- #1
Phynos
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Homework Statement
Two converging lenses having focal lengths of f1 = 12.5 cm and f2 = 19.5 cm are placed a distance d = 49.0 cm apart as shown in the figure below. The image due to light passing through both lenses is to be located between the lenses at the position x = 33.0 cm indicated. (a) At what value of p should the object be positioned to the left of the first lens? (b) What is the magnification of the final image? (c) Is the final image upright or inverted? (d) Is the final image real or virtual?
Homework Equations
1/f = 1/p + 1/q
SIGN CONVENTION:
p - positive because object is in front of mirror 1
q - positive because image is behind mirror 1
P - positive because object is in front of mirror 2
Q - negative because image is in front of mirror 2
The Attempt at a Solution
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I will be using capital letters for the quantities specific to the second lens.
For lens one:
1/f = 1/p + 1/q
Since the image of the first lens is the object of the second (q = P)
1/f = 1/p + 1/P
1/P = 1/f - 1/p
For lens two:
1/F = 1/P + 1/Q
The image of the second lens (final image) is given in the question (Q = d-x)
Also, this quantity is negative since it appears in front of the lens (-Q = x-d)
1/F = 1/P + 1/(x-d)
1/P = 1/F - 1/(x-d)
Putting together the two equations I just solved:
1/f -1/p = 1/F - 1/(x-d)
Solving for p (object for first lens):
1/p = 1/(x-d) +1/f - 1/F
p = [ 1/(x-d) + 1/f - 1/F ]^(-1)
Subbing in values:
p = -29.6cm
The magnitude seems reasonable, my problem is the sign. I followed the sign convention so why isn't this value positive? Object distance (p) is positive when the object is in front of the lens.