When two lenses are used in combination, the first one forms an image that then

Question

When two lenses are used in combination, the first one forms an image that then serves as the object for the second lens. The magnification of the combination is the ratio of the height of the final image to the height of the object. A 1.80 cm -tall object is 50.0 cm to the left of a converging lens of focal length 40.0 cm . A second converging lens, this one having a focal length of 60.0 cm , is located 300 cm to the right of the first lens along the same optic axis. Find the location and height of the image (call it I_3) formed by the lens with a focal length of 40.0 cm. I_1 is now the object for the second lens. Find the location and height of the image produced by the second lens. This is the final image produced by the combination of lenses.

Explanation / Answer

given data

y = 1.8 cm

for first lense

f1 = 40 cm

object distance, u1 = 50 cm

let v1 is image distance

Apply, 1/u1 + 1/v1 = 1/f1

1/v1 = 1/f1 - 1/u1

1/v1 = 1/40 - 1/50

v1 = 200 cm <<<<<<-------Answer

magnification, m1 = -v1/u1 = -200/50 = -4

image height, y1' = |m1|*y

= 4*1.8

= 7.2 cm <<<<<<-------Answer

b)

for second lense

f2 = 60 cm

object distance, u2 = 300 - 200 = 100 cm

let v2 is the image distance.

apply, 1/u2 + 1/v2 = 1/f2

1/v2 = 1/f2 - 1/u2

1/v2 = 1/60 - 1/100

v2 = 150 cm <<<<<<-------Answer

magnification, m2 = -v2/u2

= -150/100

= -1.5

y2' = |m2|*y1'

= 1.5*7.2

= 10.8 cm <<<<<<-------Answer

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