Question #143ab

Question

2642 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-10-20T08:57:56

$1/f=1/u+1/u^'$

enter image source here

$"triangle "" EFO "" is similar to ""triangle "OIH$

$(bar (IH))/(bar (EF))=(bar(OI))/(bar(OF))$

$g:"image height"$

$c:"object height"$

$u:"distance between object and lens"$

$u^':"distance between image and lens"$

$color(red)(g/c)=u^'/u$

$"triangle "" EFC "" is similar to ""triangle "COG$

$(bar (OG))/(bar (EF))=(bar(OC))/(bar(FC))$

$color(red)(g/c)=f/(u-f)$

$u^'/u=f/(u-f)$

$u^'(u-f)=u f$

$u^' u-u^' f=u f$

$u f+u^' f=u^' u$

$f(u+u^')=u^' u$

$f=(u^' u)/(u+u^')$

$1/f=(u+u^')/(u^' u)$

$1/f=cancel(u)/(u^' cancel(u))+cancel(u)^'/(cancel(u)^' u)$

$1/f=1/u+1/u^'$