Focal length from thin lens to object distance

  • Thread starter mikey_d
  • Start date
  • #1
5
0
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.
 

Attachments

Answers and Replies

  • #2
Philip Wood
Gold Member
1,221
78
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
5
0
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?
Only the image at P´ is diminished, original object at P is 80 mm tall.
 
  • #4
Philip Wood
Gold Member
1,221
78
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
5
0
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.
It means that there are two different focal lenghts to be figured for the same setup.
 
  • #6
Philip Wood
Gold Member
1,221
78
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
5
0
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.
Yes, there should be two different lenses.
 
  • #8
Philip Wood
Gold Member
1,221
78
From similar triangles it's easy to show that, disregarding signs, [tex]\frac{v}{u}=\frac{height\ of\ image}{height\ of\ object}[/tex]
in which u is distance of object from lens and v is distance of image from lens. So you can get the ratios [itex]\frac{v}{u}[/itex] in the two cases.
But you also have [tex]\frac{1}{u}+\frac{1}{v}=\frac{1}{f}[/tex]
(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.
 
  • Like
Likes mikey_d
  • #9
5
0
From similar triangles --
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):
[tex]f=\frac{sy'}{y+y'}[/tex]
 

Related Threads on Focal length from thin lens to object distance

  • Last Post
Replies
0
Views
3K
  • Last Post
Replies
1
Views
11K
Replies
7
Views
3K
  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
Replies
7
Views
5K
Replies
2
Views
298
Replies
4
Views
2K
Replies
1
Views
5K
Top