# Diverging and Converging lenses in a system

Tags:
1. Apr 3, 2015

### DrewHizzy

1. The problem statement, all variables and given/known data
A)Find the final position of the image (from the object, I assume?)
B)Find the size of the final image of the object.

2. Relevant equations
1/f = 1/do + 1/di

3. The attempt at a solution
1) Solved first distance: 1/5 - 1/4 = 1/di --> 4/20 - 5/20, di = -20cm
2) Use di1 for do in second calculation, but my answer is incorrect.

2. Apr 3, 2015

### Staff: Mentor

Where did you take the second lense into account?

3. Apr 3, 2015

### DrewHizzy

That's what I'm having trouble determining; what do I use for "do" in my second calculation?

4. Apr 3, 2015

### Staff: Mentor

The distance between the (virtual) image and the lense. Where in the sketch is this image?

5. Apr 3, 2015

### DrewHizzy

if di1 = -20cm, doesn't that place it behind the concave lens?

6. Apr 3, 2015

### Staff: Mentor

What is "behind", left or right?

7. Apr 3, 2015

### DrewHizzy

If I had to guess I'd say to the right of Lens 2

8. Apr 3, 2015

### DrewHizzy

Again, this is the piece I do not understand.

9. Apr 3, 2015

### DrewHizzy

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

10. Apr 3, 2015

### Staff: Mentor

For B, I think you don't have to consider the lenses separately (but it is possible) as you know the result of both together already. You just have to be careful which distances to use.