Consider a photon on approach to a electron, at a horizontal distance \(d\) 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,
\(r=\sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } } \)
\( sin(\theta )=\cfrac { d }{ r } =\cfrac { d }{ \sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } } } \)
\( F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }(d^{ 2 }+(D_{ o }-ct)^{ 2 }) } \cfrac { d }{ \sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } } } \)
\( x_{ d }=\int _{ 0 }^{ t=D_{ o }{ / }c }{ \cfrac { F_{ h } }{ m_{ p } } } dt=\int _{ 0 }^{ t=D_{ o }c }{ \cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } } d(d^{ 2 }+(D_{ o }-ct)^{ 2 })^{ -3/2 } } dt=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } } \left\{ \cfrac { (ct-D_{ o }) }{ cd\sqrt { d^{ 2 }+(D_{ o }-ct)^{ 2 } } } \right\} ^{ D_{ o }/c }_{ 0 } \)
\( x_{ d1 }=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o }c } \left\{ \cfrac { D_{ o } }{ d\sqrt{d^2+D^2_o} } \right\} \)
Deflection after being level with the electron,
\(r=\sqrt { d^{ 2 }_1+(ct)^{ 2 } } \)
where \(d_1=x_{d1}+d\)
\( sin(\theta )=\cfrac { d_1 }{ r } =\cfrac { d_1 }{ \sqrt { d^{ 2 }_1+(ct)^{ 2 } } } \)
\( F_{ h }=Fsin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }r^{ 2 } } sin(\theta )=\cfrac { q_{ p }q }{ 4\pi \varepsilon _{ o }(d^{ 2 }_1+(ct)^{ 2 }) } \cfrac { d_1 }{ \sqrt { d^{ 2 }_1+(ct)^{ 2 } } } \)
\( x_{ d }=\int _{ 0 }^{ t=D/c }{ \cfrac { F_{ h } }{ m_{ p } } } dt=\int _{ 0 }^{ t=D/c }{ \cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } }d_1 (d^{ 2 }_1+c^{ 2 }t^{ 2 })^{ -3/2 } } dt=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o } } \left\{ \cfrac { t }{ d_1\sqrt { d^{ 2 }_1+c^{ 2 }t^{ 2 } } } \right\} ^{ D/c }_{ 0 }\)
\( x_{ d2}=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o }c } \left\{ \cfrac { D }{ d_1\sqrt { d^{ 2 }_1+D^{ 2 } } } \right\} \)
\(d_1=x_{d1}+d\)
\( x_{ d2}=\cfrac { q_{ p }q }{ 4\pi m_{ p }\varepsilon _{ o }c } \left\{ \cfrac { D }{ (x_{d1}+d)\sqrt { (x_{d1}+d)^{ 2 }+D^{ 2 } } } \right\} \)
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,
\(OP=\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } } -(D_{ o }+D)\)
For destructive interference,
\(\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } } -(D_{ o }+D)=\cfrac { (2n+1)\lambda }{ 2 } \)
where \(n=0, 1, 2, 3...\)
\(\cfrac{\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } }}{D_{ o }+D} -1=\cfrac { (2n+1)\lambda }{ 2(D_{ o }+D) } \)
\(\cfrac{\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } }}{D_{ o }+D} =\cfrac {(2n+1)\lambda }{ 2(D_{ o }+D) } +1\)
\(\cfrac{1}{cos(\theta)}=\cfrac { (2n+1)\lambda }{ 2(D_{ o }+D) } +1\)
\(cos(\theta )=\cfrac { 2(D_{ o }+D) }{ (2n+1)\lambda +2(D_{ o }+D) } \)
\(\cfrac { 2(D_{ o }+D) }{ (2n+1)\lambda +2(D_{ o }+D) }<1\)
we know that \(\theta\) has solutions.
when \(n\) is small,
\(\cfrac { 2(D_{ o }+D) }{ (2n+1)\lambda +2(D_{ o }+D) }\approx 1\), \(\theta=0\)
For constructive interference,
\(\sqrt { (D_{ o }+D)^{ 2 }+(x_{ d1 }+x_{ d2 })^{ 2 } } -(D_{ o }+D)=n\lambda \)
\(n=0, 1, 2, 3,...\)
\(\cfrac{1}{cos(\theta)}=\cfrac{n\lambda}{D_{ o }+D}+1 \)
\(cos(\theta)=\cfrac{D_{ o }+D}{n\lambda+D_{ o }+D}\)
when \(n\) is small,
\(\cfrac{D_{ o }+D}{n\lambda+D_{ o }+D}\approx 1\) \(\theta=0\)
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.
