[Physics] Geometric optics- Sign conventions

conventionsoptics

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?

Best Answer

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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