# Focal length from thin lens to object distance

1. Dec 14, 2014

### mikey_d

Hello,

I've come across an problem that requires the lens to be at least about 350 mm away from the object.

Object to be examined is 80 mm tall and imaging plane is either 11 mm or 25 mm tall. I've tried to calculate required focal length, but have not succeeded. Distance from lens to imaging plane is not known.

See attached picture for clarification.

#### Attached Files:

• ###### pf1.JPG
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2. Dec 14, 2014

### Philip Wood

I'd expect one image to be magnified (if object placed between f and 2f) and the other image to be diminished (if object placed between 2f and infinity). Don't understand how both possible images can be diminished. I take it that both are real. Have you checked the data?

3. Dec 14, 2014

### mikey_d

Only the image at P´ is diminished, original object at P is 80 mm tall.

4. Dec 14, 2014

### Philip Wood

What, then, do you mean when you say that the imaging plane is either 11 mm or 25 mm tall? I assumed you meant that these were the heights of two possible images.

5. Dec 14, 2014

### mikey_d

It means that there are two different focal lenghts to be figured for the same setup.

6. Dec 14, 2014

### Philip Wood

Focal length is a characteristic property of a lens. You mean that there are two different lenses to be considered? Just checking that I understand.

7. Dec 14, 2014

### mikey_d

Yes, there should be two different lenses.

8. Dec 14, 2014

### Philip Wood

From similar triangles it's easy to show that, disregarding signs, $$\frac{v}{u}=\frac{height\ of\ image}{height\ of\ object}$$
in which u is distance of object from lens and v is distance of image from lens. So you can get the ratios $\frac{v}{u}$ in the two cases.
But you also have $$\frac{1}{u}+\frac{1}{v}=\frac{1}{f}$$
(if you're using the real-is-positive sign convention).

If you assume that u = 350 mm you have enough information to find f in each case. If you know only that u is at least 330 mm, you don't have enough information.

9. Dec 15, 2014

### mikey_d

That was the clue I was missing, thank you!

(For a while I tried to play around by myself with the thin lens equation and was banging my head to the wall as I didn't have any value for s' and for some reason I didn't notice this similarity of triangles. Which should be QUITE obvious when looking at the picture I attached... :s )

Now I can calculate the required focal lengths with given information and following formula (symbols according to the original picture):
$$f=\frac{sy'}{y+y'}$$