How Do You Calculate the Final Image Position in a Three-Lens System?

[MaysMrDolphin]

Homework Statement

. 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

Homework Equations

1/f=1/di+1/do

The Attempt at a Solution

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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?

 

[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.

 

[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.

 

[Merlin3189]

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.

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

 

[Mastermind01]

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

Can you please explain? thanks.

 

[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.

 

[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

 

[MaysMrDolphin]

Huh, didn't notice you h

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.

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

 

[Mastermind01]

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.

Thanks!