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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?

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    $\begingroup$ Possible duplicates: physics.stackexchange.com/q/70730/2451 and links therein. More on sign conventions in optics. $\endgroup$ – Qmechanic May 23 '14 at 5:31
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    $\begingroup$ 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. $\endgroup$ – dmckee May 23 '14 at 5:39
  • $\begingroup$ 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. $\endgroup$ – Swami May 23 '14 at 10:09
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In starting it is futile, but if you use many mirrors, you will get mad.

Actually, you have derived your formula in a particular case, say when image is behind the mirror. To extrapolate it to other cases, you say that you could have used opposite sign of image distance.

But this will create different formulas for different cases as distance is always positive. To avoid a large number of these formulas, we united and decided that positive is in this direction and negative in that one. Now, all of our formulas agree and we live happily together

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  • $\begingroup$ While derving the formula we reach a particluar expression that can be used as a generlized equation (one step before applying the sign convention) but instead we apply the sign convention and then call it a generalized equation. Could you please elaborate this a little more? $\endgroup$ – Swami May 23 '14 at 10:07

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