14. Two lens, each having a power of +10D, are placed 20cm apart along the same axis. If an object is 60cm from the first lens (not in between the lenses), where is the final image relative to the first lens, and what are its characteristics? Answer: 20cm on object side of the first lens. The image is virtual and inverted and the same size as the object.
Solution
lens1
object distance = s1 = 60 cm
image distance = s1\' = ?
focal length f1 = 1/P1 = 1/10 = 0.1 m = 10 cm
1/s1 + 1/s1\' = 1/f1
1/60 + 1/s1\' = 1/10
s1\' = 12 cm
magnification of converging lens m1 = -s1\'/s1 = -0.2
for lens 2:
___________
object distance s2 = 20 - 12 = 8 m
focal length f2 = f1 = 10 cm
1/s2 + 1/s2\' = 1/f2
1/8 + 1/s2\' = 1/10
image distance s2\' = -40 cm
image is at a distance 40 cm from second lens
image is at distance 40 - 20 = 20 cm from first lens
m2 = -s2\'/s2 = 40/8 = 5
total magnification M = m1*m2 = -0.2*5 = -1
image is at distance 20 cm from first lens
s2\' is negative  image is virtual
M is negative the image inverted
M = 1 , image is same size as of the object
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14- Two lens- each having a power of +10D- are placed 20cm apart along.docx
1. 14. Two lens, each having a power of +10D, are placed 20cm apart along the same axis. If an
object is 60cm from the first lens (not in between the lenses), where is the final image relative to
the first lens, and what are its characteristics? Answer: 20cm on object side of the first lens. The
image is virtual and inverted and the same size as the object.
Solution
lens1
object distance = s1 = 60 cm
image distance = s1' = ?
focal length f1 = 1/P1 = 1/10 = 0.1 m = 10 cm
1/s1 + 1/s1' = 1/f1
1/60 + 1/s1' = 1/10
s1' = 12 cm
magnification of converging lens m1 = -s1'/s1 = -0.2
for lens 2:
___________
object distance s2 = 20 - 12 = 8 m
focal length f2 = f1 = 10 cm
2. 1/s2 + 1/s2' = 1/f2
1/8 + 1/s2' = 1/10
image distance s2' = -40 cm
image is at a distance 40 cm from second lens
image is at distance 40 - 20 = 20 cm from first lens
m2 = -s2'/s2 = 40/8 = 5
total magnification M = m1*m2 = -0.2*5 = -1
image is at distance 20 cm from first lens
s2' is negative  image is virtual
M is negative the image inverted
M = 1 , image is same size as of the object
