Answer
For the first fringe: $I=5.7mW/cm^2$
For the second fringe: $I=2.9mW/cm^2$
Both values are verified by the given graph.
Work Step by Step
For the first fringe ($m_2=1$), we use equations (36-5) and (36-6) to find the intensity.
$\alpha = \frac{\pi a}{\lambda}\sin \theta$
$\alpha = \frac{\pi 5.05\mu m}{0.440\mu m}\sin 1.25^{\circ}$
$\alpha = 0.787rad$
$I=I_m (\frac{\sin \alpha}{\alpha})^2$
$I=7.0mW/cm^2 (\frac{\sin 0.787}{0.787})^2$
$I=5.7mW/cm^2$
This is consistent with the given graph.
For the first fringe ($m_2=2$), we use equations (36-5) and (36-6) to find the intensity.
$\alpha = \frac{\pi a}{\lambda}\sin \theta$
$\alpha = \frac{\pi 5.05\mu m}{0.440\mu m}\sin 2.5^{\circ}$
$\alpha = 1.57rad$
$I=I_m (\frac{\sin \alpha}{\alpha})^2$
$I=7.0mW/cm^2 (\frac{\sin 1.57}{1.57})^2$
$I=2.9mW/cm^2$
This is consistent with the given graph.
Both values of intensity agree with the given graph.
