A converging lens of focal length 8.000 cm is 19.7 cm to the left of a diverging

Question

A converging lens of focal length 8.000 cm is 19.7 cm to the left of a diverging lens of focallength -6.00 cm. A coin is placed 12.4 cm to the left of the converging lens. (a) Find thelocation of the coin's final image relative to the diverging lens.(Include the sign of each answer. Enter a negative value if theimage is to the left of the diverging lens.) (b) Find themagnification of the coin's final image. (a) Find thelocation of the coin's final image relative to the diverging lens.(Include the sign of each answer. Enter a negative value if theimage is to the left of the diverging lens.) (b) Find themagnification of the coin's final image.

Explanation / Answer

a) the image distance of converging len : q1 = 1/(1/f1-1/p1) = 22.55 cm => the object distance with diverging len : p2 = L- q1 = 19.7-22.55 = -2.85 => the image distance : q2 = 1/(1/f2 -1/p2) = 5.43 cm (the image to the right of the diverginglen) b) the magnification : M = M1*M2 = (-q1/p1) * (-q2/p2) =(-22.55/12.4)*(-5.43/-2.85) = -3.465

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