Answer
$1.78\;pm$
Work Step by Step
Each standing wave has an integral number $n$ of half wavelengths. If $L$ be the distance between two points between which standing wave occurs, then
$2n.\frac{\lambda}{2}=2L$
or, $n=\frac{2L}{\lambda}$, $[n=1,2,3,4...................]$
Partially differentiating both sides taking $L$, we obtain
$|\Delta n|=\frac{2L}{\lambda^2}|\Delta\lambda|$
or, $|\Delta\lambda|=\frac{\lambda^2}{2L}|\Delta n|$
At $\lambda=533\;nm$, the lowest possible value of $|\Delta n|=1$. Therefore,
$|\Delta\lambda|=\frac{\lambda^2}{2L}$
or, $|\Delta\lambda|=\frac{(533\times10^{-7})^2}{2\times8}\,cm$
or, $|\Delta\lambda|=1.78\times10^{-10}\;cm$
or, $|\Delta\lambda|=1.78\;pm$
Therefore, Near $\lambda=533\;nm$, the wavelengths in the standing waves are in $1.78\;pm$ apart.
