A 15 cm-focal-length converging lens is 18 cm to the right of a 8.0 cm-focal-len

Question

A 15 cm-focal-length converging lens is 18 cm to the right of a 8.0 cm-focal-length converging lens. A 4.0-cm tall object is distance L to the left of the first lens.

Question: For what value of L is the final image of this two lens system half way between the two lenses?

Explanation / Answer

Greetings... focal Length_1=8cm focal Length_2 =15cm s'2 =-18/2 = -9 cm (for converging lens) For the thin lens: 1/focal Length_2 =1/s2 + 1/s2' ----> Therefore: s2 = 5.63 Since the Image is formed at 4 cm left from the second lens. Then, the distance between object and image from the first lens. s1' = 18 cm - 5.63 cm = 12.37 cm from the thin lens equation , 1/focalLength_1 =1/s1 + 1/s1' ----> s1=22.65cm b) magnification M = M1M2 =(-s1'/s1)(-s2'/s2) .......... (3) M = -0.88 final image height h' = hM = (4 cm)(-0.88) = -3.5 cm (image is inverted) Best Regards!

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