Consider a photon on approach to a electron, at a horizontal distance dd away, what is the deflection? We break the path of the photon into 2 parts, before it is level horizontally with the electron and after it is level with the electron.
Deflection before being level with the charge,
r=√d2+(Do−ct)2
sin(θ)=dr=d√d2+(Do−ct)2
Fh=Fsin(θ)=qpq4πεor2sin(θ)=qpq4πεo(d2+(Do−ct)2)d√d2+(Do−ct)2
xd=∫t=Do/c0Fhmpdt=∫t=Doc0qpq4πmpεod(d2+(Do−ct)2)−3/2dt=qpq4πmpεo{(ct−Do)cd√d2+(Do−ct)2}Do/c0
xd1=qpq4πmpεoc{Dod√d2+D2o}
Deflection after being level with the electron,
r=√d21+(ct)2
where d1=xd1+d
sin(θ)=d1r=d1√d21+(ct)2
Fh=Fsin(θ)=qpq4πεor2sin(θ)=qpq4πεo(d21+(ct)2)d1√d21+(ct)2
xd=∫t=D/c0Fhmpdt=∫t=D/c0qpq4πmpεod1(d21+c2t2)−3/2dt=qpq4πmpεo{td1√d21+c2t2}D/c0
xd2=qpq4πmpεoc{Dd1√d21+D2}
d1=xd1+d
xd2=qpq4πmpεoc{D(xd1+d)√(xd1+d)2+D2}
If we estimate the optical path as the hypotenuse of a triangle (It is usually wrong to estimated optical path to find path difference, because the actual path difference is already small.), the optical path difference between a straight photon and a deflected photon is,
OP=√(Do+D)2+(xd1+xd2)2−(Do+D)
For destructive interference,
√(Do+D)2+(xd1+xd2)2−(Do+D)=(2n+1)λ2
where n=0,1,2,3...
√(Do+D)2+(xd1+xd2)2Do+D−1=(2n+1)λ2(Do+D)
√(Do+D)2+(xd1+xd2)2Do+D=(2n+1)λ2(Do+D)+1
1cos(θ)=(2n+1)λ2(Do+D)+1
cos(θ)=2(Do+D)(2n+1)λ+2(Do+D)
2(Do+D)(2n+1)λ+2(Do+D)<1
we know that θ has solutions.
when n is small,
2(Do+D)(2n+1)λ+2(Do+D)≈1, θ=0
For constructive interference,
√(Do+D)2+(xd1+xd2)2−(Do+D)=nλ
n=0,1,2,3,...
1cos(θ)=nλDo+D+1
cos(θ)=Do+Dnλ+Do+D
when n is small,
Do+Dnλ+Do+D≈1 θ=0
Both constructive and destructive interference mix and there is no pattern! There is no observable interference pattern due to the interaction between the straight photons and deflected photons.