Single slit diffraction formula $a\sin\theta=m\lambda$ when $m=3$ For single slit diffraction, when $a\sin\theta=3\lambda$, this means $\frac{a}{6}\sin\theta=\frac{\lambda}{2}$ so the waves from every pair of point source whose path difference is $\frac{a}{6}\sin\theta$ will destructively interfere, and this is what it says in the textbook. But what if we rewrite this as $\frac{a}{2}\sin\theta=\frac{3\lambda}{2}$, and say that the waves from every pair of point source whose path difference is $\frac{a}{2}\sin\theta$ will also destructively interfere? (Because they are out of phase by $\frac{\lambda}{2}+\lambda$). Is this also valid?
 A: The phase difference diffraction from the position $x$ of the slit onto the view angle $\theta$ is $ x\sin\theta$. Thus, the total amplitude of the diffraction wave:
$$
  A(\theta) \propto \int_0^d e^{i k x\sin\theta} dx =\int_0^d e^{i\frac{ 2\pi x\sin\theta}{\lambda}} dx
$$
if we integrate over a segment of the slit, $x_0 \to x_1$ and $(x_1 - x_0)\sin\theta = \lambda $, which renders a whole $2\pi$ phase change. Its contribution to the amplitude $\Delta A$
\begin{align*}
  \Delta A(\theta) &\propto \int_{x_0}^{x_1} e^{i\frac{ 2\pi x\sin\theta}{\lambda}} dx \\
& = \frac{\lambda}{2i\pi\sin\theta} \left\{e^{i\frac{ 2\pi x\sin\theta}{\lambda}}\right\}_{x_0}^{x_1}\\
&= \frac{\lambda}{2i\pi\sin\theta}\left\{e^{i\frac{ 2\pi x_1\sin\theta}{\lambda} } - e^{i\frac{ 2\pi x_0\sin\theta}{\lambda} }\right\}\\
&= \frac{\lambda}{2i\pi\sin\theta}\left\{e^{i\frac{ 2\pi (x_0\sin\theta + \lambda)}{\lambda} } - e^{i\frac{ 2\pi x_0\sin\theta}{\lambda} }\right\}\\
&= 0.
\end{align*}
Therefore, each segment of the slit, which gives a whole $\lambda$ optical path difference, cancels out internally,  gives zero contribution to the diffraction amplitude.
Thus, when $ d\sin\theta = n \lambda$, you can divide the silt into $n$ segments. Each segment has a $\lambda$ optical path difference, and gives zero contribution to the diffraction amplitude. This is the condition for a minimum intensity (dark) fringe.
