Past exam of the laser science and quantum information processing course of the University of Oxford / 2018 / 2

Represent a paraxial ray by its height and forward slope $(x,\theta {)}^{\mathsf{T}}$, with $\theta =dx/dz$. Free-space propagation leaves the slope unchanged and increases height by $L\theta$, givingA thin lens leaves the ray height unchanged. A parallel incident ray at height $x$ must emerge toward the focal point, so its slope changes by $-x/f$. By linearity, the same increment applies to any paraxial incident slope:These ray transfer matrices have determinant one in air and compose in reverse matrix order to the physical order of traversal.

Use the forward-slope convention established in part a, with $x^{\prime }=Ax+B\theta$ and ${\theta }^{\prime }=Cx+D\theta$, where $AD-BC=1$. Assume $B\ne 0$. Solve for the endpoint slopes:Fermat's principle makes the optical path stationary against variations at internal interfaces. Consequently only endpoint terms survive when its endpoints change. For optical path $l(x,x^{\prime })$, these areIntegrate the first equation with respect to $x$, then match the second and impose $l(0,0)=L_0$. The resulting ABCD optical path generating function isThus the geometrical kernel defined in the source would have the formThe printed minus sign is inconsistent with forward propagation under these conventions. A concrete counterexample is free space: $A=D=1$, $B=L>0$, $C=0$. Its true geometrical path iswhereas the printed expression predicts a path shorter than $L$ for nonzero transverse separation. For example, $L=1\mathrm{m}$, $x=0$, $x^{\prime }=1\mathrm{mm}$ gives $1\mathrm{m}+0.5\mu \mathrm{m}$, rather than the printed $1\mathrm{m}-0.5\mu \mathrm{m}$. Reversing the slope convention would also reverse the free-space $B$ entry and must be done consistently; it cannot justify the minus sign with $P(L)$ from part a.

The printed form can be recovered by explicitly using the opposite slope coordinate $\tilde{\theta }=-dx/dz$. Then $\tilde{B}=-B$, free space has $\tilde{B}=-L$, and the same physical path is $l=L_0-(A{x}^2+D{x}^{\prime 2}-2x{x}^{\prime })/(2\tilde{B})$. Thus the sign can be a convention choice, but the printed minus sign must not be combined with the positive-$B$ free-space ray transfer matrix used here.

For physical field propagation, the paraxial diffraction kernel retains this quadratic optical-path phase while using the leading spreading amplitude. In one transverse dimension the normalized linear canonical transform iswith a consistent square-root branch and $\lambda =2\pi /k$. The source's prefactor and $1/l$ kernel are a schematic one-dimensional version of the Fresnel-Kirchhoff diffraction integral, not this normalized general-system propagator. The quadratic denominator is normally replaced by its leading value when making the Fresnel approximation. When $B=0$, propagation is an imaging limit rather than an ordinary integral with division by $B$.

For the shown sequence of free space, thin lens, and free space, multiply the ray transfer matrices in traversal order:To obtain an ordinary Fourier transform of an arbitrary input field, neither an input quadratic phase nor an output quadratic phase may remain. Thus $A=D=0$, requiringUsing the physically consistent kernel derived in part b givesThe spatial Fourier coordinate is $x^{\prime }/(\lambda f)$. If the source's schematic integral normalization is retained instead, set its slowly varying denominator to $L_0$ to obtain the same Fourier transform with overall prefactor $ik{e}^{ik{L}_0}/(2\pi L_0)$. Keeping the source's incorrect quadratic sign would reverse the Fourier coordinate. These overall-normalization conventions do not alter the focal-plane placement, but an unapproximated transverse-dependent denominator would not give an exact Fourier transform.