Question

Two converging lenses with focal lengths of 40 cm and 20 cm are 16 cm apart. A 3.0 cm -tall object is 11 cm in front of the 40 cm -focal-length lens.

Part A- Calculate the image position. Express your answer using two significant figures.

Part B Calculate the image height. Express your answer using two significant figures.

Answer 1

Using the lens equation for first converging lens

1/f1 = 1/u1 + 1/v1

u1 = object distance = +11 cm

f1 = focal length = +40 cm

v1 = image distance = ?

1/v1 = 1/40 - 1/11

v1 = 40*11/(11 - 40) = -15.2 cm

Now this image will be -15.2 cm left from the converging lens, Image's distacne from 2nd converging lens will be

u2 = 16 - (-15.2) = +31.2 cm = object distnace for 2nd converging lens

f2 = focal length of 2nd converging lens = +20 cm

v2 = image distance = ?

1/v2 = 1/20 - 1/31.2

v1 = 20*31.2/(31.2 - 20) = +55.7 cm

v1 = +55.7 cm = +56 cm(+ve sign means image to the right of converging lens)

Part B.

Magnification is given by:

M = M1*M2 = (-v1/u1)*(-v2/u2)

M = (-15.2/11)*(-55.7/31.2)

M = +2.47 (no units)

M = hi/ho = 2.47

hi = image height = 2.47*object height

hi = 2.47*3 = +7.4 cm

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