Optics 089 · Waveguides, structured light, and modern optics

Beam Shift Microscope

Why does a reflected light beam shift slightly, both within and out of the plane of incidence? A beam of finite width contains a small spread of angles, and the reflection phase changes with angle and polarization. That phase gradient nudges the reflected beam slightly within the plane of incidence, the Goos-Hänchen shift. Circular polarization also moves it sideways, out of that plane: the Imbert-Fedorov shift.

Reflect a finite Gaussian beam at a glass–air interface on a 3D bench. Each plane-wave component gets its own polarization basis and complex Fresnel coefficient; reverse the spin, see the spot built by the Fourier sum and freeze complex detector fields.

Reflected-frame longitudinal centroid

\langle X_r\rangle

$0.50$

Transverse spin-dependent centroid

\langle Y_r\rangle

$50\%$

Integrated reflected flux

R

$0.00\pi$

Model regime $\text{valid model regime}$

Model checked against analytic results

Beam Shift Microscope

Internal angle from the normal

\theta

50^{\circ}

Incident amplitude waist radius

w_0

8\,\mu\mathrm m

Vacuum wavelength; fixed assumed indices

\lambda_0

633\,\mathrm{nm}

Polarization: zero s, one p, two positive spin, three negative spin

q

2

Detector distance along the reflected geometric axis

z_r

0\,z_R

A reflected beam does not land exactly where a ray would. Reverse its spin and the tiny sideways shift reverses too. Narrow the beam waist near the critical angle, then move the detector to tell a spatial shift apart from an angular deflection. Every spot is computed from the reflected vector angular spectrum.

Both media are lossless, nondispersive half-spaces, and the refractive index is fixed at 1.515. The input polarization is carried across the angular spectrum by minimal-rotation transport, and the Fourier window is finite and periodic. The intensity detector is ideal and paraxial, and axial distances in the apparatus are compressed. Weak-measurement amplification is not modeled, and no measured device is reproduced.

Physics tutorial

Audit the physics of Beam Shift Microscope

BackgroundA beam of finite width is a bundle of many plane waves. Each one meets the boundary at a slightly different angle, with its own polarization basis: $\widetilde{\mathbf E}_r(\mathbf q)=\mathsf R(\mathbf q)\widetilde{\mathbf E}_i(\mathbf q)$ . The reflected waves interfere into a spot that is shifted and sometimes distorted. A single geometric ray cannot predict where its centroid lands.

Why it mattersWhy does a reflected light beam shift slightly, both within and out of the plane of incidence?

Start with the essentials

Focus question: Why does a reflected light beam shift slightly, both within and out of the plane of incidence?
One-sentence intuition: Measure the shift from the field that the Lab actually builds: $\langle\mathbf r_\perp\rangle=\frac{\int\mathbf r_\perp|\mathbf E_r|^2\,d^2r_\perp}{\int|\mathbf E_r|^2\,d^2r_\perp}$ . Sliding an ideal Gaussian by a formula is not a finite-beam calculation. Reversing the spin reverses the transverse shift, while the boundary stays the same.

Core mathematical model

Coordinates and spin convention

\hat{\mathbf k}_i=(\sin\theta,0,\cos\theta),\quad\hat{\mathbf X}_r=(-\cos\theta,0,-\sin\theta),\quad(e_p,e_s)=\frac{(1,\pm i)}{\sqrt2}

The interface is the lab plane where the normal coordinate is zero, with glass below and air above. The angle is measured from the normal. For the reflected beam, the longitudinal direction lies across the beam within the plane of incidence, not along its travel. The transverse axis is the shared lab Y direction. Spin is defined by these incident Jones components, so it does not depend on the observer the way left and right labels do.

Building the reflected beam from its vector angular spectrum

\mathbf E_r(\mathbf r_\perp,z_r)=\int\mathsf R(\mathbf q)\mathbf e_i(\mathbf q)e^{-w_0^2|\mathbf q|^2/4}e^{i\mathbf q\cdot\mathbf r_\perp+i(k_z-k)z_r}\,d^2q

The incident Jones vector is carried from the central beam frame to each plane wave by the smallest rotation, which keeps the field transverse. Each wave is split into its own s and p components, reflected with complex Fresnel amplitudes and rotated into a common reflected frame. All three electric-field components are kept. The numerical Fourier synthesis uses a finite periodic window. Near the critical angle, the result comes from the sampled spectrum, not from the phase derivative, which diverges there.

Spatial shift versus angular deflection

\langle X_r\rangle(z_r)\simeq\langle X_r\rangle(0)+z_r\Theta_X,\quad z_R=\frac{\pi n_gw_0^2}{\lambda_0}

Well away from the critical region, total reflection gives no net angular deflection. Near or below the critical angle, the reflection magnitude filters the beam's finite angular spectrum, so the spot moves as the detector distance changes. The detector stands perpendicular to the geometric reflected axis. At zero distance it shows the outgoing field extrapolated back to the waist, not what a camera inside the interface would record.

The Artmann shift as a wide-beam limit

X_{{\rm A},s}=-\frac{2\sin\theta}{k\sqrt{\sin^2\theta-n_g^{-2}}},\quad k=\frac{2\pi n_g}{\lambda_0}

This is the s-polarized reference in the reflected beam's transverse coordinate. Its sign is opposite to a positive shift along the interface tangent. It assumes an angular spectrum narrow enough to lie entirely above the critical angle. A beam of finite width keeps the critical behaviour finite, and its intensity need not stay Gaussian. Weak postselection and amplification optics are not modelled.

Common difficulties

Does a sideways shift mean a photon follows a bent ray?

Typical misconceptionDrawing a slightly displaced ray, or a Gaussian shifted by hand, is enough to explain both effects.

Better mental modelNo. The shifts describe the centroids of interfering vector fields. The longitudinal shift comes from how the reflection phase varies across the spectrum; the transverse shift comes from the rotation of the polarization basis. Both calculations need the finite spectrum and a defined measurement plane.

Run the experiment

01
Scene 1: Longitudinal displacement
Compare s and p polarization well above the critical angle. Check the computed longitudinal centroid against the separate wide-beam Artmann references.
What to observe: Opposite circular inputs give transverse centroids of opposite sign and the same longitudinal centroid. Pure s or pure p light has no net transverse shift, because of reflection symmetry.
02
Scene 2: Reverse the spin
Select positive spin, then negative spin. The transverse centroid flips sign; the longitudinal centroid and the reflected flux stay the same. Read the shift from the centroid ruler, because the beam picture is exaggerated.
What to observe: If the spot moves as the detector distance changes, that can be angular filtering, not a change in the spatial shift at the interface. The centroid within the finite window and the fraction of light at the window edge show the limits of the numerical measurement.
03
Scene 3: Finite beam near critical incidence
Use the near-critical preset, change the waist, then move the detector. Because the beam mixes many angles, the spatial shift stays finite and an angular deflection appears. Capture the complex detector field before you change any control.
What to observe: The apparatus drawing is schematic and compressed along the axis. Detector images, field samples and numerical centroids are not exaggerated; only the separate ruler magnifies the centroid positions, on purpose.