Question

A converging lens (f = 13.3 cm) is located 33.0 cm to the left of a diverging lens (f = -5.88 cm). A postage stamp is placed 36.6 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.)

(b) Find the overall magnification.

Answer 1

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

1/13.3 = 1/si + 1/36.6

si = 1/ ( 1/13.3 - 1/36.6) = 20.89

the magnification of this first image is

m_1 = - (36.6/20.98) = - 1.744

For the second lens, the image is at

-1/5.88 = 1/si + 1/(33 - 20.89) = 1/si + 1/12.11

si = 1/( -1/5.88 -1/12.11) = -3.958

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

m_2 = - ( -3.958/12.11) = 0.3267

The overall magnification is then

m = m_1 m_2 = - 0.5701