A converging lens (f = 12.7 cm) is located 31.8 cm to the left of a diverging le

Question

A converging lens (f = 12.7 cm) is located 31.8 cm to the left of a diverging lens (f = -5.64 cm). A postage stamp is placed 39.6 cm to the left of the converging lens. (a) Locate the final image of the stamp relative to the diverging lens. (Include sign to indicate which side of the lens the image is on.) cm (b) Find the overall magnification. (c) Is the final image real or virtual? real virtual (d) With respect to the original object, is the final image upright or inverted? upright inverted (e) With respect to the original object, is the final image larger or smaller? larger smaller

Explanation / Answer

You can multiply the magnifications of the two lenses, but these magnifications depend on the image and object distances for each, and finding these values is a bit tricky. See the refs. The key to this kind of problem is that the 1st (left) lens's (intermediate) image is the 2nd (right) lens's object. Then using notation from ref. 1 (+uN is object distance to left of lens N and +vN is image distance to right of lens N), we have initially: u1 = 39.6 cm; f1 = 13 cm; sep = 29.7 cm; f2 = -5.94 cm; v2 = TBD Solving, v1 = 1/(1/f1-1/u1) = 19.3533834586466 cm u2 = sep-v1 = 10.3466165413534 cm v2 = 1/(1/f2-1/u2) = -3.7735831809872 cm Final Image location relative to lens 2 = v2 = -3.7735831809872 cm Final Image location relative to lens 1 = v2+sep = 25.9264168190128 cm Final Image location relative to object = v2+sep+u1 = 65.5264168190128 cm Magnification = v1v2/(u1u2) = -0.178244972577697 P.S. I don't see why you shouldn't post this but perhaps you and/or your school have rules to that effect. However, I hope you won't delete the question to hide your tracks.

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