A 15-cm-focal length converging lens is 20cm to the right of a 7.0-cm-focal leng

Question

A 15-cm-focal length converging lens is 20cm to the right of a 7.0-cm-focal length converging lens. A 1.0-cm-tall object is a distance L to the left of the first lens.

a. Suppose you want the final image of this two-lens system (after the light goes through both lenses) to be halfway between the two lenses. What kind of image would it have to be? Real or virtual?

b. Draw a ray diagram of where and how the image forms just due to the first lens (ignore the 2nd lens for this picture).

c. For what value of L is the final image of this two-lens system halfway between the two lenses? Find the value exactly using equations.

d. What is the total magnification of the object when an image is formed halfway between the two lenses?

If you could show your work, that would be great! Thanks!

Explanation / Answer

f1 = 7 cm
f2 =15 cm
h = 1 cm f1 = 7 cm
f2 =15 cm
h = 1 cm
s'2= -20 / 2 = -10 cm
1/f2=1/s2+ 1/s2' s2= f2s2'/(s2'-f2)
= 15 * -10/ (-10 - 15)
s2= 6 cm
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the distance between object and image from the first lens. s1' = 20 cm - 6 cm
= 14 cm
from the thin lens equation, 1/f1=1/s1+ 1/s1' s1= 7 * 14/(7)
s1 = L = 14 cm
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magnification M = M1M2 M = (-s1'/s1)(-s2'/s2)
M = (-14/ 14)(10 / 6)
M = -1.667
h' = hM = (1 cm)(-1.667) =-1.667 cm
image is inverted M = (-s1'/s1)(-s2'/s2)
M = (-14/ 14)(10 / 6)
M = -1.667
h' = hM = (1 cm)(-1.667) =-1.667 cm
image is inverted

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