⚠️ Disclaimer: diffraction and interference will be used interchangeably
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/0441900f-f595-4b59-84d7-44b92289d6c3/Fourier_series.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/0441900f-f595-4b59-84d7-44b92289d6c3/Fourier_series.png" width="40px" /> Fourier series: using the orthogonal basis function $\sin$ and $\cos$ we can make any function in the range $x=-L$ to $x=L$ (or periodic function with period $2L$) using the series
$$ f(x)=a_0+ \sum_n a_n \sin \left (\frac{n\pi x}{L} \right ) + b_n \cos\left ( \frac{n\pi x}{L} \right ) $$
where
$$ \begin{aligned} a_0&=\frac{1}{2L}\int^L_{-L}f(x)\,\text dx \\ a_n&=\frac{1}{L}\int^L_{-L}f(x)\sin\left ( \frac{n \pi x}{L} \right )\,\text dx \\ b_n&=\frac{1}{L}\int^L_{-L}f(x)\cos\left ( \frac{n \pi x}{L} \right )\,\text dx \end{aligned}
$$
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<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/202bf7c1-5420-4baf-abcc-dac8c46ff85a/Orthogonal.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/202bf7c1-5420-4baf-abcc-dac8c46ff85a/Orthogonal.png" width="40px" /> Orthogonal: two function $f(x)$ and $g(x)$ are orthogonal if
$$ \int^L_{-L}f^*(x)g(x)\,\text dx=0 $$
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🗒️ Notes:
$\int f(x)g(x)\,\text dx$ tells you how much $f$ looks like $g$
we can write everything in exponential form aswell
$$ f(x)=\sum^\infin_{-\infin}a_n e^{in\pi x/L} \qquad a_n=\frac 1{2L}\int^L_{-L}f(x)e^{-in\pi x/L}\,\text dx $$
<aside> <img src="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/ff231438-958c-440a-a074-a86ae717e3f4/Fourier_transforms.png" alt="https://prod-files-secure.s3.us-west-2.amazonaws.com/369dfa6b-d4d9-4cf2-a446-e369553b6347/ff231438-958c-440a-a074-a86ae717e3f4/Fourier_transforms.png" width="40px" /> Fourier transforms: if we take the limit of the Fourier series as $L\to\infin$ we get
$$ f(x)=\frac{1}{2\pi} \int^\infin_{-\infin}F(\xi)e^{i\xi x}\,\text dx \qquad F(\xi)=\int^\infin_{-\infin} f(x)e^{-i\xi x}\,\text dx $$
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🗒️ Notes:
we can write the $\frac{1}{2\pi}$ in either side, or any combination where the product = $\frac{1}{2\pi}$
The minus signs in the exponent can be inverted
The Fourier transform of a small spiky function is a wide function and vise versa, 💃 Example:
$$ \text{square} \lrarr \operatorname{sinc} \quad \text{Gaussian} \lrarr \operatorname{Gaussian} \quad \text{Constant} \lrarr \delta \quad \text{2 }\delta \lrarr \cos $$
🗒️ Note: in this section will assume large distance
💼 Case: consider a single slit plane where we define the axis such that the slit is at $-a/2<x<a/2$
The path delay and phase delay are
$$ x\sin \theta \qquad kx\sin\theta $$
🗒️ Note: this is for the same reason as 2 slits
Adding all the waves from $-a/2$ to $a/2$ we get
$$ \begin{aligned} E_\text{tot} &=(E_0\,\text dx/a)e^{i(kz-\omega t)}(e^{-\frac{ika\sin\theta}2}+\ldots +e^{\frac{ika\sin\theta}2})\\ &=(E_0/a)e^{i(kz-\omega t)} \int^{a/2}_{-a/2}e^{-ikx\sin\theta}\,\text dx \end{aligned} $$
Solving we get
$$ \begin{aligned} E_\text{tot}&=\frac{E_0 e^{i(kz-\omega t)}}{\pi a \sin \theta/\lambda}\left ( \frac{\exp[-i\pi a \sin \theta /\lambda]-\exp[i\pi a \sin \theta /\lambda]}{-2i} \right ) \\ &=E_0 e^{i(kz-\omega t)}\operatorname{sinc} \left ( \frac{\pi a \sin \theta}{\lambda}\right ) \end{aligned} $$
Calculating intensity with $E_\text{tot}^* E_\text{tot}$ we get
$$ I=E^2_0 \operatorname{sinc}^2\left ( \frac{\pi a \sin \theta}{\lambda} \right ) $$
🗒️ Note: to get this answer we assumed the following:
The phase varies linearly across the aperture
Amplitude of arriving waves at the screen is the same
ignored obliquity factor
source is far away
Repeating the same process with phasors:
each phasor is at an angle
$$ k\,\text dx\sin\theta\equiv 2\pi \,\text dx \sin \theta/\lambda $$
with the last one $2\pi a \sin \theta /\lambda$ from first
Considering OAPQ we get
$$ E_\text{tot}=2r\sin \left ( \frac{\pi a \sin \theta}{\lambda} \right ) $$
Length is total amplitude $E_0=2\pi r(2\pi a \sin \theta )/2\pi \lambda =2\pi r a \sin \theta/ \lambda$ thus
$$ E_\text{tot}=E_0 \frac{\sin[\pi a \sin (\theta)/\lambda]}{\pi a \sin (\theta)/\lambda} $$
