Single Slit Diffraction issue with derivation

When deriving the minima, the classical approach is to say

$$\frac d 2 \sin(\theta) = \frac{\lambda}2$$

therefore, $$d \sin(\theta) = \lambda$$ It can then be shown for any integer value of $$\lambda$$. But why couldn't you say that the 2 point sources that interfere destructively are the full width of the slit apart, so $$d\sin(\theta) = \lambda/2$$, therefore $$d\sin(\theta) = \lambda/2$$ for destructive interference, which wouldn't be an integer. Maybe I've misunderstood something.

• What about the sources in between the first and last sources? They are not interfering destructively.
– user258881
Commented Apr 25, 2020 at 12:06

you really are misunderstanding. You should consider the slit as the source of many or n elementary waves, so every elementary wave of the first half has the Difference $$\lambda/2$$ to one of the second half, while in your picture only two of the n waves have the difference of 1/2 all the rest have less, so the amplitude will be smaller, but never zero

• That makes sense. Then my textbook says that you can split the slit into 4, 6, 8, ... parts and claims all $\theta$ such that $\sin\theta=\frac{\lambda m}{d}$ for positive integer $m$ would be destructively interference. So my question would be why can't you split it into $3$ parts and say $\sin\theta=\frac{3}{2}\cdot \frac{lambda}{d}$? Commented Apr 22, 2023 at 15:56
• if $sin𝜃= \frac{3}{2} \frac{\lambda}{𝑑}$ the first and third part have difference of $\lambda/2$ and interfere to $0$ the rest gives a small max. Commented Apr 23, 2023 at 21:01
• Oh that makes sense. Thanks! Commented Apr 24, 2023 at 1:31