Calculating Minimum Distance p+q

  • Context: Undergrad 
  • Thread starter Thread starter Callisto
  • Start date Start date
  • Tags Tags
    Minimum
Click For Summary

Discussion Overview

The discussion revolves around calculating the minimum distance \( p + q \) between an object and its image in the context of the thin lens formula \( \frac{1}{p} + \frac{1}{q} = \frac{1}{f} \), where \( p \) is the object distance, \( q \) is the image distance, and \( f \) is the focal length. Participants explore mathematical relationships and implications of the lens equation, including the behavior of the distances involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant seeks clarification on how to determine the minimum distance \( p + q \) given a fixed focal length.
  • Another participant suggests that since \( p \) and \( q \) are distances, they can simply be added together.
  • A different participant proposes that the relationship between \( p \) and \( q \) can be expressed as \( \frac{1}{p} + \frac{1}{q} = k \) and suggests solving for one variable and graphing the result, speculating it may be logarithmic.
  • Another participant notes that the thin lens relationship is a hyperbola.
  • One participant shares their recollection of a related presentation given by a sibling.
  • A participant presents a mathematical derivation leading to the conclusion that the minimum distance \( p + q \) is \( 4f \) and questions why it cannot be zero.
  • Another participant agrees with the derivation but advises checking whether \( 2f \) is indeed a minimum.
  • There is a discussion about the inadmissibility of zero solutions for \( p \) and \( q \), with one participant asserting that \( 1/0 \) is undefined and cannot be used in the lens equation.
  • A participant concludes that \( 2f \) must be the minimum distance since if the object is at infinity, \( q \) equals \( f \).

Areas of Agreement / Disagreement

Participants express differing views on the interpretation of the minimum distance and the implications of zero values for \( p \) and \( q \). There is no clear consensus on the correctness of the mathematical derivations or the nature of the minimum distance.

Contextual Notes

Participants highlight the importance of excluding zero values for \( p \) and \( q \) in the lens equation, indicating that assumptions about the definitions of these variables are crucial for the discussion.

Callisto
Messages
41
Reaction score
0
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 can't figure it out.
any tips please

Callisto
 
Mathematics news on Phys.org
Arent p and q distances...? Just add them.
 
If your saying given a fixed focal length, find the relationship between p and q, that case

[tex]\frac{1}{p} + \frac{1}{q} = k[/tex]

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.
 
What math tools do you have at your disposal?

Actually the thin lens relationship is a hyperbola.
 
I should know that, my brother did a presentation on it a week ago.
 
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

am i completely wrong?any comments.

Callisto
 
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:
1/0+1/0=0 , 0 doesn't = 1/f
therefore 0's are excluded from the problem, correct?
Callisto
 
1/0 + 1/0 is 0? Are you sure?
 
  • #10
1/0+1/0=1/0, 1/0 doesn't = 1/f
is this why 0's are excluded?
 
  • #11
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
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.
 

Similar threads

Replies
2
Views
2K
  • · Replies 22 ·
Replies
22
Views
4K
Replies
2
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
3K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 31 ·
2
Replies
31
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K