Recent content by DrewHizzy

Chet, I'm sure this is an algebra based physics question, not calculus.

You shouldn't have to explain this scientifically to know it's a bad idea..

Mind = blown

About the di1 being do2, that's what I was thinking I was just unsure which notation was more accurate

Two calculations are required, hi1/ho = -di/do for the first lens yields 5cm, then hi2/hi1 = - di2/di1 = 1.0 cm, correct answer

I answered Part A correctly with 9.6cm, hi/ho = di/do for part B? Do I make two calculations like Part A?

Thanks for all your help. So for final position, it's relative to the original object correct?

POSITIVE 32cm! --> 1/-8 - 1/32 = 1/di2 --> -4/32 - 1/32 = 1/di2 --> di2 = 32/-5 = -6.4cm

-20cm - 12cm(distance b/w lenses), -32cm?

20cm left of the first lens

20 - 4 since do was already 4cm away from the first lens? Clearly I do not understand, any real explanation you could give would help..

16cm left?

Would it be to the left of the first lens?

If the lens equation gives a negative image distance, then the image is virtual on the same side of the lens as the object?

Again, this is the piece I do not understand.