# Converging Lenses and image position

1. Apr 30, 2007

### Burner

1. The problem statement, all variables and given/known data
Two converging lenses with focal lengths of 40 cm and 20 cm are 10 cm apart. A 2.0-cm-tall object is 60 cm in front of the 40-cm-focal-length converging lens. a. Use ray tracing to find the position and height of the image. do this accurately with a ruler. Determine the answers by taking direct measurements on your diagram. b. Calculate the image position (with respect to the second lens, the one with focal length 20 cm) and height. Compare with your answers to part a.

2. Relevant equations
M = h'/h = s'/s
1/s + 1/s' = 1/f

3. The attempt at a solution
So far, I haved tried many times at the ray tracing part. This is just one page of my attempts: <link not available>.

As far as the calculation part is concerned, I have so far treated the the first, 40-cm focal length lens as if it were by itself. By applying the thin lens equation, the real image would be 120 cm away. However, since the second lens is in the way, that must be treated as a virtual image. Treating that as a the object for the second lens tells me that the it is now 24.4 cm away from the second lens, which is 14.4 cm away from the first lens. Using this as the object for the first lense gets me at -22.6 cm. And that's where I'm stuck at.

I have asked my professor about this problem, and he said that it was difficult, and upon trying it on the spot himself, seemed to have the same difficulties in ray tracing that I and other students were having.

I've also tried using this applet to no avail (since it's a web-based homework submission system, I get immediate feedback). I have only one try left on the web-based homework submission system before it'll prevents me from submitting any more.

Any tips would be greatly appreciated. Thanks!

Last edited by a moderator: Dec 28, 2010
2. Apr 30, 2007

### hage567

I agree with this.
I don't understand what you did here. If the image from the first lens is 120 cm to the right of the first lens, then what is the distance of the image to the second lens? It is not 24.4 cm. You can use the information given in the question (all those distances) to figure this out. Also, since the image from the first lens is to the right of the second lens, you must use -s in your equation when finding the image due to the second lens.

Remember that M=h'/h = -s'/s. Don't forget the negative sign there.

For the magnification, you must find the magnification due to each lens. The product of the two will be the total magnification. Be careful to keep the signs straight.

Hope that helps.

3. Apr 30, 2007

### Burner

All right, thanks.

The image from the first lens is 110 cm to the right from the second lens. Applying the thin lens equation:

1/-110 + 1/x = 1/20 -> x = 220/13 ~~ 16.92 cm
OR
1/110 + 1/x = 1/20 -> x = 220/9 ~~ 24.4 cm (which is where I got the 24.4 from before)

So, the image from the second lens is either 24.4 or 16.92 cm to the left of the second lens, which means it is either 14.4 or 6.92 cm to the left of the first lens. Assuming 6.92 cm as correct, let that be the object of the first lens. Applying the thin lens equation again, back through the first lens:

1/6.92 + 1/x = 1/40 -> x = -8.37 cm

This tells me that the image is a virtual image and is therefore still on the left side of the first lens, correct? Anyway, carrying this on:

1/8.37 + 1/x = 1/40 -> x = -10.58
1/10.58 + 1/x = 1/40 -> x = -14.38
1/14.38 + 1/x = 1/40 -> x = -22.45
1/22.45 + 1/x = 1/40 -> x = -51.17
1/51.17 + 1/x = 1/40 -> x = 183.24

This just puts me in an endless loop as far as I can tell. :-/

Where am I going wrong?

4. Apr 30, 2007

### hage567

After you get the 16.9 cm, you stop. That is your final image distance. I'm not sure why you are applying the lens equation again, you only do it twice (once for each lens).

Also, if it is 16.9 cm to the right of the second lens, the distance relative to the first lens is not 6.92 cm. The distance given between the first and second lens was 10 cm, right? So 10 + 16.9 = 26.9 cm (although this isn't needed to finish the question).

5. Apr 30, 2007

### Burner

Oh my goodness, thank you so very much! You just made my day! :-D

Last edited: Apr 30, 2007
6. Apr 30, 2007

### hage567

You're welcome.