Total Magnification of Two Converging Lenses

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SUMMARY

The total magnification of two identical converging lenses, each with a focal length of 15 cm and separated by 120 cm, is calculated using the thin lens equation and magnification formula. The first lens produces an image with a distance of 52.5 cm, resulting in a magnification of -2.5. The second lens, using the image from the first lens as its object, yields a magnification of -0.286. The final total magnification is determined by multiplying the individual magnifications, resulting in a total magnification of approximately 0.715.

PREREQUISITES
  • Understanding of the thin lens equation (1/f = 1/do + 1/di)
  • Knowledge of magnification equation (m = di/do)
  • Familiarity with the concept of object and image distances in lens systems
  • Ability to perform calculations involving negative magnification
NEXT STEPS
  • Study the effects of lens separation on total magnification in multi-lens systems
  • Explore the impact of varying focal lengths on image formation and magnification
  • Learn about ray diagrams for converging lenses to visualize image formation
  • Investigate advanced lens systems, such as compound lenses and their applications
USEFUL FOR

Students studying optics, physics educators, and anyone interested in understanding the principles of lens systems and magnification calculations.

aChordate
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Homework Statement



Two identical converging lenses have focal lengths of 15 cm and are aligned on the
same principal axis. They are separated by a distance of 120 cm. An object sits a distance
of 21 cm to the left of the left lens. What is the total magnification of the two lenses?


Homework Equations



Thin lens equation
1/f=1/do+1/di

magnification equation
m=di/do

The Attempt at a Solution



Lens 1:
1/.15m=1/.21+1/di

d-i=0.525m

m=0.525m/0.21m=2.5

Lens 2:
1/.15=1/(.21+1.20m)+1/di

d-i=0.168m

m=0.80


m-total=0.80+2.5 ?
 
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aChordate said:

Homework Statement



Two identical converging lenses have focal lengths of 15 cm and are aligned on the
same principal axis. They are separated by a distance of 120 cm. An object sits a distance
of 21 cm to the left of the left lens. What is the total magnification of the two lenses?


Homework Equations



Thin lens equation
1/f=1/do+1/di
That's the right equation. :smile:

magnification equation
m=di/do
You're missing a negative sign in there somewhere.

The Attempt at a Solution



Lens 1:
1/.15m=1/.21+1/di

d-i=0.525m
I think you mean "di" or di instead of "d-i". But whatever the case, your number is correct.

m=0.525m/0.21m=2.5
You're missing a negative sign somewhere.

Lens 2:
1/.15=1/(.21+1.20m)+1/di
Here is where things start to go south (read: here is where things start to go wrong).

You don't want to use the actual distance to the original object when evaluating the second lens. Instead, when evaluating the second lens, treat the image produced by the first lens as the "object."

It might help to draw a diagram.

Draw the first lens and the original object. Then draw the image created by the first lens. That image is the "object" that you will be using for the second lens.

d-i=0.168m

m=0.80
Of course you'll have to redo the image produced by the second lens. Use the image produced by the first lens as the second lens' object.

m-total=0.80+2.5 ?
When it does come time to combine the magnifications, you'll need to multiply the respective magnifications from the individual lenses, not add them. :wink:
 
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On my second attempt I got an mtotal of 0.357
 
aChordate said:
On my second attempt I got an mtotal of 0.357

I arrived at a different answer. :frown:

Could you show how you got your answer? Maybe I can help then.
 
Lens 1:
1/.15m=1/.21+1/di

d-i=-0.525m

m=-0.525m/0.21m=-2.5

Lens 2:
1/.15=1/(.21+1.20m)+1/di

d-i=0.171m

m=-0.171/1.2m=-0.143


m-total=-2.5*-.143=0.357
 
aChordate said:
Lens 1:
1/.15m=1/.21+1/di

d-i=-0.525m

m=-0.525m/0.21m=-2.5
Yes, so far so good. :approve:

Lens 2:
1/.15=1/(.21+1.20m)+1/di
Here's the problem. For the second lens, you are using the total distance all the way back to the original object for dO. Essentially, this is as through the first lens isn't even there! :eek:

Instead, you need to use the image produced by the first lens as the second's lens' object. You've already calculated that it lies 52.5 cm to the right of the first lens. This image of the first lens is what you want to use as the object of the second lens. :wink:
 
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collinsmark said:
Yes, so far so good. :approve:


Here's the problem. For the second lens, you are using the total distance all the way back to the original object for dO. Essentially, this is as through the first lens isn't even there! :eek:

Instead, you need to use the image produced by the first lens as the second's lens' object. You've already calculated that it lies 52.5 cm to the right of the first lens. This image of the first lens is what you want to use as the object of the second lens. :wink:

Oops, I meant to correct that on the last attempt. Let me try again.

Lens 1:
1/.15m=1/.21+1/di

d-i=-0.525m

m=-0.525m/0.21m=-2.5

Lens 2:

do=1.20m-.525m=0.675m

1/.15=1/0.675m+1/di

d-i=0.193

m=-0.193m/1.2m=-0.161


m-total=-2.5*-0.161=0.402


I hope that's correct!
 
aChordate said:
Oops, I meant to correct that on the last attempt. Let me try again.

Lens 1:
1/.15m=1/.21+1/di

d-i=-0.525m

m=-0.525m/0.21m=-2.5
So far so good, again. :approve:

Lens 2:

do=1.20m-.525m=0.675m

1/.15=1/0.675m+1/di

d-i=0.193
Also good! :smile:

m=-0.193m/1.2m=-0.161
Instead of using 120 cm as the distance to the object, use the distance of the first lens' image to the second lens. (Hint: you've already calculated this. This is the second lens' "object" distance. :wink:)
 
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m=-0.193m/0.675m=-0.286

mtotal=-2.5*-.286=.715

thank you!
 
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aChordate said:
m=-0.193m/0.675m=-0.286

mtotal=-2.5*-.286=.715

thank you!

There you go. :smile:

(If you use more precision in your intermediate calculations, you might obtain a slightly better answer due to rounding; but your above answer is roughly correct.)
 

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