Classical derivation of first and second minima in Single slit diffraction

Question

I was reading about single slit diffraction in Fundamentals of Physics by Halliday and Resnick. So, to derive the equation for first minima, they divided the slit into two halves. If the width of the slit is $a$ and the observation screen is placed far away, for the fist minima we get, $$\frac{a}{2}\sin\theta=\frac{\lambda}{2}\implies a\sin\theta=\lambda\tag{Eq.1}$$ as the condition for the first minima.

Now for the second minima, they divided the slit into four equal parts and $$\frac{a}{4}\sin\theta=\frac{\lambda}{2}\implies a\sin\theta=2\lambda\tag{Eq.2}$$

Now, my doubt is, What if I divide the slit into FOUR equal parts while I am trying to find the FIRST minima and analyze it in the same way? Then I will get Eq.2 for the first minima itself. What is my mistake here?

Answer 1

Your not anyalising for the first minima because we are showing when the path length between the 2 distances is $\lambda /2$ so when the distance between are 2 points is a/2 we find that asinθ=2λ so it just means we are comparing when the path length is λ/2 for the points we are mesuing where the waves come from