anaucen Part b fo-16 cm grven bocat lengeh S 28cm 우-후 -우 28 -16 &= -10-18cm -G10-18)/28 10-13/28 is the magnibicabion of imaga Created by diverging tens obiect distance b Conenging lens Rart d SI 13 +1O 18 = 23.18 Cm 14 f16cm = 51.65 Cm - 51-65/23-13 The magnibiabon M2 =-2.228 there bee magnibiabon ot biral Mmxm2 image is 36 x-0228 M -0.802