Answer
$9\mu m$
Work Step by Step
We can find the required diameter as follows:
$\theta=3.7^{\circ}(\frac{\pi rad}{180^{\circ}})=3.7\times 0.01745rad=0.06456rad$
We know that
$\theta=\frac{1.22\lambda}{Dn}$
This can be rearranged as:
$D=\frac{1.22\lambda}{\theta n}$
We plug in the known values to obtain:
$D=\frac{(1.22)(630\times 10^{-9}m)}{(1.336)(0.06456rad)}$
$D=9\times 10^{-6}m=9\mu m$
