Chapter 36 - Diffraction - Problems - Page 1109: 13c

$I(\theta)/I_m =0.93$

Using the value for alpha we found in part b ($\alpha=0.46rad$), we can use the given values in equation (36-5) to find the ratio of intensity: $I(\theta)=I_m (\frac{\sin \alpha}{\alpha})^2$ $I(\theta)/I_m =(\frac{\sin \alpha}{\alpha})^2$ $I(\theta)/I_m =(\frac{\sin 0.46rad}{0.46rad})^2$ $I(\theta)/I_m =0.93$