Chapter 36 - Diffraction - Problems - Page 1114: 91

$m\approx2$

First, we find $d$: $d=\frac{0.012}{8900}=1.3483\times 10^{-6}m$ We know that $dsin{\theta}=m\lambda$ This can be rearranged as: $m=\frac{dsin{\theta}}{\lambda}$ We plug in the know values to obtain: $m=\frac{1.3483\times 10^{-6}\times sin{90^{\circ}}}{500\times 10^{-9} }\approx2$