Answer
The phase at r = 3.5 m is $\frac{\pi}{2}$
The phase at r = 4.5 m is $\frac{3\pi}{2}$
Work Step by Step
Let $\phi_1 = \pi$ be the phase at $r_1 = 4.0~m$ and let $\phi_2$ be the phase at $r_2 = 3.5~m$. Therefore;
$\Delta \phi = \frac{2\pi~\Delta r}{\lambda}$
$\phi_2-\phi_1 = \frac{2\pi~(r_2-r_1)}{\lambda}$
$\phi_2 = \phi_1+\frac{2\pi~(r_2-r_1)}{\lambda}$
$\phi_2 = \pi+\frac{2\pi~(3.5~m-4.0~m)}{2.0~m}$
$\phi_2 = \frac{\pi}{2}$
The phase at r = 3.5 m is $\frac{\pi}{2}$.
Let $\phi_1 = \pi$ be the phase at $r_1 = 4.0~m$ and let $\phi_3$ be the phase at $r_3 = 4.5~m$. Therefore;
$\Delta \phi = \frac{2\pi~\Delta r}{\lambda}$
$\phi_3-\phi_1 = \frac{2\pi~(r_3-r_1)}{\lambda}$
$\phi_3 = \phi_1+\frac{2\pi~(r_3-r_1)}{\lambda}$
$\phi_3 = \pi+\frac{2\pi~(4.5~m-4.0~m)}{2.0~m}$
$\phi_3 = \frac{3\pi}{2}$
The phase at r = 4.5 m is $\frac{3\pi}{2}$.
