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phunphysics2
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Homework Statement
A converging lens of focal length 8.100 cm is 20.3 cm to the left of a diverging lens of focal length 6.63 cm. A coin is placed 12.4 cm to the left of the convering lens.
Is the location of the coins final image to the left of the convering lens, between the lenses, or to the right of the divering lens?
What is the location and magnification of the coin's final image?
Homework Equations
1/do + 1/di= 1/f
m=hi/ho= -di/do
The Attempt at a Solution
first coin goes through converging lens
1/i + 1/o = 1/f
1/i + 1/12.4 = 1/8.1
i=23.358
this acts as object for next lens
o = 20.3-23.358
o=-3.058
1/i + 1/o = 1/f
1/i + 1/-3.058 = 1/-6.63
i=5.676
so to the right of the diverging lens
m = i1/o1 * i1/o2 = (23.358/12.4)*(5.676/-3.058)=-3.496
***Please help me with which bubble to choose. I think it is between the lenses, but I am not absolutely sure