- #1

JulienB

- 408

- 12

## 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.