# Geometric optics- Sign conventions

Why do we need to use the sign convention again in the mirror equation while solving numericals when we know we have already used a convention while deriving the mirror equation? The question is not about the importance of sign conventions but why do we need to use the sign conventions twice? Doesn't the equation already include a sign convention?

1/u + 1/v = 1/ f

Or say, in the magification formula: height of image/ height of object = - (v/u)

Why do I need to bother about the signs ( -u or - v or +f) while using the above equations?

• Possible duplicates: physics.stackexchange.com/q/70730/2451 and links therein. More on sign conventions in optics. – Qmechanic May 23 '14 at 5:31
• Er ... the point is that you must use the same rules when applying a formula as when you derived it, no? It's like saying that you must check that the conditions under which a result was derived apply to your problem before you use the result. – dmckee --- ex-moderator kitten May 23 '14 at 5:39
• Exactly! we must not apply the sign convention to generalize something. Say, Magnification, in any case should be: Image_height/Object_height = Image_distance/Object_distance. So, now, why cant i use this particular expression for solving numericals and use the sign conventions accordingly. – Swami May 23 '14 at 10:09