- #1
jgpenn12
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Homework Statement
A +15 cm focal length lens and a flat mirror are 45 cm apart (the mirror is to the right of the lens and can be treated as a concave mirror with an infinite focus length). There is an intermediate image 30 cm to the right of the lens (15 cm to the left of the mirror) with a height of -2mm.
A) Find the height and location of the original object.
B) Find the location of the FINAL image produced by your object in part a.
Homework Equations
1/do + 1/di = 1/f
m=di/do
The Attempt at a Solution
So the solutions (in the pdf) say that there are two ways to do this. I chose solution one, but either way i choose to do this...I get stuck at the same part. So I understand that this intermediate image can first be just from the lens so I do the lens equation to find the do = 30 cm...meaning it's to the left of the lens (positive). So in the second part I first reflect the image off of the mirror and get -15cm (to the right of the mirror). NOW is the part I get confused on...this IMAGE is now the OBJECT for the lens...it is 60 cm away from the lens BUT IT IS A VIRTUAL OBJECT AND IS TO THE RIGHT OF THE LENS...therefore I believe it should be -60 cm...so I did the lens equation with -1/60 + 1/di = 1/15...which results in a different answer than the one in the solutions. I'm not sure why my professor uses a positive object distance for both solutions in the last part...it's confusing me.