Double Lens System: Converging/Diverging

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Homework Help Overview

The discussion revolves around a double lens system involving a converging lens and a diverging lens. Participants are tasked with determining the position and size of the final image produced by the system, given specific parameters such as focal lengths and object distances.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore calculations for image distances using the lens formula, questioning the accuracy of their results and the interpretation of object distances for the second lens.
  • Some participants express confusion regarding the sign conventions for image and object distances, particularly in relation to virtual images.
  • There are discussions about the need for diagrams to visualize the setup and clarify the relationships between the lenses and images.
  • Questions arise about how to correctly apply the lens formula for the second lens based on the image produced by the first lens.

Discussion Status

The conversation is ongoing, with participants providing insights and corrections to each other's calculations. There is a collaborative effort to clarify concepts and ensure understanding of the relationships between the lenses and their respective images. Some guidance has been offered regarding the sign conventions and the treatment of images as objects for subsequent lenses.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may impose specific requirements for how answers should be presented. There is also a noted confusion regarding the interpretation of distances in relation to the lenses, which may affect their calculations.

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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