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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^'##