More Geometric optics

  • Thread starter leolaw
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  • #1
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Two lenses, one converging with focal length 20cm and one diverging with focal length -10cm, are placed 25cm apart. An object is placed 60cm in front of the converging lens. Determine (a) the position and (b) the magnification of the final image formed.

I have a problem with this becasue when i find the image distance after the first len:
[tex]\frac{1}{d_i} + \frac{1}{d_o} = \frac{1}{f}[/tex]

[tex]\frac{1}{d_i} + \frac{1}{60} = \frac{1}{20} [/tex]

[tex] d_i = 30cm [/tex], which is located behind the second len. But i am wondering if the ray pass through the second len, is it going to change its path (See attachment)? If this is so, how can we find the location of the image formed by the first len?
 

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  • #2
learningphysics
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Deal with the first lens alone (as if the second isn't there). Find the magnification and the image distance.

Then deal with the image created by the first lens as the object for the second lens.... If the image created by the first lens lies on the opposite side of the second lens then the object distance for the second lense is negative.... find the image distance for the second lens... and the magnification due to the second lens.

Multiply the two magnifications to get the magnification of the final image.

I'm not exactly sure how to show why this works... It almost seems like the first lens does not form an image because the second lens is in the way... But the math works out.
 
  • #3
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so when I am loooking for the image position from the first len, i can basically ignore the existence of the second len. Is that what you mean?
 
  • #4
Doc Al
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Yes. As learningphysics explained, treat the action of each lens separately. Since the first lens produces an image on the far side of the second lens, that image becomes a virtual object as far as the second lens is concerned. In the usual sign conventions, virtual objects have a negative object distance.
 
  • #5
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now i have [tex] \frac{1}{-5} + \frac{1}{d_i} = \frac{1}{-10} [/tex] and after calculation, i have d_i = 10cm
it means that the final position of the image is 10 cm behind of the second len right? (when i say behind it, i mean the image would be in between the first and second len), and it is also a real image since that it is coming out from the other side of the second len

Just want to make sure if I understand the sign conversion
 
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  • #6
Doc Al
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Your calculation is correct, but your interpretation is not. In the usual sign convention, light is assumed to move from left to right. (Just like your diagram shows.) A positive image distance means the image is to the right of the lens, just like it did in your first calculation. So the image is 10 cm to the right of the second lens and is real.

Sanity check: If the image were between the two lenses, it couldn't be real, since the light goes through both lenses.
 
  • #7
your rays on your drawing is wrong too. You should try drawing it to scale, so you know for sure that the location of the images are correct. In optics, don't just sketch the diagrams cause the drawings help you with your calculations.
 
  • #8
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Doc Al said:
Your calculation is correct, but your interpretation is not. In the usual sign convention, light is assumed to move from left to right. (Just like your diagram shows.) A positive image distance means the image is to the right of the lens, just like it did in your first calculation. So the image is 10 cm to the right of the second lens and is real.

Sanity check: If the image were between the two lenses, it couldn't be real, since the light goes through both lenses.
So when we are using the sign conversion, we only concern where the light is coming from in relative to the beginning right?
 
  • #9
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I dont know how the light would travel after entering the second len, the attachment shows where the rays are after the first len.
It is drawn in scale, and I should expect the final image, which is inverted, have the same size as the original (becasue I got the mangification to be -1), and located right under its focal point.
 

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  • #10
Doc Al
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leolaw said:
So when we are using the sign conversion, we only concern where the light is coming from in relative to the beginning right?
For the lens equation you are using, the sign convention says that:
(1) a real object to the left of the lens has a + object distance; a virtual object would have a - object distance;
(2) a + image distance is a real image to the right of the lens; a - image distance is a virtual image to the left of the lens.​

Treat each lens separately, one at at time.
 
  • #11
Doc Al
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leolaw said:
I dont know how the light would travel after entering the second len, the attachment shows where the rays are after the first len.
Your diagram correctly shows the position of the image formed by the first lens, if the second lens were not there. Now use that image as the object for the second lens and draw where its image would be. That's the final image produced by this two-lens system.
It is drawn in scale, and I should expect the final image, which is inverted, have the same size as the original (becasue I got the mangification to be -1), and located right under its focal point.
Right. The final image will be the same size, inverted, and real. And it will be 10cm to the right of the second lens (which is at one of its focal points).
 

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