When a thin converging lens is oriented perpendicular to sunlight, the beam is f
ID: 2018789 • Letter: W
Question
When a thin converging lens is oriented perpendicular to sunlight, the beam is focused on a piece of paper if the paper is held 10 cm from the lens. A) Suppose now that a Buzz Lightyear toy with height 5 cm is placed 15 cm in front of the lens. Find the image location and image height. State whether the image is real or virtual, and whether it is upright or inverted. B) In the previous setup, where would a second thin converging lens (same focal length) need to be placed in order to produce a final inverted, virtual image of Buzz that has a combined magnification of -6? C) Where is the final image in this case? D) Include a ray diagram (only 2 rays required per lens).Explanation / Answer
Hello first lens: focal length f = 10 cm, do = 15 (distance of object) use 1/do + 1/di = 1/f 1/15 + 1/di = 1/10 di = 30 cm (distance of image) Image is inverted, enlarged, real Magnification = di/do = 30/15 = 2. -------------- second lens: To reach a total magnification of 6, the image by the second lens must be enlarged 3 times. It must be upright (relative to the image of the first lens, but inverted relative to the object of the first lens) , and virtual --> it follows, that the object (= image of the first lens) must be within the focus length of the second lens (f = 10 cm). Again, the image size = 3 = - di/do = (- di + f)/f (with "- di " because the virtual image is on the object side) 3 = (-di + 10)/10 ---> di = - 20 cm and to find 'do' use magnification = -di/do 3 = 20/do do = 6,67 cm So the image of the first lens must be 6,67 cm before the second lens. Since the image of the first lens is 30 cm behind the first lens, --->it follows, that the second lens is 36,67 cm behind the first lens. Total magnification = 2*3 = 6, the image is inverted and virtual. Regards
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