Two identical convergent lenses

In summary, the figure shows a combination of two identical lenses. There is a 2cm tall object that is 36 cm away from the first lens (f=9 cm). The second lens is 15 cm away from the first lens. It also has a focal length of 9 cm. So, the first question was "find the position of the final image of the 2 cm tall object". I've solved that one, using 1/f=1/s+1/s' for each lens, making sure to use the "image" from the first lens as the "object" for the second lens. The next question is "find the size of the final image of the 2 cm tall object". I am getting -4.
  • #1
Niki4444
9
0

Homework Statement


the figure shows a combination of two identical lenses. There is a 2cm tall object that is 36 cm away from the first lens (f=9 cm). The second lens is 15 cm away from the first lens. It also has a focal length of 9 cm. So, the first question was "find the position of the final image of the 2 cm tall object". I've solved that one, using 1/f=1/s+1/s' for each lens, making sure to use the "image" from the first lens as the "object" for the second lens. The next question is "find the size of the final image of the 2 cm tall object".

Homework Equations



I thought I could use h'/h=s'/s, although that didn't seem to work out.
I already used 1/f=1/s+1/s'

The Attempt at a Solution



My attempt was to use h'/h=s'/s. I thought I would solve for h' (height of image).
I tried to do this in a 2 step process
a. used original object size & lens #1 h'/2=12/36
b. used image from 1st lens as "object" for second lens h'/0.67=45/3
c. got 1.01 cm which is incorrect.

Any helpful thoughts to move me in the right direction would be hugely appreciated.
Thanks in advance.
 
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  • #2
OK. You are going about this in the right way, but I think one of your numbers may be off. Are you sure the distances you found are correct?

I am getting -4.5cm as the image distance from the second lens, not 45cm. Check your work from the first part of the problem.
 
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  • #3
Sure... for the first part of the problem, my work is:
(I drew the picture so the object is on the far left of my paper, with the lenses to the right)

1/f=1/s+1/s'
1/9=1/36+1/s'
4/36-1/36=3/36=1/12
so s' would be 12 cm
so the image is 12 cm to the right of the first lens

for the second lens, I am supposed to use the image as the new "object", so
1/f=1/s+1/s'
1/9=1/3+1/s'
1/9-3/9=1/s'
-2/9=1/s'
so I think s' would =-4.5 cm
meaning that the final image is 4.5 cm to the left of the second lens

so then I thought h'/h=-s'/s
I thought I'd use h'/2=-(-4.5)/(-3)
h'=(-1.5)(2)=-3 cm

?
 
  • #4
You have to use the magnification formula in two steps as you did in your first post.

Use the magnification formula with the first lens, to find the height of the first image. Then, use that height to find the height of the second image.

Your numbers look right. (meaning your initial answer of 1.01cm) How do you know your answer is wrong?
 
  • #5
yes, 3 cm and -3 cm are definitely wrong. I'm stumped.
 
  • #6
Niki4444 said:
yes, 3 cm and -3 cm are definitely wrong. I'm stumped.

Do you know the correct answer? If so, post it. Maybe I'll be able to see the mistake we're making if I knew the answer we were looking for.
 
  • #7
nope, the program just tells me that my answer is incorrect. Oh man.
 
  • #8
Niki4444 said:
nope, the program just tells me that my answer is incorrect. Oh man.

Well. It could always be that the solution the program has is incorrect. Or, we could both be missing something obvious. I'll keep thinking about it. If I figure anything out, I'll let you know. In the mean time, I suggest you bring this up with your instructor. They would be able to help you if the program was marking you incorrect, even when you were correct.

Another thing to check is if you are using the correct numbers. Are you sure you copied the numbers from the original problem correctly? (I know this sounds stupid, but always worth double checking something like this.:smile:)

Also, it could be a rounding error. I was coming up with an answer of 1.00, where you were coming up with an answer of 1.01. Most computer programs for physics problems will allow errors this small, but if your using some obscure program, them maybe this could be it?
 
  • #9
1.00 is correct with the mastering physics website.
 

Related to Two identical convergent lenses

What is the purpose of two identical convergent lenses?

The purpose of using two identical convergent lenses is to create a more powerful lens system that can magnify an object or focus light rays to a specific point. This is often used in telescopes, microscopes, and other optical instruments.

How do two identical convergent lenses work together?

When two identical convergent lenses are placed close together, they act as a single, stronger lens. The first lens refracts the light rays, and the second lens further focuses the rays to create a more intense image or a sharper focal point.

What is the focal length of two identical convergent lenses?

The focal length of two identical convergent lenses depends on the distance between them. If the lenses are placed close together, the focal length will be shorter, resulting in a more powerful lens system. However, if the lenses are placed further apart, the focal length will be longer, resulting in a weaker lens system.

Can two identical convergent lenses be used to correct vision?

Yes, two identical convergent lenses can be used to create corrective lenses for individuals with certain vision impairments. By placing the lenses at a specific distance from each other, they can help focus light onto the retina, correcting nearsightedness or farsightedness.

What are some advantages of using two identical convergent lenses?

One advantage is that they can create a more powerful lens system compared to using a single lens. They also allow for greater flexibility in adjusting the focal length and magnification. Additionally, using two identical lenses can reduce chromatic aberration and improve image quality.

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