Question
In the figure below, a pinecone is at distancep1 = 1.6 m in front of a lens of focal lengthf1 = 0.58 m; a flat mirror is at distanced = 2.1 m behind the lens. Light from the pinecone passesrightward through the lens, reflects from the mirror, passesleftward through the lens, and forms a final image of the pinecone.What are (a) the distance between the lens andthat image and (b) the overall lateralmagnification of the pinecone? Is the image (c)real (enter 1) or virtual (enter 0. If it is virtual, it requiressomeone looking through the lens toward the mirror),(d) to the left (enter 1) or right (enter 0) ofthe lens, and (e) inverted relative to thepinecone (enter 1) or not inverted (enter 0)?
(a)
Units
(b)
Units
(c)
Units
(d)
Units
Chapter 34, Problem 119
Explanation / Answer
Given : from the book P1 = 1.6m focul length f1 = 0.58m distance d = 2.1m suppose that the lens is placed to the left of themirror . the image formed by the converging lens is located at adistance i = (1/f - 1/p)-1 = (1/0.58 - 1/6) -1 =----m to the right of the lens, or 2.1m- 1.6m = 0.5m in frontof the mirror . the image formed by the mirror for this realimage is then at 1.0m to the right of the mirror , 2.1m+ 1.6 m = 3.7m to the right of the lens. this image then results inanother image formed by the lens, located at a distance. i1 = (1/f - 1/p1)-1 = (1/0.58m - 1/3.7m ) -1 = ---m to the left of the lens that is 2.6cm from themirror. (b) the laterial magnification is m = (- i/p) ( - i1/p1) = --- (c) the final image is real since i1> 0. (d) the image is to the left of the lens. (e) it has also the same oriention because m>0 , so imageis inverted Solve it I hope it helps you focul length f1 = 0.58m distance d = 2.1m suppose that the lens is placed to the left of themirror . the image formed by the converging lens is located at adistance i = (1/f - 1/p)-1 = (1/0.58 - 1/6) -1 =----m to the right of the lens, or 2.1m- 1.6m = 0.5m in frontof the mirror . the image formed by the mirror for this realimage is then at 1.0m to the right of the mirror , 2.1m+ 1.6 m = 3.7m to the right of the lens. this image then results inanother image formed by the lens, located at a distance. i1 = (1/f - 1/p1)-1 = (1/0.58m - 1/3.7m ) -1 = ---m to the left of the lens that is 2.6cm from themirror. (b) the laterial magnification is m = (- i/p) ( - i1/p1) = --- (c) the final image is real since i1> 0. (d) the image is to the left of the lens. (e) it has also the same oriention because m>0 , so imageis inverted Solve it I hope it helps you