# Optics - combination of thin lenses

1. Feb 22, 2009

### jakey

What if two converging or diverging lenses were placed side by side, in contact with each other. Where would be the image formed for the converging lens, would it be to the right of the lens combined or to the left? Why? Also, where would it be if both lens were diverging? thanks!:)

2. Feb 22, 2009

If this is a homework question please show what you have tried already.

3. Feb 22, 2009

### jakey

i have actually tried. btw, the original problem is: if 2 thin lenses come in contact with each other, show that 1/fT = 1/f1 + 1/f2. i considered 4 cases, diverging-converging and the reverse, and div-div, and conv-conv. But i'm having trouble with the last 2.

1) conv-conv.
So, when rays from infinity strike the first lens, the image formed will be at the focus of that lens, so the image is at the right. this image will then serve as the object of the 2nd lens, and the image produced will be to the left. if this is the case, 1/f2 = -1/f1 - 1/fT. but, this won't prove the above equation.

2) div-div.
Same principle, when the rays strike the first lens, the image formed will be at the left, specifically at the lens' focus. so, this will be the object for the 2nd lens which wouls produce the image still at the left. so, 1/f2= -1/f1 - 1/fT. but, this won't prove the above equation.

do you know where i got wrong?

4. Feb 22, 2009

### Staff: Mentor

Why do you think the image produced by the virtual object (formed by the first lens) would be on the left? (I assume that light goes from left to right in your analysis.)

The first lens produces an image at its focal point. That becomes the "virtual object" for the second lens. Since the virtual object is on the right of the lens, the object distance is negative. Find the image distance for the second image--that tells you the image position of the lens combination and thus its combined focal length.
Similar considerations apply. Realize that the focal length of a diverging lens is negative when used in the thin lens equation.

5. Feb 22, 2009

### jakey

1) conv-conv: the image produced by the virtual image is to the left, but why should fT be positive? it's in the same side as the incoming light...so it should be negative.

2)div-div: same thing, why should be fT positive? in the first place, it's at the left side and in the same side as the incoming light.

thanks for the help!:)

ps, f1 represents focal length of first lens, f2 by the second, fT the combined.

6. Feb 22, 2009

### Staff: Mentor

Let's say that the light moves from left to right towards the lens. Do you agree that the image formed by the first lens will be on the right of the lens?

There are several sign conventions used, but in the one I use that means that the image distance is positive. When analyzing the second lens, the virtual image is on the right, thus its object distance is negative.

Here's the version of the thin lens equation that I'm using: http://hyperphysics.phy-astr.gsu.edu/hbase/geoopt/lenseq.html#c1". (It's the form used in most intro textbooks.)

They are just talking about the magnitude of the combined focal length. When you use it in the thin lens equation, it will be negative.

Last edited by a moderator: Apr 24, 2017
7. Feb 22, 2009

### jakey

i agree that the image formed by the first lens will be at the right. this is because we assume that lens 1 is stronger than lens 2, and thus, lens 1 would be able to form the image before light rays enter lens 2.

if you're saying that the magnitude is what we're just after, then all of these have been correct all along?

Last edited by a moderator: Apr 24, 2017
8. Feb 22, 2009

### Staff: Mentor

It has nothing to do with one lens being stronger than another. For a converging lens, the image (from parallel rays from the left) is on the right.

To find the combined effect of two lenses, treat their images one after another--the image from the first lens becomes the (virtual) object for the second lens.
More accurately, the focal length of a converging lens is positive and a diverging lens is negative. In any case, two thin lenses together will satisy 1/f = 1/f1 + 1/f2.

9. Feb 22, 2009

### jakey

doc al, sorry if i'm starting to annoy you, but i've been considering those things which you have said - consider the image as the virtual object, etc., i understood your suggestions. but i'm confused as to how i went wrong, or if my answers are really wrong.

do i still need to change anything that i left uncorrected, especially the 2 equations? or would it be satisfactory to claim that the magnitude etc...?

thanks, very much.

10. Feb 22, 2009

### Staff: Mentor

To avoid going around in circles, let's start over. Why don't you show me each step in your analysis of the convex-convex case? The first lens has a focal length of +f1; the second, +f2.

(a) Analyze the action of the first lens. Where is the image? (Show how you applied the thin lens equation.)
(b) Analyze the action of the second lens. Where is the object (from the first lens)? Where is the final image? (Show how you applied the thin lens equation.)

11. Feb 22, 2009

### jakey

Oh! i did things differently. i kinda followed the solution on how to get the focal length of a diverging lens when a converging lens is placed beside it. This is the reason why i have 1/f2=... because I'm trying to compute for the focal length of the 2nd lens, knowing where the image distance and object distance are located thru ray diagraming.

so, if i follow your suggestions, then i'd use the equation 1/f1 = 1/do1 + 1/di1.
then, 1/f2 = -1/di1 -1/fT. now if i add both equations, with do1=infinity and 1/do1=0,
then 1/f1 + 1/f2 = -1/fT.

this is where i'm stuck.for the image.please refer to the attached. thanks, really!

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12. Feb 23, 2009

### Staff: Mentor

OK.
Why the second minus sign? Just apply the same lens equation a second time:
1/f2 = -1/di1 + 1/di2
1/f1 + 1/f2 = 1/di2 = 1/fT

Why do you have image 2 on the left of the lenses?

13. Feb 23, 2009

### jakey

oh! i just needed to do that. well, it's just a matter of sign, but it's that vital.

isnt the second image supposed to be on the left side because the object, though virtual, is on the right side? would this still hold or its supposed to be at the right because it doesn't matter if the object is virtual? i think i'm starting to understand what you're trying to explain. thanks doc al!!!

14. Feb 23, 2009

### Staff: Mentor

Not at all. Both images are to the right of the lenses, since the lenses are converging. In any case, you solve for the location of the 2nd image using the lens equation. The only thing that changes is that the object distance is negative since the (virtual) object is on the right of the lens.