PX275 - I4 - key geometry

an aperture and a detector screens are in parallel, separated by a distance, $D$
calling the direction connecting the screens $z$ , so $\hat{z} = (0, 0, 1)$
using $\vec{y} = (y_{1}, y_{2}, y_{3})$ to describe points on the aperture screen, such that $y_{3}$ is in $\hat{z}$ direction
using $\vec{x} = (x_{1}, x_{2}, x_{3})$ to describe points on the detector screen, such that $x_{3}$ is in $\hat{z}$ direction
$\vec{r}$ connects $\vec{x}$ with $\vec{y} :$ $\vec{r} = \vec{x} - \vec{y}$
holes in the aperture screen will act as secondary sources
intensity pattern on the detector, $I (\vec{x}) = | \vec{u} (\vec{x}) |^{2}$ is the superposition of all secondary sources
considering a simple case of a single pinhole aperture
so, one secondary point source emanating from aperture screen
for convenience, setting $y_{3} = 0$ , then $\vec{y} = (y_{1}, y_{2}, 0)$ , where, $(y_{1}, y_{2})$ denotes the position on the screen
if pinhole is at the centre, ${\vec{y}}_{p i n h o l e} = (0, 0, 0)$ , the origin
then, for the detector screen, $\vec{x} = (x_{1}, x_{2}, D)$ , $x_{3} = D$ , as screens are parallel
the amplitude is just that of a single point source:

u (x, t) = \frac{A}{r} \exp (i k (r - c t)) = \frac{A}{| \vec{x} - \vec{y} |} \exp (i k (| \vec{x} - \vec{y} | - c t))

u = \frac{A}{D} \exp (i k (D - c t))

away from the pinhole, the path length is longer, so the intensity falls as $1 / r^{2}$
now, considering two pinholes

u (\vec{x}, t) = \frac{A^{1}}{| \vec{x} - {\vec{y}}^{1} |} \exp (i k (| \vec{x} - {\vec{y}}^{1} | - c t)) + \frac{A^{2}}{| \vec{x} - {\vec{y}}^{2} |} \exp (i k (| \vec{x} - {\vec{y}}^{2} | - c t))

the path length differences lead to phase differences between the two secondary sources, causing interference
different path lengths are seen in ${\hat{y}}_{1}$ , but there is no difference in ${\hat{y}}_{2}$