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 1.20 cm -tall object is 56.0 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.
a) Find the location and height of the image (call it I 1) formed by the lens with a focal length of 40.0 cm
140,3.00
b) 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.
Solution
Part A)
Using the formula: 1/f = 1/p + 1/q
Putting the values
1/40 = 1/60 + 1/q
q = 120 cm
We know that h\'/h = -q/p
h\'/1.2 = -120/60
h\' = -2.4 cm (Here negative means its inverted)
Part B)
From the question the lenses are 300 cm apart, the object distance for the second lens is 300 - 120 = 180 cm
1/f = 1/p + 1/q
1/60 = 1/180 + 1/q
q = 90 cm
h\'/h = -q/p
h\'/-2.4 = -90/180
h\' = 1.2cm
.
When two lenses are used in combination- the first one forms an image.docx
1. 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 1.20 cm -tall object is 56.0 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.
a) Find the location and height of the image (call it I 1) formed by the lens with a focal length of
40.0 cm
140,3.00
b) 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.
s ?1, | y ?1| = 140,3.00 cm
Solution
Part A)
Using the formula: 1/f = 1/p + 1/q
Putting the values
1/40 = 1/60 + 1/q
q = 120 cm
We know that h'/h = -q/p
h'/1.2 = -120/60
h' = -2.4 cm (Here negative means its inverted)
Part B)
From the question the lenses are 300 cm apart, the object distance for the second lens is 300 -
