Question 11.3: (High-energy scattering from a black disk) Discuss the scatt...

A black disk is totally absorbing (i.e.,\eta _l(k)=0) .  Assuming the values of l do not exceed a maximum value l_{max}(l\leq l_{max})  and that k is large (high-energy scattering), we have l_{max}=ka  where a is the radius of the disk. Since \eta _l=0,  equations (11.113) \sigma _{el}=4\pi \sum\limits_{l=0}^{\infty }{(2l+1)|f_l|^2}=\frac{\pi }{k^2}\sum\limits_{l}{(2l+1)(1+\eta ^2_l-2{\eta _l}\cos 2\delta _l)}.  and (11.114) \sigma _{inel}=\frac{\pi }{k^2}\sum\limits_{l=0}^{\infty }{(2l+1)}\left(1-\eta ^2_l(k)\right) .   lead to

\sigma _{inel}=\sigma _{el}=\frac{\pi }{k^2}\sum\limits_{l=0}^{ka }{(2l+1)}=\frac{\pi }{k^2}(ka+1)^2\simeq \pi a^2;              (11.117)

hence the total cross section is given by

\sigma _{inel}=\sigma _{el}+\sigma _{inel}=2\pi a^2.                  (11.118)

Classically, the total cross section of a disk is equal to \pi a^2.   The factor 2 in (11.118) is due to purely quantum effects, since in the high energy limit there are two kinds of scattering: one corresponding to waves that hit the disk, where the cross section is equal to the classical cross section \pi a^2. ,  and the other to waves that are diffracted. According to Babinet’s principle, the cross section for the waves diffracted by a disk is also equal to \pi a^2.