# 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?

• Yes they will since one distance is an integer multiple of the other. But you won't catch all of the nodes. You'll miss two out of three. Commented Jul 17, 2021 at 13:00

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.