What Is the Focusing Range for a 45 mm Lens on an SLR Camera?

Click For Summary

Homework Help Overview

The discussion revolves around the focusing range of a 45 mm lens on a 35 mm single lens reflex (SLR) camera, specifically how the lens must adjust to focus on objects from infinity down to 1.4 m. The original poster presents a problem involving the thin-lens formula and seeks clarification on handling the concept of infinity in equations.

Discussion Character

  • Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to apply the thin-lens formula to find the distance the lens must move to focus on objects at different distances. They question how to interpret the term involving infinity in the equation.

Discussion Status

Some participants provide insights into the mathematical interpretation of infinity, suggesting that it can be treated as zero in the context of the equation. There is an ongoing exploration of terminology related to this concept, with no explicit consensus reached on preferred phrasing.

Contextual Notes

The original poster indicates they received assistance from a tutor, which may influence their understanding of the problem. The discussion includes varying interpretations of mathematical concepts, particularly regarding the treatment of infinity in equations.

HenryHH
Messages
12
Reaction score
0

Homework Statement



In a 35 mm single lens reflex camera (SLR) the distance from the lens to the film is varied in order to focus on objects at varying distances. Over what range must a lens of 45 mm focal length vary if the camera is to be able to focus on objects ranging in distance from infinity down to 1.4 m from the camera?

Homework Equations



1/f = 1/p +1/q

The Attempt at a Solution



First, the thin-lens formula is applied for the case of focusing on objects at a distance of infinity:

1/45 = 1/∞ + 1/q
q = 45 mm

Then, the same formula is applied for the case of focusing on objects at a distance of 1.4 mm:

1/45 = 1/1400 + 1/q
q = 46.49 mm

Therefore, the range of movement is 46.49 - 45 = 1.49 mm.

I actually had help from a tutor with solving this problem. I understand the problem itself, but what I don't understand is how q = 45 was derived from the equation 1/45 = 1/∞ + 1/q since the equation contains ∞ as a variable.

Generally speaking, how do you solve equations with infinity (∞) as one of the variables? I am really stumped...
 
Physics news on Phys.org
1/∞ is a very tiny number, called zero. :smile:

So 1/p = 1/q, giving p=q
 
OH... so 1/infinity basically just cancels out, then? Duh... thanks!
 
HenryHH said:
OH... so 1/infinity basically just cancels out, then? Duh... thanks!
I wouldn't use the phrase "cancels out".
mVVtk.gif
 Perhaps "fades away", "disappears", or "vanishes", would be appropriate though.
 

Similar threads

  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 20 ·
Replies
20
Views
4K
Replies
15
Views
8K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
4
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
5K
Replies
9
Views
3K
  • · Replies 8 ·
Replies
8
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K