Off-axis holography

Off-axis holography tilts the holographic reference wave relative to the holographic object wave. Its interference fringes acquire a carrier spatial frequency; the two cross terms occupy different spatial frequency regions when the carrier is sufficiently large compared with the object's signal bandwidth. This permits filtering a single image term.

Gabor's lecture credits Emmett Leith and Juris Upatnieks with the later skew-reference method using helium–neon laser illumination, separating reconstructed images angularly.^[ref]

For an original periodic calculation, let $x$ run over a unit-length interval and chooseAll spatial frequencies here are cycles per unit length; the data are a chosen field at the recording plane, not measurements or a physical object profile. Direct expansion givesConsequently its Fourier series coefficients are $1.05$ at zero, $0.02$ at $\pm 8$, $0.2$ at $\pm 17$ and $0.1$ at $\pm 25$. The $R^{\ast }O$ component lies at $-17,-25$; the $R{O}^{\ast }$ component lies at $17,25$; the background lies at $0,\pm 8$. Filtering the negative-frequency cross term and multiplying by $R$ recovers $O$, including its relative wave phase.

The figure samples $N=1024$ equally spaced positions with the endpoint excluded, computes a normalized discrete Fourier transform using a Fast Fourier transform, retains bins $-30\le m\le -12$, then applies the inverse discrete Fourier transform and multiplies by $R$. The recovered real and imaginary parts coincide with the chosen $O$ to numerical precision. This exact periodic example avoids leakage by placing every frequency on an integer bin; finite nonperiodic records need allowance for broadening and an aperture.

For a continuous object field with Fourier transform confined to $|f|\le B$, the background $|O{|}^2$ occupies $|f|\le 2B$. A carrier $f_c>3B$ is a sufficient separation condition for the sideband centred at $-f_c$, and $f_c+B$ must remain below the Nyquist spatial frequency. Here $B=5$, $f_c=20$ and the Nyquist spatial frequency is $512$. This sufficient bound describes a whole bandwidth, rather than exploiting the gaps in this particular two-component field.

The same example also makes optical playback explicit. Choose $t_0=0.3$ and $\beta =0.01$. Since $|O|\le 0.3$, the bounds $0.49\le I\le 1.69$ imply $0.3049\le t\le 0.3169$, and $|\beta I|/t_0\le 0.0564$. The affine response is stipulated; its weak-modulation interpretation is approximate. In $Rt$, the desired field $0.01O$ has frequencies $3,-5$; the conjugate term $0.01R^2{O}^{\ast }$ has frequencies $37,45$; the background has frequencies $20,12,28$. Thus the same carrier that separates the recorded cross terms separates the playback components. Reillumination restores a field through diffraction; computational filtering extracts the corresponding encoded field through Fourier analysis.