The Goos–Hänchen effect, named after Hermann Fritz Gustav Goos (1883 – 1968) and Hilda Hänchen (1919 – 2013), was first theorized by Isaac Newton (1643 – 1727), and is an optical phenomenon in which a finite-width beam of light undergoes a small lateral shift when totally internally reflected. The shift arises because a bounded beam comprises a continuous distribution of plane wave components with differing wave vectors. The Fresnel reflection coefficients are both polarization and angle dependent, so each plane wave component acquires a different phase shift upon reflection. The superposition of these phase-shifted components displaces the reflected beam's centroid along the interface. The magnitude of the shift is small, and it depends on the beam's polarization state, the wavelength, and the angle of incidence. It is among the most studied non-specular reflection phenomena in optics. Acoustic analog of the Goos–Hänchen effect is known as Schoch displacement.
Description This effect occurs because the reflections of plane wave components of a finite-sized beam undergo different phase shifts. A finite-width beam can be expressed as a superposition of plane waves via a Fourier decomposition.
ψ ( x , z ) = 1 2 π ∫ − ∞ ∞ Φ ( k x ) e − i [ k x x + z ( k 1 2 − k x 2 ) ] d k x {\displaystyle \psi (x,z)={\frac {1}{\sqrt {2\pi }}}\int \limits _{-\infty }^{\infty }\Phi (k_{x})e^{-i[k_{x}x+z{\sqrt {(k_{1}^{2}-k_{x}^{2})}}]}dk_{x}}
where Φ ( k x ) {\displaystyle \Phi (k_{x})} is the angular spectrum of the beam. Each value of k x {\displaystyle k_{x}} represents a plane wave in the direction of k x {\displaystyle k_{x}} . Without loss of generality k 1 {\displaystyle k_{1}} lies in the x-z plane. Under total internal reflection, each plane-wave component reflects according to the Fresnel equations where | r ( k x ) | = 1 {\displaystyle |r(k_{x})|=1} but acquires a phase shift χ ( k x ) {\displaystyle \chi (k_{x})} where
χ ( k x ) = − 2 tan − 1 [ m ( k 1 2 − k x 2 ) 1 / 2 ( k x 2 − k 2 2 ) 1 / 2 ] {\displaystyle \chi (k_{x})=-2\tan ^{-1}\left[{\frac {m(k_{1}^{2}-k_{x}^{2})^{1/2}}{(k_{x}^{2}-k_{2}^{2})^{1/2}}}\right]}
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