Double Lens System: Converging/Diverging

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SUMMARY

The discussion focuses on solving a double lens system problem involving a 1.0-cm-tall object, with a converging lens (f1 = 5 cm) and a diverging lens (f2 = -8 cm). The image distances calculated were di1 = -20 cm for the first lens and di2 = -6.4 cm for the second lens. The final image position relative to the original object was determined to be 9.6 cm. Participants emphasized the importance of correctly identifying object distances and applying the lens formula accurately.

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  • Understanding of the lens formula: 1/f = 1/do + 1/di
  • Knowledge of sign conventions for image and object distances in optics
  • Familiarity with magnification calculations for lenses
  • Basic principles of converging and diverging lenses
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  • Study the application of the lens formula in double lens systems
  • Learn about the sign conventions for object and image distances in optics
  • Explore magnification calculations for multiple lenses
  • Review practical examples of double lens systems in optical devices
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Venerable R
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Homework Statement



Image: http://i.imgur.com/NMS7BUK.jpg?1

a) Find the position of the final image of the 1.0-cm-tall object.
b) Find the size of the final image of the 1.0-cm-tall object.

h = 1 cm
f1 = 5 cm
do = 4 cm
f2 = -8 cm
L = 12 cm
di = ?

Homework Equations



1/f = 1/do + 1/di

The Attempt at a Solution



a) di1 = 1/(1/f1 - 1/do) = 1/(1/5cm - 1/4cm) = -20 cm (virtual image through a converging lens?)
di2 = 1/(1/f2 - 1/(l - do)) = 1/(1/-8cm - 1/(12-(-2))) = -6.4 cm

This answer is not right. I would really appreciate any help with double lens systems. We're covering them now in class, and I'm very confused by optics!
 
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Venerable R said:
a) di1 = 1/(1/f1 - 1/do) = 1/(1/5cm - 1/4cm) = -2 cm

Check the above calculation.

[EDIT: I think this might just be a typo. I also get -6.4 cm for the image distance of the second lens. Did the question state how the answer should be given? That is, are you supposed to give the answer relative to the second lens or relative to the first lens?]
 
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TSny said:
Check the above calculation.

[EDIT: I think this might just be a typo. I also get -6.4 cm for the image distance of the second lens. Did the question state how the answer should be given? That is, are you supposed to give the answer relative to the second lens or relative to the first lens?]

Oh! It's the distance from the object! So, I added 4 cm + (12 cm - 6.4 cm) which gave me the correct answer of 9.6 cm. (Also: Yep, that was a typo!)

Thank you! :D
 
I have this same problem, and I also get di1 as -20cm, what am I missing? I'm having trouble determining what "do2" I should use in the second calculation? How do I get there?
 
-20 cm is correct for di1. The image of the first lens is treated as the object of the second lens.
 
TSny said:
-20 cm is correct for di1. The image of the first lens is treated as the object of the second lens.
I get that far but when I do the calculation, the answer is wrong. 1/f - 1/di1 = 1/di2 --> 1/-8 - 1/-20 = 1/di2 --> -5/40 - (-2/40) = 1/di2 --> 40/-3 = di2 =.13.3333cm , which is incorrect according to previous replies in this forum.
 
When you treat the image of the first lens as the object of the second lens, it doesn't mean that the image distance of the first lens equals the object distance for the second lens.

Draw a diagram. Locate the image of the first lens in your diagram. If that image is treated as the object of the second lens, then what should you use for the object distance for the second lens?
 
Again, this is the piece I do not understand.
 
Can you describe in words what it means to say that the image distance of the first lens is -20 cm?
 
  • #10
If the lens equation gives a negative image distance, then the image is virtual on the same side of the lens as the object?
 
  • #11
Yes. So, where did you draw the image of the first lens?
 
  • #12
TSny said:
Yes. So, where did you draw the image of the first lens?
Would it be to the left of the first lens?
 
  • #13
Yes. How far to the left of the first lens?
 
  • #14
16cm left?
 
  • #15
No. How did you get that?
 
  • #16
TSny said:
No. How did you get that?
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..
 
  • #17
You need to review the basic definition of "image distance of a lens (di)". It's defined to be the distance from the lens to the image (with images on the left defined as having negative image distances).

So, when you find that di1 = -20 cm, where is the image of the first lens located in relation to the first lens?
 
  • #18
TSny said:
You need to review the basic definition of "image distance of a lens (di)". It's defined to be the distance from the lens to the image (with images on the left defined as having negative image distances).

So, when you find that di1 = -20 cm, where is the image of the first lens located in relation to the first lens?
20cm left of the first lens
 
  • #19
Yes. Good. So, next you treat this image of the first lens as the object of the second lens. What should you use for do2?
 
  • #20
-20cm - 12cm(distance b/w lenses), -32cm?
 
  • #21
Almost. What is the sign convention for object distances?
 
  • #22
TSny said:
Almost. What is the sign convention for object distances?
POSITIVE 32cm! --> 1/-8 - 1/32 = 1/di2 --> -4/32 - 1/32 = 1/di2 --> di2 = 32/-5 = -6.4cm
 
  • #23
That's it. Good.
 
  • #24
TSny said:
That's it. Good.
Thanks for all your help. So for final position, it's relative to the original object correct?
 
  • #25
Apparently the question asks for the distance between the image of the second lens and the object of the first lens. See post #3. Did you get an answer for that?
 
  • #26
TSny said:
Apparently the question asks for the distance between the image of the second lens and the object of the first lens. See post #3. Did you get an answer for that?
Two calculations are required, hi1/ho = -di/do for the first lens yields 5cm, then hi2/hi1 = - di2/di1 = 1.0 cm, correct answer
 
  • #27
DrewHizzy said:
Two calculations are required, hi1/ho = -di/do for the first lens yields 5cm,
The 5 cm that you are getting represents the height of the image of the first lens? If so, OK.

then hi2/hi1 = - di2/di1 = 1.0 cm, correct answer
The di1 should be do2. Also, note that the ratio hi2/hi1 is dimensionless, so it does not have units of cm.
 
  • #28
About the di1 being do2, that's what I was thinking I was just unsure which notation was more accurate
 
  • #29
If you want to make sure you are thinking correctly about the second part of the question, try answering the following questions?
(1) What is the magnification of the first lens alone?
(2) What is the magnification of the second lens alone?
(3) What is the magnification of both lenses together?
(4) How do you use the result of question (3) to determine the height of the final image?
 

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