Answer to Question #193454 in Optics for Anas

Question

For a wave that travels only in directions that have small angles with respect to the optical axis, the general from of the complex field may be approximated by

U(x, y, z) ≈ A(x, y, z)e^jkz

 where A(x, y, z) is a slowly varying function of z

(a) Show that for such a wave the Helmholtz equation can be reduced to ∇t^2 * A + j2k ∂A/∂z = 0

 where ∇t^2 * A = ∂^2/ ∂x62 + ∂^2/ ∂y^2 is the transverse portion of the Laplacian. This equation is known as the paraxial Helmholtz equation.

(b) Show that a solution to this equation is given by A(x, y, z) = A1/q(z) * (e^jk * ((x^2+y^2)/2*q(z)) for any complex q(z) having dq(z)/dz = 1.

Answer 1

(a)

from the given information

substitute $U(x,y,z)\approx A(x,y,z)e^{jkz}$ in to the Helmholtz equation

$(\Delta ^2 + k^2)U=0$

$[\frac{\eth^2}{\eth x^2}+\frac{\eth^2}{\eth y^2}+\frac{\eth^2}{\eth z^2}+ k^2]e^{jth}=0$

then

$[\frac{\eth^2}{\eth x^2}+\frac{\eth^2}{\eth y^2}]Ae^{jkz} + \frac{\eth}{\eth z}[ \frac{\eth}{\eth z}Ae^{jkz}+jkAe^{jkz}] + k^2Ae^{jkz}=0$

$\Delta^2_tAe^{jkz} + \frac{\eth^2A}{\eth z^2}e^{jkz} + 2jk \frac{\eth A}{\eth z}e^{jkz}+(jk)^2Ae^{jkz} +k^2Ae^{jkz}=0$

Dividing by $e^{jkz}$ on both sides

$\Delta^2_tAe^{jkz} + \frac{\eth^2A}{\eth z^2}e^{jkz} + 2jk \frac{\eth A}{\eth z}e^{jkz}+(jk)^2Ae^{jkz} +k^2Ae^{jkz}=0 \over e^{jkz}$

$\Delta^2_tA +2jk\frac{\eth A}{\eth z} + \frac{\eth^2 A}{\eth z^2} - k^2 + k^2 = 0$

$\Delta^2_tA +2jk\frac{\eth A}{\eth z} + \frac{\eth^2 A}{\eth z^2}=0$

the slowly varying function approximation for A implies that

$\frac{\eth^2}{\eth z^2}A<< 2jk \frac{\eth A}{\eth z}$

so by neglecting the terms $\frac{\eth ^2}{\eth z^2}A$

$\Delta^2A+2jk\frac{\eth A}{\eth z}=0....................(1)$

Hence proved

(b)

First we can evaluate a number of different derivatives

$A(x,y,z)= \frac{A_1}{q}e^{jk(x^2+y^2)\over2q(z)}$

$\frac{\eth}{\eth z}A(x,y,z)=-\frac{A_1}{q}\frac{dq}{dz}e^{jk(x^2+y^2)\over2q}-\frac{A_1}{q}{(jk(x^2+y^2)\over 2q^2}\frac{dq}{dz}e^{jk(x^2+y^2)\over2q}$

$\frac{\eth}{\eth z}A(x,y,z)=-(\frac{1}{q}+{(jk(x^2+y^2)\over 2q^2})\frac{dq}{dz}A(x,y,z)$

$\frac{\eth}{\eth x}A(x,y,z)=jk\frac{xA_1}{q^2}e^{jk(x^2+y^2)\over2q}$

$\frac{\eth^2}{\eth x^2}A(x,y,z)=jk\frac{xA_1}{q^2}e^{jk(x^2+y^2)\over2q} + (jk\frac{x}{q})^2.\frac{A_1}{q}e^{jk(x^2+y^2)\over2q}$

$\frac{\eth^2}{\eth x^2}A(x,y,z)=(jk\frac{1}{q}-k^2.\frac{x^2}{q^2})A(x,y,z)$

similarly

$\frac{\eth^2}{\eth y^2}A(x,y,z)=(jk\frac{1}{q}-k^2\frac{1}{q}-k^2\frac{y^2}{q^2})A(x,y,z)$

Now substitute the partial derivative of A into the paraxial Helmholz equation.

Noting that $\frac{dq}{dz}$ is equal to (1) equation

$\Delta ^2_tA+2jk\frac{\eth A}{\eth z}=(jk \frac{1}{q}-k^2\frac{x^2}{q^2}+jk\frac{1}{q}-k^2\frac{y^2}{q^2}-2jk\frac{1}{q}\frac{dq}{dz}+k^2{(x^2+y^2)\over q^2}\frac{dq}{dz})$

$\Delta ^2_tA+2jk\frac{\eth A}{\eth z}=0$

hence proved