Editing Cross section (physics) (section)

=== Scattering from a 2D circular mirror ===

The following example deals with a beam of light scattering off a circle with radius {{math|''r''}} and a perfectly reflecting boundary. The beam consists of a uniform density of parallel rays, and the beam-circle interaction is modeled within the framework of [[geometric optics]]. Because the problem is genuinely two-dimensional, the cross section has unit of length (e.g., metre). Let {{math|''α''}} be the angle between the [[ray (optics)|light ray]] and the [[radius]] joining the reflection point of the ray with the center point of the mirror. Then the increase of the length element perpendicular to the beam is
: <math>\mathrm dx = r \cos \alpha \,\mathrm d \alpha.</math>
The reflection angle of this ray with respect to the incoming ray is {{math|2''α''}}, and the scattering angle is
: <math>\theta = \pi - 2 \alpha.</math>
The differential relationship between incident and reflected intensity {{math|''I''}} is
: <math>I \,\mathrm d \sigma = I \,\mathrm dx(x) = I r \cos \alpha \,\mathrm d \alpha = I \frac{r}{2} \sin \left(\frac{\theta}{2}\right) \,\mathrm d \theta = I \frac{\mathrm d \sigma}{\mathrm d \theta} \,\mathrm d \theta.</math>
The differential cross section is therefore ({{math|dΩ {{=}} d''θ''}})
: <math>\frac{\mathrm d \sigma}{\mathrm d \theta} = \frac{r}{2} \sin \left(\frac{\theta}{2}\right).</math>
Its maximum at {{math|''θ'' {{=}} π}} corresponds to backward scattering, and its minimum at {{math|''θ'' {{=}} 0}} corresponds to scattering from the edge of the circle directly forward. This expression confirms the intuitive expectations that the mirror circle acts like a diverging [[lens (optics)|lens]]. The total cross section is equal to the diameter of the circle:
: <math>\sigma = \int_0^{2 \pi} \frac{\mathrm d \sigma}{\mathrm d \theta} \,\mathrm d \theta = \int_0^{2 \pi} \frac{r}{2} \sin \left(\frac{\theta}{2}\right) \,\mathrm d \theta = 2 r.</math>