Considering a Hermite-Gaussian (HG) Beam as a perfectly plane wave and ignoring the phase due to curvature(exp(ik(x^2+y^2/2R)), the phase of lobes is pi phase shifted from its neighbor. When we consider the beam at any one plane (z=const), how the neighboring lobes may have a pi phase shift? what is term that contributing this phase shift?
I have simulated HG beam field and observed that, when i include phase due to curvature of wave front, the total phase map is having number of concentric circles and if i ignore this term i am getting a clear pattern. In both cases i can identify the pi phase shift between the lobes. what could be the possible reason for this??
If you look at the Hermite polynomials in amplitude, you will see that these will evaluate to values less than 0 (zero). Generally, this is regarding as a phase change of pi (180 degs) if the complex amplitude is expressed as a phasor as one does when using complex beam curvature notation.
I will try to make you understand in simple way..
Take Sin or cosine function as an example. the values oscillates between +1 and -1 .
Sin is positive for theta = 0 to 180,i.e 1st and 2nd quadrant. it is negative in 3rd and 4rth quadrant..i.e theta between 180 to 360. now , you can write, -1= exp(i*pi), 1=exp(i*0). now it is directly indicates the PI phase difference between +1, -1... it is the same way with Hermite polynomials , if the value alters between positive and negative , the phase difference of PI is obvious.
for your second question regarding the phase curvature. the phase curvature is added if the beam propagates to another plane Z=Z1, this phase curvature is the consequence of Fresnel propagation in free space. it is added to the beam cross section apart from the already existed PI phase difference between the lobes. now the values may not be exactly 0,pi but pi/3, pi+pi/3. for a particular concentric ring the total phase difference between the lobes is again pi.
Thank you for your explanation. But to my knowledge the phase curvature is transverse phase i.e., x y dependent. So it should also change in a plane from point to point. I was thinking that this transverse change of phase is creating circles. please see the patterns i got with and without phase curvature at a plane for HG01.
yes the phase curvature is transverse, the image that you got is due to more curvature you added. try to give less curvature..If you provide the entire equation i can suggest which parameter makes your pattern look better. without curvature the pattern is correct.
example exp(-i*(r.^2)/2) and exp(-i*(r.^2)/4) will have different curvatures. its just like lens equation.
TRY this CODE in MATLAB to see the effect of curvature on image
close all
clear all
L=321;
N=321;
dx=L/N;
[x1,y1] = meshgrid((-N/2 : N/2-1)*(L/N));
[th,r]=cart2pol(x1,y1);
z1=cos(th).*exp(-i*r.^2/20);
z2=cos(th).*exp(-i*r.^2/50);
z3=cos(th).*exp(-i*r.^2/150);
figure(),imshow(angle(z1),[])
figure(),imshow(angle(z2),[])
figure(),imshow(angle(z3),[])
Thank you sir. Now i understood.
exp(-i*k*((x(i)^2+y(j)^2)/2*R_z(l))). I am using this expression for phase curvature in HG field eqn. Here R_z is the wave front curvature which has usual meaning. Basically it depends on waist of the beam. I should take care of the beam waist then. Thank you so much sir.
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