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

Question

24.41

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 55.0 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.

Part A Find the location and height of the image (call it I1) formed by the lens with a focal length of 40.0 cm . Enter your answer as two numbers separated with a comma. s1, |y1| = m cm

Part B I1 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. Enter your answer as two numbers separated with a comma. s2, |y2| = m cm

Explanation / Answer

A) for the first lense

y = 1.8 cm

s1 = 55 cm

f1 = 40 cm

let s1' is the image distance.

apply, 1/s1 + 1/s1' = 1/f1

1/s1' = 1/f1 - 1/s1

1/s1' = 1/40 - 1/55

s1' = 146.6 cm

= 1.466 m <<<<<<<<<----------------Answer

magnification, m1 = -s1'/s1

= -146.6/55

= -2.67

so, |y1'| = |m|*y

= 2.67*1.8

= 4.8 cm <<<<<<<<<----------------Answer

B) for second lens

s2 = 300 - 146.6 = 153.4 cm

f2 = 60 cm

let s2' is the image distance.

apply,

1/s2 + 1/s2' = 1/f2

1/s2' = 1/f2 - 1/s2

1/s2' = 1/60 - 1/153.4

s2' = 98.5 cm

= 0.985 <<<<<<<<<----------------Answer

magnifiction, m2 = -s2'/s2

= -98.5/153.4

= -0.64

|y2'| = m2*y1'

= 0.64*4.8

= 3.08 cm <<<<<<<<<----------------Answer

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