Question

An object is placed 20.5 cm from a screen.

Part A

At what two points between object and screen may a converging lens with a 2.40 cm focal length be placed to obtain an image on the screen?

Enter your answer as two numbers, separated with a comma. Count the distances from the object.

Part B

What is the magnification of the image for each position of the lens?

Enter your answer as two numbers, separated with a comma.

Answer 1

f-+240 cm. The positions of the object, lens, and screen are sketched in the figure below. Since the image is projected onto a screen, the image is real and s'>0. The figure shows that s'+s 20.5 cm and Set Up: s' = 20.5 cm-s. Screen Lens Object 20.5cm 20.5cm Solve: (a) -. s( 20.5 cm-s) = 49.2 cm2. S2-1 20.5cm)s+ -+-=--+ s =_ f s 20.5 cm-s f s( 20.5cm-s) 2.40 cm 49.2 cm. 17.7 cm or s 2.8 cm The lens can be cm or cm from the object. (b) s- 17.7cm: s'-20.5 cm-17.7 cm-28 cm. In 2 80cm-- 0.160 s-28 cm: s'-20.5 cm-28 cm = s 17.7 cm , 17.7 cm, m=-s= 17.7 cm 17.7 cm. m 2.8 cm 6.32

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