Question

Chapter 26 - Introductory Physics II, section Phy1520, Summer 1 2020 WebAssign 10. 478 Points DETAILS PREVIOUS ANSWERS SERVUIPS 2D, HE DO degree to Y NOTES Example 26.6 Where is the Final Image? 4 Two thin converging lenses of focal lengths Lens ! Lens 2 = 11.0 cm and 12 = 20.0 cm are separated by 20.0 cm as illustrated in the figure 1 An object is placed 32.0 cm to the left of lens 1. Find the position and the magnification of 4 the final image. 20.0 cm A combination of two converging lenses. The ray diagram shows the location of the SOLVE IT final image (1) due to the combination of lenses. The black dots are the focal points of lens i and the red dots are the focal points of lens 2 Conceptualize Imagine light rays passing through the first lens and forming a real image (because p > in the absence of a second lens. The figure shows these light rays forming the inverted image ,. Once the light rays converge to the image point, they do not stop. They continue through the image point and interact with the second lens. The rays leaving the image point behave in the same way as the rays leaving an object. Therefore, the image of the first lens serves as the object of the second lens. Categorize We categorize this problem as one in which the thin lens equation is applied in a stepwise fashion to the two lenses. 1 Analyze Find the location of the image formed by lens 1 from the thin lens equation: 91 1 91 91 1 1 F Pi 1 11.0 cm 16 67 1 32.0 cm cm Find the magnification of the image! -- PL 9: 32.0 cm - -0.52

plz help w the blank ones!!

Answer 1

Criven h = 11.00 CM. solution Exple-206 showrot point and 2 -20cm - Pi X gi ag Suhl to est fjal lens separation = 20 cm, f = +20cm. P = -32cm, for lens 1 use dens formula 2 + f + % f Pros 1 t11 + I - 32 u an 32 32-11 11X32 11832 11732 - 116 762 cm / itve image will be behind the lens 1 Copposite side of Li

magnification='m,= 1 / 1 16 • 762 Mi = -0.52 The image formed by Lens f is object for lensa 2 sign indicates infront of lens 22. 2 +0.524 -32.00 .-ve sign for object P2 V P2 I 99 + 20 -3.24 10 3.24-20 16.76 64.8 64.8 16.76 ga -3.87 cm So, P2 = - (20cm - 9). (20-16-762) 3.238 cm JP 2 ~ - 3.24 cm Here I use as per my sign convension, it shows object present deft side of the lens. use lol lens formula for lens I fr P2 I znalo 20 x 3:24 > qa --3.87 cm. -ve ber

92 Magnification My P2 -3.87 for 3r24 21-1944 N 1.20 Mg 2 1.20 vo overall magnification of the systery M = M, Mg -0.52% 1.20 M = -0.624 Again if lens separation is 22.5cm. ther from egr (22.5 - 16.762) 5.738 0 - 5.74 cm. use lens formula. ㅗ I 92 P2 I + fa t 82 ten 92 1 92 go -t5.74

amb to 5.74 5.74 5.74-20 ga 2005.74 2 : 20x 5.74 114.26 = -14.26 20x574 8.05 cm object dinge same side of Magnification of lens 2 92 8.05 Pg - 1.4024 5.74 M2 = 1.40 Overall ( ន. ) ១ ៤១ M-MIM2 1 -0-52x264 - 0.73 M 8.18 (b). Given Radius of canvature R = 44CM f = 8/3 22 см for convex spherical mirror f=+22cm. object distance. Po = 49 cm. use mirror formula Itt at $+ q f

22x49 71 q > -- L 1 22 -49 12 t +49 49+22 q = 22849 = 15.18 image location 19 q - 15. 18 cm i sign is the image will forem opposite side of object on behind the mirror Magnification M -49 = 0.3097 No.31 (M = 0.31 15.18 10 Р