Why Does the Area Under a Diffraction Curve Equal This Value?

[albertrichardf]

Hi,
consider the following curve:
$$f(\theta) = \frac {I_0sin^2(n\theta/2)}{sin^2(\theta/2)}$$

When the area over a cycle from $0$ to $2π$ is evaluated it gives $(2πnI_0)$. This is exactly $\frac {I_{max} + I_{min}}{2}$ , since
$I_{min}$ is $0$. Is this a coincidence, or is there a reason behind the area under the curve is the same as this value?
Thank you for your answers.

 

[RUber]

Does this mean that $I_{max} = 4\pi n I_0$ all the time? If so, then that would be all the coincidence needed.

 

[Charles Link]

$I_{max}=n^2 I_o$. The maximum occurs periodically when both the numerator and denominator equal (i.e. approach) zero. This is in the limit $\frac{\theta}{2} \rightarrow m \pi$. I don't know if you have the integral evaluated correctly for a single cycle. I would need to try to look that one up. I don't even know that it has a simple closed form. $\\$ Usually the letter $\phi$ is used instead of $\theta$ in this diffraction theory integral for $N$ slits, where the phase $\phi=\frac{2 \pi d \sin{\theta}}{\lambda}$, and the $N$ is designated with a capital letter.