Question

Photos - Sereenshot (131).png X N See all photos + Add to 0 Search Edil & Creale Slate Two converging lenses with respective focal lengths f1 = 8.00 cm and f2 = 4.00 cm are separated by 20.0 cm. The lens on the left has the longer focal length. An object stands 24.0 cm to the left of the first lens in the combination a) On a graph paper, trace both lenses, their principal axis, their focal lengths, the object and then trace the ray diagram to show the image. [4 points] b) Determine the final image distance obtained by the system relative to the lens on the right. [2 points] c) Calculate the overall magnification. [2 points] d) Is the final image real or virtual with respect to the original object? Justify (2 points] e) Is the final image upright or inverted relative to the object? Justify [2 points] f) is the image larger or smaller? Does this agree with your ray diagram? Explain (2 points] 10:31 AM Type here to search O 5/16/202026

Answer 1

f = 8.00cm f2=4.00cm (a) В 311 다. LAIL IP 24.0cm Principal axis 12cm ст som! 20.0 cm → do (b) Fix left lens (fi = 8.00cm) do = distance of object lens, die distance of image from lens. do=-24 cm, fi=8cm. using t - to - 12/17 - 12/24 - 12 = t - - 8 - 12 24 = 3 2 1 1 2 - 142 Idi= 12cm & Magnification MiddeF% = For second lens ( f2=4cm) So distance of this image from ngut lens = 20-12 = 8cm do = -8 cm, fp=4cm, doo -a a co - 20 = :- 33 = (di=8cm => Final image 8cm Right of Right magnification M2 =d8%20- .-1=1 m2=-1 ② overall magnification M= M M = )(-1) = 5 (d) Final image is real because rays are actually meet to Magnification & pacitive => Final image is erect wat original object or can seen dur derechy(using ray diagram). (A) Final image is smaller than original object Cactually just half) yes may ray diagram agrees this as Image AVBI L AB(object) Lens Form image I2 (e)