Question

Please help with this practice question about optics!

6. A certain convex lens is used to create an image of an object that is 2.0 m in front of the lens on a white screen that is 15.0 cm behind the lens. For the sake of clarity in the description that follows, assume that the object, lens, and screen are on the x axis, the lens sits at the origin, and the object is in the positive x direction (so the object, lens, and screen are at x = 2.0 m, x = 0.0, and x = -15 cm respectively. A second lens is then placed in front of the first at x = 5.0 cm to create an image of an object located at x = 5.0m on the screen. 1. What is the focal length of the first lens? 2. Is the second lens convex or concave? 3. What is the focal length of the second lens? 4. Before the second lens is placed in front of the first, the image of the object at x = 5.0 m exists, it is just not at the plane of the screen. Is this image in front of or behind the screen? 5. How far is the image of the object at x = 5.0 m moved to place it on the screen?

Answer 1

0 0 12.00 x = = 0 e object distance = +2.om. & Image distance - 5.0m - 5+15 in using len's formula 1 1 1 1 1 1 1 1 1 1 2 À = 0.5+20 - 2015 i focal length of first lens = = 0.049m 20-5 = 4.9cm concave Lepas. We need real image on screen, Hent internal Hence second lens should form virtual image exactly at 2 m fram fost law Hence second lens is e Concave $14.9cm x=5.om 2= Sem X--5-0cm For second reus object distance - 495cm Let object distance is then focal length fo = 2 (on Image distance de la This o should be equal to 2.om from x on 185cm from dens fa, then 195= f₂ x 495 0 t 495-f, u lee selaku 2-321.75cm f2 = -3-22 m Concare lans. 495x195 focal length of second lens

C If second lens not placed, then image distance & form object r = 50min - 500cm o l. f 4.9 500 - 4.946m 21 of 500-49 Hence image is infront of screen. f. = 4.0cm 4= 5.0cm then object position re of = 0.1 object s.om to be shifted to 2-47mine, it is to be shifted by = 2 of 5.0r4 % 247.2cm