Understanding the Mirror Formula: Actual Measurements vs Sign Convention

  • Context: Undergrad 
  • Thread starter Thread starter Ashu2912
  • Start date Start date
  • Tags Tags
    Formula Mirror Optics
Click For Summary

Discussion Overview

The discussion revolves around the mirror formula in optics, specifically addressing the treatment of image distance (v), object distance (u), and focal length (f) in terms of actual measurements versus sign conventions. Participants explore the implications of using signs in calculations and whether these distances can be treated solely as magnitudes.

Discussion Character

  • Debate/contested
  • Technical explanation

Main Points Raised

  • One participant asserts that v, u, and f should represent actual measurements and questions why signs are used in numerical applications.
  • Another participant emphasizes that signs are necessary to distinguish between real and virtual images and maintains that v, u, and f represent actual distances.
  • A participant suggests that while signs are used in derivations, it raises the question of whether v, u, and f can be treated as magnitudes in numerical problems.
  • Another participant argues that avoiding signs would lead to nonsensical equations, except in cases where all signs are positive.
  • One participant expresses confusion about the necessity of signs, suggesting that they have already been accounted for in derivations.

Areas of Agreement / Disagreement

Participants express differing views on the necessity and treatment of signs in the mirror formula. There is no consensus on whether v, u, and f can be treated as magnitudes without signs, indicating an unresolved debate.

Contextual Notes

Participants reference various sign conventions and their implications in deriving the mirror formula, highlighting the complexity of applying these conventions in practical scenarios.

Ashu2912
Messages
106
Reaction score
1
In the derivation of the mirror and magnification formula, v, u and f represent image distance, object distance and focal length resp. While deriving, we take v,u and f as the actual measurements of distances, apply sign convention to these, and come to the mirror formula. Thus, in the mirror formula, v,u and f must represent the actual measurements. Then, in numericals, why do we replace v,u and f with the signs and why not consider them as the magnitudes of distances only?
 
Science news on Phys.org
I'm sorry, but I don't really understand the question. The signs are needed to distinguish real and virtual images (and objects). In all cases, v, u, and f represent actual distances.
 
Doc Al said:
I'm sorry, but I don't really understand the question. The signs are needed to distinguish real and virtual images (and objects). In all cases, v, u, and f represent actual distances.

I mean that we derive the relation between the distances, substitute these distances with v, u, f, h.etc. with signs, for eg. -v, -u, +f, -h.etc. Thus, v, u, f.etc. represent the magnitude of distances like 25cm, 35cm, rather than -25cm or +35cm.etc., as we have substituted them with the signs in the equation and then derived the mirror formula. But, in numericals, in mirror formula application, we represent v, u, f.etc. with +25cm or -35 cm, i.e. with signs. This is what I mean...
 
I think the reason for using signs is that, in a situation with non-plane mirrors, the final image may be either side of the mirror. Whatever sign convention you use (I did some geometric optics about 100 yrs ago and I remember two others, in additional to the 'schoolboy' one), as long as you stick to it, you will get the right answer. But you can't avoid signs - implicit or explicit.
 
sophiecentaur said:
as long as you stick to it, you will get the right answer. But you can't avoid signs - implicit or explicit.
True, I too use signs and get the answer in numericals. But is it valid if we avoid them and treat v,u,f.etc. as modulus of the distances?
 
Ashu2912 said:
True, I too use signs and get the answer in numericals. But is it valid if we avoid them and treat v,u,f.etc. as modulus of the distances?
If you ignore the signs, the equations will not make sense. Except where the signs are all positive, of course.
 
How could you possibly think that the signs were not vital?
 
Because we have already used the signs in the derivation, and treated v,f,u.etc. as modulus of distances!
 

Similar threads

  • · Replies 11 ·
Replies
11
Views
3K
  • · Replies 13 ·
Replies
13
Views
11K
  • · Replies 2 ·
Replies
2
Views
8K
  • · Replies 17 ·
Replies
17
Views
4K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 3 ·
Replies
3
Views
5K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 10 ·
Replies
10
Views
8K