Question

A diverging lens with a focal length of -15 cm is placed 16 cm to the right of a converging lens with a focal length of 18 cm . An object is placed 33 cm to the left of the converging lens. Part A Where will the final image be located? to the left of the diverging lens to the right of the diverging lens Submit Previous Answers Correct Part B Express your answer using two significant figures. V AEO ? d; = cm from the diverging lens Submit Request Answer
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Answer 1

distance nce a 33cm Yren &a= -15cm, object -) 4,2 33 cm f = 18 un L2 1 formula, By lens € 1-1 ܩܟ 11 <!- -(-) 18 33 a) 1 ta HEHE 1 18 -1 33 0.0 25 V, 2 +39.59 an on sign Now this image formed day converting lenish works as an object for diverging lens. Positive shows that image is formel right alde of line Now object distance for da"

M2 = (39.59 - 16 cm ) u :-(23:59 am) right of leus 1 2 115=tissa) 5) - 23.59 1 -15 V₂= a. 169 cm And Idi= =) Negative sign shows that image is formed on same side the object is placed And image will be right of lens 9.17 cm from diverging lens e Given distance between object Object distances = d image distance = 88.4-d f = 11.0cm and screen = 68.4 cm 9 Scrum 68-4-0) 7 کا۔ V by lens formula ao I 68.4.d a) 11

1 1 68.4-d + 1 n 11 d 5 d +6814-d (684-d) (d) 11 -> (68.4-d)d (11)( 68 8.4) =) 68.4d-d2 = 952-4 - d² - 68.4d +752.4=0 d = 68.4+ 168.0)2- 4(1)( 152.4) 2137 & 2 13.77 54cm on sa d = 13:47 So, lens should be placed con 13.77 cm from the object ar Id=54.67 an dz ㄷ 13.77, 54.63 cm