Lens equation

1. Apr 30, 2005

Callisto

Hi Peaple :!!)

How do you determine the minimum distance p+q between the object and image for given focal length given

1/p + 1/q = 1/f

this may seem trivial, but i cant figure it out.

Callisto

2. Apr 30, 2005

whozum

Arent p and q distances....? Just add them.

3. Apr 30, 2005

whozum

If your saying given a fixed focal length, find the relationship between p and q, that case

$$\frac{1}{p} + \frac{1}{q} = k$$

Solve it for either p or q (the result is the same) and graph it. I believe its a logarithmic relation, but I'm not sure.

4. Apr 30, 2005

Integral

Staff Emeritus
What math tools do you have at your disposal?

Actually the thin lens relationship is a hyperbola.

5. Apr 30, 2005

whozum

I should know that, my brother did a presentation on it a week ago.

6. Apr 30, 2005

Callisto

solving 1/p+1/p=1/f for p
i get
-qf/(f-q)
now i substitute this into p+q, then i get -q^2/(f-q).
now if i use calculus to find the minimum value i differentiate -q^2/(f-q) and let it equal zero so
d/dq= -q(2f-q)/(f-q)^2 =0 when q = 0 or 2f

Repeating this process for p then p = 0 or 2f
so the minimum distance is
p+q=4f. why is it not zero

Callisto

7. May 1, 2005

Data

Looks fine to me. You might want to check that 2f is a minimum (not a maximum or point of inflection), to be thorough. Looking at the lens equation itself do you see why the 0 solutions are inadmissible?

Last edited: May 1, 2005
8. May 1, 2005

Callisto

1/0+1/0=0 , 0 doesn't = 1/f
therefore 0's are excluded from the problem, correct?
Callisto

9. May 1, 2005

Data

1/0 + 1/0 is 0? Are you sure?

10. May 1, 2005

Callisto

1/0+1/0=1/0, 1/0 doesn't = 1/f
is this why 0's are excluded?

11. May 1, 2005

Data

1/0 is undefined, as is 1/0+1/0, and no other arithmetic with these quantities is defined either, so if p or q were 0, you could not rearrange the equation the way that you did (without first defining a new arthmetic). In order for the lens equation to make sense with usual definitions, the assumption p, q not zero must be made.

12. May 1, 2005

Callisto

Thanks
2f must be minimum since if the object p is at infinity then 1/p=0 so q = f for an object at infinity.