# Double Lens System: Converging/Diverging

• Venerable R
In summary, the homework statement is that find the position of the final image of the 1.0-cm-tall object. The size of the final image of the 1.0-cm-tall object is found to be h = 1 cm, f1 = 5 cm, do = 4 cm, f2 = -8 cm, L = 12 cm, and di = ?
Venerable R

## 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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Venerable R
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?

If the lens equation gives a negative image distance, then the image is virtual on the same side of the lens as the object?

Yes. So, where did you draw the image of the first lens?

TSny said:
Yes. So, where did you draw the image of the first lens?
Would it be to the left of the first lens?

Yes. How far to the left of the first lens?

16cm left?

No. How did you get that?

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..

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?

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

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?

-20cm - 12cm(distance b/w lenses), -32cm?

Almost. What is the sign convention for object distances?

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

That's it. Good.

TSny said:
That's it. Good.
Thanks for all your help. So for final position, it's relative to the original object correct?

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?

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

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.

About the di1 being do2, that's what I was thinking I was just unsure which notation was more accurate

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?

## 1. What is a double lens system?

A double lens system is a combination of two lenses placed close to each other. The first lens is called the objective lens and the second lens is called the eyepiece lens.

## 2. What is the difference between a converging and diverging double lens system?

A converging double lens system, also known as a convex lens, has two lenses that bend light rays towards each other, causing them to converge at a focal point. A diverging double lens system, also known as a concave lens, has two lenses that bend light rays away from each other, causing them to diverge.

## 3. How does a converging double lens system work?

In a converging double lens system, the objective lens collects and focuses light rays from an object onto a focal point. The eyepiece lens then magnifies the image produced by the objective lens, making it appear larger to the viewer.

## 4. What are the applications of a diverging double lens system?

A diverging double lens system is commonly used in corrective eyeglasses, as it helps to spread out light rays and correct nearsightedness. It is also used in cameras and telescopes to produce a wider field of view.

## 5. How is the magnification of a double lens system calculated?

The magnification of a double lens system is calculated by dividing the focal length of the objective lens by the focal length of the eyepiece lens. For example, if the objective lens has a focal length of 10 cm and the eyepiece lens has a focal length of 5 cm, the magnification would be 10/5 = 2x.

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