Why use signed quantities in the mirror formula if it's derived from magnitudes

[Ashu2912]

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?

 

[Doc Al]

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.

 

[Ashu2912]

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...

 

[sophiecentaur]

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.

 

[Ashu2912]

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?

 

[Doc Al]

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.

 

[sophiecentaur]

How could you possibly think that the signs were not vital?

 

[Ashu2912]

Because we have already used the signs in the derivation, and treated v,f,u.etc. as modulus of distances!