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Two converging lenses

  • Thread starter Jared944
  • Start date
  • #1
10
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A coin is located 20.0 cm to the left of a converging lens (f=16.0 cm). A second identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Find the separation between the lenses.

So,

f= 16 cm
do = 20 cm

m=-di/do, since the object and image have the same size and orientation,
m=1, so -di=do > -di=20cm > di= -20cm

And that's about where I left it. Does anyone have any input they can contribute?

Much appreciated...
 

Answers and Replies

  • #2
13
0
Real image of the first lens becomes the virtual object to the second lens, vice versa. f = 16.0cm , o=20.0cm. The resulting image of the first lens = V/R?

Treat the real image of the first lens as the virtual object(o = negative) for the second lens. Treat the virtual image of the first lens as the real object(o = positive) for the second lens. Using the equation 1/o + 1/i = 1/f, and there you go.
 

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