I'm sitting and trying to solve the equation of the phase difference given by:
$\Delta \phi = k (\sqrt{a^2+d^2} -d ) \approx \frac{ka^2}{2d}$
Where $a$ is the size of an aperture and $b$ is the distance the point at the aperture's center as shown in the figure below.
I'm not a math expert here, so I wondering if anyone can explain the approximation that has been done above here.
Reference:
Applications of Classical Physics by Roger D. Blandford and Kip S. Thorne - Chapter 8 - Diffraction
$$k\sqrt{a^2 +d^2} -kd = d k\left(\sqrt{1+\frac{a^2}{d^2}} -1 \right)$$
Assuming $\frac {a^2}{d^2}$ is small, you can binomially approximate (for small $x$, $\sqrt {1+x} \approx 1+\frac 12 x$):
$$\approx dk \left( (1+\frac{a^2}{2d^2}) -1 \right)\\ =\frac{ka^2}{2d}$$