Modal approximation for the electromagnetic field of a near-field optical probe

Robert D. Grober; Todd Rutherford; Timothy D. Harris

doi:10.1364/AO.35.003488

Modal approximation for the electromagnetic field of a near-field optical probe

Robert D. Grober, Todd Rutherford, and Timothy D. Harris

Applied Optics
Vol. 35,
Issue 19,
pp. 3488-3495
(1996)
•https://doi.org/10.1364/AO.35.003488

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Robert D. Grober, Todd Rutherford, and Timothy D. Harris, "Modal approximation for the electromagnetic field of a near-field optical probe," Appl. Opt. 35, 3488-3495 (1996)

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History
- Original Manuscript: September 21, 1995
- Revised Manuscript: December 7, 1995
- Published: July 1, 1996

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Abstract

A formalism is given in which the optical field generated by a near-field optical aperture is described as an analytic expansion over a complete set of optical modes. This vectoral solution preserves the divergent behavior of the near field and the dipolar nature of the far field. Numerical calculation of the fields requires only evaluation of a well behaved, one-dimensional integral. The formalism is directly applicable to experiments in near-field scanning optical microscopy when relatively flat samples are evaluated.

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Transverse Magnetic Field Modes		Transverse Electric Field Modes
$E_{ρ} = i κ \frac{\partial ψ}{\partial ρ}$	$B_{ρ} = - \frac{i k_{0}}{ρ} \frac{\partial ψ}{\partial θ}$	$E_{ρ} = \frac{i k_{0}}{ρ} \frac{\partial ψ}{\partial θ}$	$B_{ρ} = i κ \frac{\partial ψ}{\partial ρ}$
$E_{θ} = \frac{i κ}{ρ} \frac{\partial ψ}{\partial θ}$	$B_{θ} = i k_{0} \frac{\partial ψ}{\partial ρ}$	$E_{θ} = - i k_{0} \frac{\partial ψ}{\partial ρ}$	$B_{θ} = \frac{i κ}{ρ} \frac{\partial ψ}{\partial θ}$
E_z = k ²ψ	B_z = 0	E_z = 0	B_z = k ²ψ
ψ _n,k (ρ, θ, z > 0, t) = exp(inθ)J_n (kρ)exp(iκz − iωt)

Electric Field Components	Magnetic Field Components
ρ/a = ξ < 1
$E_{ρ} = - \frac{8 i k_{0} a}{3 π} \frac{(1 - 0.5 ξ^{2})}{{(1 - ξ^{2})}^{1 / 2}} cos θ$	B _ρ = sin θ
$E_{θ} = - \frac{8 i k_{0} a}{3 π} {(1 - ξ^{2})}^{1 / 2} sin θ$	B _θ = cos θ
E_z = 0	$B_{z} = - \frac{4}{π} \frac{ξ}{{(1 - ζ^{2})}^{1 / 2}} sin θ$
ρ/a = ξ > 1
E _ρ = 0	$B_{ρ} = \frac{2}{π} [arccsc ξ - \frac{(1 + ξ^{2})}{ξ^{2} {(ξ^{2} - 1)}^{1 / 2}}] sin θ$
E _θ = 0	$B_{θ} = \frac{2}{π} [arccsc ξ - \frac{{(ξ^{2} - 1)}^{1 / 2}}{ξ}] cos θ$
$E_{z} = \frac{4 i k_{0} a}{3 π} \frac{1}{ξ {(ξ^{2} - 1)}^{1 / 2}} cos θ$	B_z = 0

Transverse magnetic field modes ^b
$E_{ρ} (k, ρ, θ) = i κ cos θ \frac{\partial J_{1} (k ρ)}{k \partial ρ}$	$B_{ρ} (k, ρ, θ) = i k_{0} sin θ \frac{J_{1} (k ρ)}{k ρ}$
$E_{θ} (k, ρ, θ) = - i κ sin θ \frac{J_{1} (k ρ)}{k ρ}$	$B_{θ} (k, ρ, θ) = i k_{0} cos θ \frac{\partial J_{1} (k ρ)}{k \partial ρ}$
E_z (k, ρ, θ) = k cos θJ ₁(kρ)	B_z (k, ρ, θ) = 0
Transverse electric field modes ^b
$E_{ρ} (k, ρ, θ) = - k_{0} cos θ \frac{J_{1} (k ρ)}{k ρ}$	$B_{ρ} (k, ρ, θ) = - κ sin θ \frac{\partial J_{1} (k ρ)}{k \partial ρ}$
$E_{θ} (k, ρ, θ) = k_{0} sin θ \frac{\partial J_{1} (k ρ)}{k \partial ρ}$	$B_{θ} (k, ρ, θ) = - κ cos θ \frac{J_{1} (k ρ)}{k ρ}$
E_z (k, ρ, θ) = 0	B_z (k, ρ, θ) = ik sin θJ ₁(kρ)

Transverse Magnetic Field Modes		Transverse Electric Field Modes
$E_{ρ} = i κ \frac{\partial ψ}{\partial ρ}$	$B_{ρ} = - \frac{i k_{0}}{ρ} \frac{\partial ψ}{\partial θ}$	$E_{ρ} = \frac{i k_{0}}{ρ} \frac{\partial ψ}{\partial θ}$	$B_{ρ} = i κ \frac{\partial ψ}{\partial ρ}$
$E_{θ} = \frac{i κ}{ρ} \frac{\partial ψ}{\partial θ}$	$B_{θ} = i k_{0} \frac{\partial ψ}{\partial ρ}$	$E_{θ} = - i k_{0} \frac{\partial ψ}{\partial ρ}$	$B_{θ} = \frac{i κ}{ρ} \frac{\partial ψ}{\partial θ}$
E_z = k ²ψ	B_z = 0	E_z = 0	B_z = k ²ψ
ψ _n,k (ρ, θ, z > 0, t) = exp(inθ)J_n (kρ)exp(iκz − iωt)

Electric Field Components	Magnetic Field Components
ρ/a = ξ < 1
$E_{ρ} = - \frac{8 i k_{0} a}{3 π} \frac{(1 - 0.5 ξ^{2})}{{(1 - ξ^{2})}^{1 / 2}} cos θ$	B _ρ = sin θ
$E_{θ} = - \frac{8 i k_{0} a}{3 π} {(1 - ξ^{2})}^{1 / 2} sin θ$	B _θ = cos θ
E_z = 0	$B_{z} = - \frac{4}{π} \frac{ξ}{{(1 - ζ^{2})}^{1 / 2}} sin θ$
ρ/a = ξ > 1
E _ρ = 0	$B_{ρ} = \frac{2}{π} [arccsc ξ - \frac{(1 + ξ^{2})}{ξ^{2} {(ξ^{2} - 1)}^{1 / 2}}] sin θ$
E _θ = 0	$B_{θ} = \frac{2}{π} [arccsc ξ - \frac{{(ξ^{2} - 1)}^{1 / 2}}{ξ}] cos θ$
$E_{z} = \frac{4 i k_{0} a}{3 π} \frac{1}{ξ {(ξ^{2} - 1)}^{1 / 2}} cos θ$	B_z = 0

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