Why Fresnel zones are needed

Fresnel zones are the areas on which the surface of a sound or light wave is broken for calculating the results of diffraction of sound or light. This method was first applied by O. Frenel in 1815.

Historical reference

Augustin Jean Fresnel (10.06.1788-14.07.1827) - French physicist. He devoted his life to studying the properties of physical optics. In 1811, under the influence of E. Malius, he began to study physics independently, soon he was carried away by experimental research in the field of optics. In 1814, "rediscovered" the principle of interference, and in 1816 supplemented the widely known Huygens principle, which introduced the idea of coherence and interference of elementary waves. In 1818, based on the work done, he developed a theory of diffraction of light. He introduced the practice of considering diffraction from the edge, as well as from a circular hole. Conducted experiments that later became classical, with biprism and biserkalami on interference of light. In 1821 he proved the fact of the transversality of light waves, in 1823 he discovered circular and elliptical polarizations of light. He explained on the basis of wave representations the chromatic polarization, as well as the rotation of the polarization plane of light and birefringence. In 1823 he established the laws of refraction and reflection of light on a fixed plane interface between two media. Along with Jung is considered the creator of wave optics. He is the inventor of a number of interference devices, such as Fresnel mirrors or Fresnel biprism. It is considered the founder of a fundamentally new way of lighting.

A bit of theory

Determine Fresnel zones can be either for diffraction with a hole of arbitrary shape, and generally without it. However, from the point of view of practical expediency, it is best to consider it on a circular hole. In this case, the light source and the observation point must be on a straight line that is perpendicular to the plane of the screen and passes through the center of the hole. In fact, Fresnel zones can break any surface through which light waves pass. For example, the surface of an equal phase. However, in this case it will be more convenient to break a flat hole into zones. To do this, consider an elementary optical problem that will allow us to determine not only the radius of the first Fresnel zone, but also the subsequent ones with arbitrary numbers.

The problem of determining the dimensions of rings

To begin with, it should be imagined that the surface of a flat hole is between the light source (point C) and the observer (point H). It is perpendicular to the line CH. The segment CH passes through the center of the circular hole (point O). Since our problem has an axis of symmetry, the Fresnel zones will have the form of rings. And the solution will be reduced to determining the radius of these circles with an arbitrary number (m). The maximum value is called the radius of the zone. To solve the problem it is necessary to make an additional construction, namely: to select an arbitrary point (A) in the plane of the hole and connect it by segments of straight lines with the observation point and with the light source. As a result, we obtain the triangle SAN. Then you can make it so that the light wave coming to the observer along the SAN path will go a long way than the one that will follow the CH path. From this it follows that the path difference CA + AN-CH determines the difference of the wave phases that passed from the secondary sources (A and O) to the observation point. From this value depends the resulting interference of waves from the position of the observer, and hence the light intensity at this point.

Calculation of the first radius

We get that if the path difference is equal to half the length of the light wave (λ / 2), then the light will come to the observer in the antiphase. Hence, we can conclude that if the path difference is less than λ / 2, then the light will come in the same phase. This condition CA + AN-CH≤ λ / 2 by definition is the condition that the point A is in the first ring, that is, this is the first Fresnel zone. In this case, for the boundary of this circle, the path difference will be equal to half the length of the light wave. So this equality allows us to determine the radius of the first zone, we denote it by P ₁ . With the path difference corresponding to λ / 2, it will be equal to the segment OA. In the case when the distances of CO significantly exceed the diameter of the hole (usually such variants are considered), then for geometric reasons the radius of the first zone is determined by the following formula: P ₁ = √ (λ * CO * OH) / (CO + OH).

Calculation of the radius of the Fresnel zone

The formulas for determining the subsequent values of the radii of the rings are identical to those discussed above, only the numerator of the number of the sought-for zone is added. In this case, the equality of the path difference will have the form: CA + AN-CH≤m * λ / 2 or CA + AN-CO-ON≤m * λ / 2. It follows that the radius of the desired zone with the number "m" determines the following formula: P _m = √ (m * λ * CO * OH) / (CO + OH) = P ₁ √ m

Summarizing intermediate results

It can be noted that splitting into zones is the separation of the secondary light source into sources having the same area, since P _m = π * P _m ² - π * P _m-1 ² = π * Р ₁ ² = П ₁ . Light from neighboring Fresnel zones comes in the opposite phase, since the path difference of the neighboring ring will by definition be equal to half the length of the light wave. Generalizing this result, we find that breaking a hole into circles (such that light from neighboring ones comes to an observer with a fixed phase difference) will mean breaking into rings with the same area. This assertion can easily be proved by means of the problem.

Fresnel zones for a plane wave

Consider the breakdown of the hole area into thinner rings of equal area. These circles are secondary sources of light. The amplitude of the light wave coming from each ring to the observer is approximately the same. In addition, the phase difference from the neighboring circle at the point H is also the same. In this case, the complex amplitudes at the observer's point, when added to a single complex plane, form part of a circle-an arc. The total amplitude is a chord. Now let us consider how the picture of summation of complex amplitudes changes in the case of a change in the radius of the hole, provided that the remaining parameters of the problem are preserved. In the event that the hole opens for the observer only one zone, the picture of addition will be represented by a part of the circle. The amplitude from the last ring will be rotated by an angle π with respect to the central part, since the path difference of the first zone, as defined, is λ / 2. This angle π will mean that the amplitudes are half the circumference. In this case, the sum of these values at the observation point will be zero - the zero length of the chord. If three rings are opened, then the picture will represent a half circle and so on. The amplitude at the observer's point for an even number of rings is zero. And in the case when an odd number of circles are used, it will be the maximum and equal to the value of the length of the diameter in the complex plane of amplitude addition. The above problems fully disclose the Fresnel zone method.

Briefly about special cases

Consider the rare conditions. Sometimes, when solving a problem, it is said that a fractional number of Fresnel zones is used. In this case, under the half of the ring is meant the quarter of the circle of the picture, which will correspond to half the area of the first zone. Similarly, any other fractional value is calculated. Sometimes the condition assumes that a certain fractional number of rings is closed, and so many are open. In this case, the total amplitude of the field is found as the vector difference of the amplitudes of the two problems. When all zones are open, that is, there are no obstacles in the path of light waves, the picture will look like a spiral. It is obtained because when opening a large number of rings, it is necessary to take into account the dependence of the light emitted by the secondary light source to the observer's point and the direction of the secondary source. We get that light from the zone with a large number has a small amplitude. The center of the obtained spiral is in the middle of the circle of the first and second rings. Therefore, the amplitude of the field, in the case when all the zones are open, is half that of the open one first circle, and the intensity differs by a factor of four.

Diffraction of light in the Fresnel zone

Let us consider what is meant by this term. Fresnel diffraction is a condition where several zones open at once through a hole. If many rings are open, then this parameter can be neglected, that is, we are in the approximation to geometric optics. In the case when the opening for the observer opens significantly less than one zone, this condition is called Fraunhofer diffraction. It is considered fulfilled if the light source and the observer's point are at a sufficient distance from the hole.

Comparison of a lens and a zone plate

If you close all odd or all even Fresnel zones, then at the observer's point there will be a light wave with a larger amplitude. Each ring gives half the circumference on the complex plane. So, if you keep odd zones open, then only halves of these circles will remain from the common spiral, which contribute to the total amplitude "from the bottom up". The obstruction to the passage of a light wave, in which only one type of ring is open, is called an area plate. The intensity of light at the observer's point will repeatedly exceed the intensity of light on the plate. This is because the light wave from each open ring hits the observer in the same phase.

A similar situation is observed with the focusing of light with a lens. Unlike the plate, it does not cover any rings, but shifts the light in phase by π * (+ 2 π * m) from those circles that are covered by the zone plate. As a result, the amplitude of the light wave doubles. Moreover, the lens eliminates the so-called mutual phase shifts that pass inside one ring. It unfolds on the complex plane half of the circle for each zone in a straight line segment. As a result, the amplitude increases by a factor of π, and the entire spiral on the complex plane is rotated by a lens into a straight line.