Answer
$I(\theta)=I_m (\frac{\sin \alpha}{\alpha})^2$
$I_m/2=I_m (\frac{\sin \alpha}{\alpha})^2$
$1/2=(\frac{\sin \alpha}{\alpha})^2$
$1/2=(\frac{\sin^2 \alpha}{\alpha^2})$
$\alpha^2*1/2=\sin^2 \alpha$
We show that $\alpha^2*1/2=\sin^2 \alpha$ when the intensity is one-half that at the center of the pattern
Work Step by Step
We require that the intensity is one-half that at the center of the pattern, ($I=I_m/2$).
Now we use this is equation (36-5):
$I(\theta)=I_m (\frac{\sin \alpha}{\alpha})^2$
$I_m/2=I_m (\frac{\sin \alpha}{\alpha})^2$
$1/2=(\frac{\sin \alpha}{\alpha})^2$
$1/2=(\frac{\sin^2 \alpha}{\alpha^2})$
$\alpha^2*1/2=\sin^2 \alpha$
So it is true that $\alpha^2*1/2=\sin^2 \alpha$ when the intensity is one-half that at the center of the pattern
