$\lambda=b sin \theta=b sin \frac{\phi}{2}$

$b=\frac{\lambda}{sin\frac{\phi}{2}}$

$\phi=30^0$, $b=\frac{550nm}{sin\frac{30^0}{2}}=2.125\mu m$

$\phi=45^0$, $b=\frac{550nm}{sin\frac{45^0}{2}}=1.437\mu m$

$\phi=90^0$, $b=\frac{550nm}{sin\frac{90^0}{2}}=0.778\mu m$

$\phi=180^0$, $b=\frac{550nm}{sin\frac{180^0}{2}}=0.55\mu m$