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1. Mar 14, 2016

### MaysMrDolphin

1. The problem statement, all variables and given/known data
. Three thin lenses, each with a focal length of 40.0 cm, are aligned on a common axis; adjacent lenses are separated by 52.0 cm. Find the position of the image of a small object on the axis, 80.0 cm to the left of the first lens.

d=52cm between
f=40cm
d of object=80cm

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

3. The attempt at a solution

by the equation:
1/40cm = 1/80cm + 1/di (1st lens)

di= 80cm

52-80cm= -28cm

2nd lens
1/40 = -1/28 + 1/di

di (lens 2) = 16.47

52-16.47= 35.53cm

3rd lens

1/40 = 1/35.33+1/di

di=-302.6 ---> This is where I'm stuck, the answer is supposed to be 134 cm to the left according to the book. Even if I add 104 cm to the answer, it still doesn't come close enough. Any thoughts on this?

Last edited by a moderator: Mar 14, 2016
2. Mar 14, 2016

### bigguccisosa

In your last calculation you used 35.33 instead of 35.53. If you use the correct value you obtained earlier, you get di = -317.94 instead. Now add the 2 lens distances (+104) and add the distance of the first object (+80), this gives you -133.94 cm. So you book just wants the distance relative to the first object. Hope this helps.

3. Mar 14, 2016

### Mastermind01

You didn't the sign convention. All distances to the left of the optical centre are taken as negative and to the right as positive. So the 80 cm to the left of the first lens that you took as $$d_o$$ should be -80 cm.

4. Mar 14, 2016

### Merlin3189

But it depends what convention he's using. The calculations matched the "real positive" sign convention.

5. Mar 14, 2016

### Mastermind01

6. Mar 15, 2016

### Merlin3189

Real images, real focal lengths (convex lens) and real objects have positive values, virtual images, virtual focal lengths (concave lens) and virtual objects have negative values. Summarised here.
I believe there are other conventions which work. So long as you use a consistent sign convention, you can use whichever.

7. Mar 15, 2016

### Merlin3189

You might have meant, why did I think OP used this convention.

1/40cm = 1/80cm + 1/di (1st lens) real f, real object

di= 80cm real image

52-80cm= -28cm virtual object

2nd lens
1/40 = -1/28 + 1/di real f, virtual object

di (lens 2) = 16.47 real image

52-16.47= 35.53cm real object

3rd lens

1/40 = 1/35.33+1/di real f, real object

di=-302.6 virtual image

8. Mar 15, 2016

### MaysMrDolphin

Huh, didn't notice you h
Thanks, I was kind of in a rush doing this homework. The prof gave us too much and even had a quiz the next day

9. Mar 15, 2016

Thanks!