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 1.90cm -tall object is 51.0cm to the left of a converging lens of focal length 40.0cm . A second converging lens, this one having a focal length of 60.0cm , is located 300cm 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 40.0cm .

Enter your answer as two numbers separated with a comma.

s'1,ly'1l=??

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.

s'2,ly'2l=??

Answer 1

1) 1/f = 1/u + 1/v

1/40 = 1/51 + 1/v

v = 185.4545 cm

m = -v/u

m = -(185.45/51) = -3.6

h1 = m*h = 1.9*3.636 = 6.909 cm

distance s1 = 185.4545 cm right to th elens

height = 6.909 cm (real and inverted)

2) distance between the lenses = 300 cm

distance of the object of lens2 is 300-185.4545 = 114.54 cm

f2 = 60 cm

1/f2 = 1/u2+ 1/v2

1/60 = 1/114.54 + 1/v2

v2 = 125.99 = 126 cm right to the second lens

m2 = -(v2/u2) = -(125.99/114.54) = 1.1

total magnification = m1*m2 = 6.909*1.1 = 7.6

final height = 7.6*1.9 = 14.44 cm (real and upright

final distance s2 = 125.99 cm

final height = 14.44 cm

Answer 2

s1 = 51

1/s1 + 1/s1' = 1/f1

(1/51)+(1/s1') = 1/40

s1' = 185.45 cm

y1'/y1 = -s'/s = 97.61 cm

B)

s2 = 300-s1' = 300-185.45 = 114.55 cm

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

s2' = 61.46 cm

y2'/y1' = -s2'/s2

y2' = (61.46*97.61)/114.55 = 52.37 cm