Raileigh diffraction by circular aperture

Question

I am a beginner to physics and would need an explaination on a statement in a book "Karttunen, Fundamental astronomy". In a section named "Raileigh diffraction by circular aperture", author states:

The sum of the amplitudes of the waves from a small surface element is proportional to the area of the element $\operatorname{d}x$ $\operatorname{d}y$

And then author writes down the equation for amplitude vector $\boldsymbol A$

$$ \boldsymbol A = \operatorname{d}x \operatorname{d}y (\cos{\delta} \boldsymbol{\hat{i}} + \sin{\delta} \boldsymbol{\hat{j}}), $$

where $\cos{\delta} \boldsymbol{\hat{i}} + \sin{\delta} \boldsymbol{\hat{j}}$ is a dirrection of an amplitude vector $\boldsymbol A$ and vectors $\boldsymbol{\hat{i}}$ and $\boldsymbol{\hat{j}}$ are unit vectors. And my question is, how do the author know that amplitude is proportional to element $\operatorname{d}x$ $\operatorname{d}y$ ???

Answer 1

Amazingly that section in Karttunen's book is available here. Truly the students of today are spoilt!

Anyhow, assuming the illumination is even, the amount of light through some area element is just proportional to that area. However the phase of the light varies across the circular aperature. What Karttunen is saying is that suppose we take an element $(dxdy)_0$ at the centre, then the intensity at some other element $(dxdy)_r$ at a radius $r$ is the same because the illumination is even, but you have to allow for the phase shift of $\delta$. The $cos\delta\vec{i} + sin\delta\vec{j}$ is just shifting a phase $\delta$ along the wave.

Putting this another way, if the light coming from the central element is $Ae^{i\omega t}$, where A is some constant, then the light coming from the element at $r$ is $Ae^{i(\omega t + \delta)}$, where $A$ is the same constant and $\delta$ is the phase difference between the centre and the point at $r$.