Fresnel zones of a planar wavefront

[JulienB]

Homework Statement

Hi everybody! I'd like to ask you guys some details about the following problem:

Determine the area of the l-th Fresnel zone of a planar wavefront considered from a point P. Give the numerical result for the 1st Fresnel zone for a wavelength $\lambda=600$nm and a distance wavefront-P of $0.5$m. How big is the error, if the term proportional to $\lambda^2$ would be removed?
Hint: use the result from the previous problem by changing it from a spherical to a planar wavefront.

Homework Equations

The previous problem mentioned in the hint was about determining the area of the l-th Fresnel zone for a spherical wave. Integration gave me the result $A_l = \frac{\lambda \pi \rho}{\rho + r_0} \big(r_0 + \frac{(2l - 1) \lambda}{4}\big)$ and I concluded that the ratio $A_l/r_l$ is independent of l (where $r_l$ is the average distance to P from the Fresnel zone).

The Attempt at a Solution

I am stuck right at the beginning because I am not really sure how to "change" $A_l$ for a planar wavefront situation. When I look at the elements of:

$A_l = \frac{\lambda \pi \rho}{\rho + r_0} \big(r_0 + \frac{(2l - 1) \lambda}{4}\big)$,

I see the following:

- the denominator $\rho + r_0$ refers to the distance between the source and the observation point. This stays the same with a planar wavefront I believe;
- the $\pi$ element comes from having a "ring" on the sphere. But I think this element also stays since the Fresnel zones are still rings for a planar wavefront (right?);
- $\rho$ was constant for a spherical wave, and it remains constant for a planar wavefront if we consider it to be the distance between the wavefront and the source (a plane). However the distance (say $x$) to a theoretical source point (so that the distance between S on the source plane and P would be minimal) would not be constant anymore. Using the law of cosines as I did for the previous problem, I find the expression $r^2 = x^2 + (\rho + r_0)^2 - 2 x (\rho + r_o) \cos \varphi$ but it does not really take me anywhere I'm afraid.

I must be thinking something wrong, or misunderstanding the "hint". Do you guys have another hint?

Thank you in advance for your answer, I'm looking forward to reading them.Julien.

 

[JulienB]

One idea I just had would be to take the expression $\frac{A_l}{r_l} = \frac{\lambda \pi \rho}{\rho + r_0}$ which should hold for a planar wavefront since $r_l = r_0 + \frac{(2l-1) \lambda}{4}$ is independent of $\rho$. Then the rest of the work would be to find an expression for $\rho$ corresponding to a planar wavefront, in which I am unsuccessful so far. Does that make sense though?

 

[Charles Link]

Hello again. The Fresnel zone theory is basically used with plane waves in a plane that can be used to compute the intensity at an on-axis point P. (It considers each point in the original plane as a Huygen's source and diffraction theory is used to compute the intensity at point P=again, usually on-axis.) The Fresnel zones (circular rings) have places where the interference is constructive, (phase between $-\pi/2$ and $+ \pi/2$),and regions where it is destructive.(You can look at it as a set of striped rings=white for "constructive" and dark for "destructive").( Constructive or destructive is relative to the phase of the center zone). Buy using only the constructive regions, something called a zone plate can be constructed (by blocking out the regions that contribute to destructive interference) to boost the signal at the observation point. The phasor diagram that you can construct for the individual zones will show this in more detail. $\\$ Alternatively, you can block out the center region and just allow the first outer dark ring or even the entire rest of the pattern to be open to produce Poisson's bright spot=I believe Poisson originally used this as an argument against the diffraction theory. He did not expect the bright spot to appear in the shadow of the obstacle.

 

[JulienB]

Hi again @Charles Link and thank you for answering me again! I've actually seen that interpretation of the Fresnel zones (circular rings on a flat plane) several times while searching on the internet. However the surface I considered in the previous problem was curved. I've attached a picture to make every step clearer (and maybe my confusion as well!).

Note: in the planar section $\rho$ refers to the radius of the circle.

 

[Charles Link]

I think the planar case is actually simpler. The phasor diagram for each of the zones is then precisely a semi-circle, apart from an obliquity factor (e.g. $cos(\theta)$ is often used.) $\\$ editing: Each area element $dA$ contributes an arc length $ds$ to the phasor diagram. When $d \phi / ds$ is constant, it becomes a circle with $d \phi/ds=1/r$. I calculated this a few years ago, but I remember the result and it should be easy enough to repeat the calculation, if necessary.

 

[JulienB]

On this page (http://www2.physics.ox.ac.uk/sites/default/files/optics-yr2-5-6.pdf, page 10), the surface element seems to be $dS = 2 \pi r_0 dr$ and after integration I'd get:

$A = 2 \pi r_0 \int_{r_0 + \frac{l-1}{2} \lambda}^{r_0 + \frac{l}{2} \lambda} dr \\ = 2 \pi r_0 (r_0 + \frac{l}{2} \lambda - r_0 - \frac{l-1}{2} \lambda) \\ = \lambda \pi r_0$

That seems a little too simple though.

Note: I've been starting this post before your last post, so I'm going to try with $\theta$ now. Does it refer to the angle between the optical axis and $r$ in my drawings?Julien.

 

[Charles Link]

Please read my edited version of post #5.

 

[Charles Link]

The obliquity factor causes the otherwise circular phasor diagram to spiral inward with a final value equal to 1/2 that of the first positive zone. Yes, the $cos(\theta)$ is the angle from the source to the on-axis point. $\\$ editing... And yes, your calculation that the area of the zones is independent of zone number $l$ is what you could expect for the planar case. That's all consistent with $d \phi /ds$ and $d \phi/dA$ being constant.

 

[Charles Link]

@JulienB Please see edited version of post #8.

 

[JulienB]

I've performed a new calculation which includes 2nd order terms (and l) which is (maybe) both coherent with my previous answer and with the problem (which mentions a 2nd order term). I'm now getting:

$A_l = \pi \lambda \big(r_0 + \frac{(2l-1) \lambda}{4} \big)$

I've attached a picture detailing my calculations and with a picture. I haven't used the obliquity factor though. Do you think it is correct?

 

[Charles Link]

I managed to find my old calculations, so it won't even be necessary to regenerate anything. The integral at hand is basically $E= \int \frac{e^{-i k d }}{r} \, dA$. (I'm using $d$ in place of $r$ and on-axis it will be $d_o$. Now $E=e^{- i k d_o} \int \frac{e^{-i k(d-d_o)}}{d} \, dA$ and $\phi=- \frac{2 \pi}{\lambda}(d-d_o)=-\frac{2 \pi}{\lambda}d_o(\frac{1}{cos(\theta)}-1)$. $\\$ Now $ds=\frac{dA}{d}=\frac{dA}{d_o}cos(\theta)$. $\\$ (I'm just copying verbatim from my calculations of a few years ago.) $\\$ $\frac{d \phi}{ds}=(-\frac{2 \pi}{\lambda} d_o) \frac{(sin(\theta)/cos^2(\theta)) \, d \theta)}{(dA \, cos(\theta)/d_o)}$. (Numerator is $d \phi$, denominator is $ds$ ) . $\\$ But $dA=2 \pi r \, dr$ and $\theta=tan^{-1}(\frac{r}{d_o})$ . $\\$ $\frac{d \theta}{dr}=(\frac{1}{r^2+d_o^2})(\frac{1}{d_o})$. $\\$ $tan(\theta)=\frac{r}{d_o}$ and $cos^2(\theta)=\frac{d_o^2}{r^2+d^2}$. $\\$ Finally, I get $\\$ $\frac{d \phi}{ds}=-1/\lambda$.

 

[JulienB]

Thanks @Charles Link you're so dedicated! I am not sure I understand yet the relevance of the last line ($d\phi / ds = -1/\lambda$), I just started with Fresnel zones today actually. Did you happen to calculate a value of $A$ (the area) as well?

 

[Charles Link]

Thanks @Charles Link you're so dedicated! I am not sure I understand yet the relevance of the last line ($d\phi / ds = -1/\lambda$), I just started with Fresnel zones today actually. Did you happen to calculate a value of $A$ (the area) as well?

The $d \phi / ds$ is the phasor diagram curvature. $\phi$ is the angle relative to the horizontal of a contribution of the phasor diagram that computes the electric field $E$ in diffraction theory. Each area element $dA$ weighs equally with an arc length $ds$ (for uniform illumination of the plane). When $d \phi /ds$ is a constant, (in this case that constant is $-1/\lambda$), it means the phasor diagram will be a circle rather than a spiral, or some other shape. ( $d \phi /ds =1/R$ where $R$ is the radius of the circle of the phasor diagram.) $\\$ Most likely you have seen phasor diagrams for multiple equally spaced slits where you get a polygon for the phasor diagram as the $\phi$ increments equally in going from one slit to the next. You may also have seen the continuous phasor diagram (again a circle) for the case of the diffraction pattern of a single slit of finite width. ( $\phi=(\frac{2 \pi}{\lambda})x \, sin(\theta)$. Notice $d \phi/dx =constant$ ( $x$ is position along the slit)). $\\$ For the Fresnel zone theory, you have a continuous phasor diagram (not a polygon), as the radius $r$ out from the center varies continuously. Once again, it is the $dA$ that is proportional to the $ds$ of the phasor diagram, ie. $ds$ is the arc length of the curve. To get the resultant $E$ , you connect the origin to the endpoint.

 

[TSny]

Julien, Your result of post #10 looks good to me. How does this result compare to the spherical case in the limit ρ → ∞?

 

[JulienB]

@TSny Yes it is the same... I didn't really consider the planar wavefront as a spherical wavefront from far away, I've realized that in between. Thanks for pointing that out.Julien.

 

[Charles Link]

Thanks @Charles Link you're so dedicated! I am not sure I understand yet the relevance of the last line ($d\phi / ds = -1/\lambda$), I just started with Fresnel zones today actually. Did you happen to calculate a value of $A$ (the area) as well?

Since $dA= \frac{d_o}{cos(\theta)} \, ds$, and $\frac{d \phi}{ds}=-\frac{1}{\lambda}$,(so that $ds=- \lambda \, d \phi$), $A_o= d_o \lambda \pi$. (approximately=assumes $\theta=0$). $\\$ And, in fact, yes, the zones get slightly larger as you get farther away from center to exactly account for the $1/d$ falloff of the electric field intensity, making for a phasor diagram that is a precise circle when no additional obliquity factor is included.