Question

A converging lens with a focal length of 40 cm and a diverging lens with a focal length of -40 cm are 160cm apart. A 3.0-cm-tall object is 60 cm in front of the converging lens.

Part A

Calculate the image position.

Part B

Calculate the image height.

Answer 1

40 60 = 0025-00167 0.0083 - 120.48 cm This image acts as object for the diverging lens And the object distance for the diverging lens is p2 160 cm -120.48cm 39.52 cm Hence the image distance is given by using the lens formula as (-10) (39.52) --0.025-0.0253 =-0.0503 42-1988 cm Thus the final image is at a distance of 19.88 cm fro the áiver ging lens andis at a ditamie t dthe image is between 140.12 cm from the convex lens and the image is between the two lenses. Magnification of the combination of the lenses is given by -120.5 ㎝)(-(-19.88 cm) 60 cm 39.5 cm =-101 cm

image height = 3.0 *-1.01 = - 3.03 cm

image is inverted

Answer 2

the thin lens equation

1/f = 1/do + 1/di

where do = object distance and di = image distance

Objects located within the focal point for converging lenses will produce virtual upright images. (in fact if you plug in (b) 2.75 into the equation

1/11 = 1/2.75 = 1/di you will see that the di = negative (indicating a virtual image)