When two lenses are used in combination, the first one forms an image that then

Question

When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 2.00 cm -tall object is 60.0 cm to the left of a converging lens of focal length 40.0 cm. A second converging lens, this one having a focal length of 60.0 cm. is located 300 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I_1) formed by the lens with a focal length of 40.0 cm. Enter your answer as two numbers separated with a comma. I_1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses. Enter your answer as two numbers separated with a comma.

Explanation / Answer

Part A) Thin-lens equation using Cartesian sign convention is:

1/v = 1/f + 1/u

=> 1/v = 1/40 + 1/(-54)

=> v = (54 * 40) / (54 - 40) = 154.3 cm

Height of image, hi = ho(v/u) = 1.90 * 154.3 / (-54) = -5.43 cm

Negative sign indicates that the image is inverted.

Part B) Object distance for the second lens, u = -(300 - 154.3) = -145.7 cm

Now, 1/v = 1/f + 1/u

=> 1/v = 1/60 + 1/(-145.7)

=> v = 60 * 145.7 / (145.7 - 60) = 102 cm

Height of image, hi = ho(v/u) = (-5.43) * 102 / (-145.7) = 3.8 cm

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