A converging lens ( f = 12.2 cm) is located 29.4 cm to the left of a diverging l

Question

A converging lens (f = 12.2 cm) is located 29.4 cm to the left of a diverging lens (f = -5.64 cm). A postage stamp is placed 35.1 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?

(d) With respect to the original object, is the final image upright or inverted?

(e) With respect to the original object, is the final image larger or smaller?

Explanation / Answer

The converging lens will project a real image of the stamp at

1/12.2 = 1/si + 1/35.1

si = 1/ ( 1/12.2 - 1/35.1) = 18.7

the magnification of this first image is

m_1 = - (35.1 / 18.7) = - 1.877

For the second lens, the image is at

-1/5.64 = 1/si + 1/(29.4-18.7)

si = -3.693

So, the image is virtual and located -145 cm from the diverging lens. The magnification of the image is

m_2 = - ( -3.693/(29.4-18.7) ) = 0.345

The overall magnification is then

m = m_1 m_2 = - 0.64

the image is virtual and located 3.693 cm to the left of the diverging lens.

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