- #1
hale2bopp
- 21
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When we derive the relations between image distance(v), object distance(u) ond focal length(f) for mirrors or lenses (1/v+1/u=1/f and 1/v-1/u=1/f respectively), using the concept of similar triangles and alternate exterior angles, in the last step we apply standard sign conventions to get the above formulae.
But why do we need to apply the sign conventions to the standard formulae again while solving problems based on lenses and mirrors? Isn't that sort of cancelling out the effect of applying sign conventions at all?
I thought it might be because ultimately distances are always positive and this was to nullify the effect of taking distances measured opposite to direction of incident ray as negative(according to standard sign conventions). But then why apply sign conventions in the first place then?
Thanking you in advance
Hale2bopp
But why do we need to apply the sign conventions to the standard formulae again while solving problems based on lenses and mirrors? Isn't that sort of cancelling out the effect of applying sign conventions at all?
I thought it might be because ultimately distances are always positive and this was to nullify the effect of taking distances measured opposite to direction of incident ray as negative(according to standard sign conventions). But then why apply sign conventions in the first place then?
Thanking you in advance
Hale2bopp