Two beam interference general question Suppose we have two intensity outputs proportional to a harmonic oscillation. Let,
$$I_{1} = I_{0}*cos(kx)$$ be the first intensity output and $I_{2}$ be the second intensity output with a slightly different wavelength with respect to $I_{1}$.  Then, $$I_{2} = I_{0}*cos((k+Δk)x) = I_{0}*cos(\frac{2\pi}{\lambda + \Delta\lambda}x)$$
Question is, why $$k+Δk = \frac{2\pi}{\lambda + \Delta\lambda}$$ rather than, $$k+Δk = \frac{2\pi}{\lambda} + \frac{2\pi}{\Delta\lambda}$$
My logic is that since, $k = \frac{2\pi}{\lambda}$ then, $Δk = \frac{2\pi}{\Delta\lambda}$.
Many thanks,
 A: The units are not even the same on both sides of the equation $\Delta k=2\pi/\Delta \lambda$ so we should never expect it to hold.

Here's an argument based on definitions:
We know the formula $k=2\pi/\lambda$ always holds, so this means that it holds for both $k_1=2\pi/\lambda_1$ and $k_1=2\pi/\lambda_2$, regardless of what $k_1$ and $k_2$ or $\lambda_1$ and $\lambda_2$ are.
We can thus always say
$$k_2-k_1=2\pi\left(\frac{1}{\lambda_2}-\frac{1}{\lambda_1}\right)=2\pi\frac{\lambda_1-\lambda_2}{\lambda_1\lambda_2}.$$ Defining $\Delta k=k_2-k_1$ and similarly for the wavelengths, we always have
$$\Delta k=-\frac{2\pi\Delta \lambda}{\lambda_1\lambda_2}=-\frac{2\pi\Delta \lambda}{\lambda_1\left(\lambda_1+\Delta\lambda\right)}.$$ Then, if we want to know what $k_1+\Delta k$ is, we simply calculate:
$$k_1+\Delta k=\frac{2\pi}{\lambda_1}-\frac{2\pi\Delta \lambda}{\lambda_1\left(\lambda_1+\Delta\lambda\right)}=\frac{2\pi\left(\lambda_1+\Delta\lambda-\Delta \lambda\right)}{\lambda_1\left(\lambda_1+\Delta\lambda\right)}=\frac{2\pi}{\lambda_1+\Delta \lambda}.$$ This all really just follows from the definitions of $\Delta k$ and $\Delta \lambda$:
$$k_1+\Delta k\equiv k_2=\frac{2\pi}{\lambda_2}\equiv\frac{2\pi}{\lambda_1+\Delta\lambda}.$$

Now to address propagation of uncertainty to understand this "logically:" we begin with
$$\frac{\partial k}{\partial \lambda}=-\frac{2\pi}{\lambda^2}=-\frac{k}{\lambda}.$$ For infinitesimal $dk$ and $d\lambda$ we can thus write
$$\frac{dk}{k}=-\frac{d\lambda}{\lambda}.$$ This tells us that
$$k+dk=k\left(1-\frac{d\lambda}{\lambda}\right).$$ If we do a series expansion of the correct expression, we find
$$\frac{2\pi}{\lambda+d\lambda}=\frac{2\pi}{\lambda}\left(1-\frac{d\lambda}{\lambda}+\mathcal{O}(d\lambda^2)\right).$$ These two expressions clearly match to first order in $d\lambda$, so we see that the expression
$$k+\Delta k=\frac{2\pi}{\lambda+\Delta \lambda}$$ is necessary for error propagation to be respected, which is not the case for a formula like $\Delta k=2\pi/\Delta \lambda$.
