$$\frac{1}{f} = \frac{n_2-n_1}{n_1\left(\frac{1}{R_1}-\frac{1}{R_2}\right)}$$

How do I isolate $R_2$ onto the left side of the equation? So far, I know that I have to cross multiply first and then subtract $1/R_1$ on both sides. Then subtract $1/R_1$ from $f(n_2-n_1)/n_1$. However I'm confused as to what to do next. Could someone please explain step by step on how to get the answer? Thank you!

  • $\begingroup$ I have edited your question to make the formula clearer (based on what I guessed it was). Is that what was intended? $\endgroup$
    – Aryabhata
    Aug 1, 2017 at 7:13
  • $\begingroup$ Welcome to MathSE. Always use MathJax for math formatting- meta.math.stackexchange.com/questions/5020/… $\endgroup$
    – Naive
    Aug 1, 2017 at 7:14
  • $\begingroup$ @Aryabhata Yes, that was the equation I was going for. Thanks for editing it! $\endgroup$
    – LLL
    Aug 1, 2017 at 7:16

1 Answer 1


Take it one step at a time. At each step, see what the first ("outermost") obstacle to $R_2$ being alone is, and deal with it. First get $R_2$ out from under the large fraction ("cross multiply") (If you are confident here, you can leave $n_1$ in the denominator and just multiply the bracket; I chose to remove the entire fraction) $$ n_1\left(\frac{1}{R_1}-\frac{1}{R_2}\right) = f(n_2-n_1) $$ Then get $R_2$ out from the brackets $$ \frac{n_1}{R_1}-\frac{n_1}{R_2}= f(n_2-n_1) $$ Then get all the non-$R_2$ terms away from the side of the equation where $R_2$ is $$ -\frac{n_1}{R_2}= f(n_2-n_1) - \frac{n_1}{R_1} $$Then get $-n_1$ away from the fraction where $R_2$ is $$ \frac1{R_2} = \frac{1}{R_1}-\frac{f(n_2-n_1)}{n_1} $$ Then get $R_2$ out from under the fraction (invert the whole shebang): $$ R_2 = \frac{1}{\frac1{R_1} - \frac{f(n_2-n_1)}{n_1}} = \frac{R_1n_1}{n_1 - R_1f(n_2-n_1)} $$


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