I am given that:

$$2k(L-x_\text{min}) = 2n\pi$$ $$k=\frac{2\pi}{\lambda}$$

the expected result is: $$x_\text{min} = L-\frac{n\lambda}{2}$$

Could someone help to get this result?

This is what I tried, but it s not going in the expected direction so far...

$x_\text{min}= \frac{2n\pi-2kL}{-2k}$

$=\frac{n\pi-kL}{k}$ $=\frac{n\pi-\frac{2\pi}{\lambda}L}{\frac{2\pi}{\lambda}}$ $=(n\pi-\frac{2\pi L}{\lambda})\frac{2\pi}{\lambda}$

And this iw where I am stuck. I think I am now way too far away from the expected result

  • $\begingroup$ Typo. You mean $2 k (L-x_{min})=2 n \pi.$ $\endgroup$ – DanielWainfleet Apr 16 '16 at 13:01
  • $\begingroup$ @user254665, my bad. Corrected $\endgroup$ – J.Doe Apr 16 '16 at 13:02
  • $\begingroup$ Welcome to MathSE! You are more likely to get a good answer to your question if you follow a few guidelines. In particular, what have you tried so far, and just where are you stuck? This is not a homework-answering site: many of us want to see that you have put significant work into the problem. You have shown us the expected result, but please show us some work of your own. $\endgroup$ – Rory Daulton Apr 16 '16 at 13:31
  • $\begingroup$ @RoryDaulton, edited $\endgroup$ – J.Doe Apr 16 '16 at 13:36
  • $\begingroup$ Your first line is correct. Then you dropped a minus sign, should have been $\frac{n\pi-kL}{-k}$. At the end of the second line you multiplied by $\frac{2\pi}{\lambda}$ instead of dividing by it. $\endgroup$ – almagest Apr 16 '16 at 13:47

I think you're making it harder by taking some inefficient steps in the alegbra. You want to isolate $x$, so instead of distributing, divide through by $2k$.

$$2k(L-x_\text{min}) = 2n\pi$$ $$L-x_\text{min} = \frac{n\pi} k$$


$$L - \frac{n\pi} k= x_\text{min}$$

Since $k=\frac{2\pi}{\lambda}$, then $\frac 1k = \frac{\lambda}{2\pi}$. Substitute that

$$L - \frac{n\pi\lambda}{2\pi}= x_\text{min}$$ $$L - \frac{n\lambda}{2}= x_\text{min}$$


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