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How to prove the thin lens equation, geometrically?
##1/f=1/u+1/u^'##
##"triangle "" EFO "" is similar to ""triangle "OIH##
##(bar (IH))/(bar (EF))=(bar(OI))/(bar(OF))##
##g:"image height"##
##c:"object height"##
##u:"distance between object and lens"##
##u^':"distance between image and lens"##
##color(red)(g/c)=u^'/u##
##"triangle "" EFC "" is similar to ""triangle "COG##
##(bar (OG))/(bar (EF))=(bar(OC))/(bar(FC))##
##color(red)(g/c)=f/(u-f)##
##u^'/u=f/(u-f)##
##u^'(u-f)=u f##
##u^' u-u^' f=u f##
##u f+u^' f=u^' u##
##f(u+u^')=u^' u##
##f=(u^' u)/(u+u^')##
##1/f=(u+u^')/(u^' u)##
##1/f=cancel(u)/(u^' cancel(u))+cancel(u)^'/(cancel(u)^' u)##
##1/f=1/u+1/u^'##