Chapter XVIII Diffraction of light

Chia sẻ: Thanh An | Ngày: | Loại File: PDF | Số trang:38

lượt xem

Chapter XVIII Diffraction of light

Mô tả tài liệu
  Download Vui lòng tải xuống để xem tài liệu đầy đủ

“Diffraction” of light can be understood as any deviation of light rays from their geometrical propagation line (that is straight in a homogeneuos material) An example (shown in the picture):The edge of shadow is never perfectly sharp. Some light appears in the geometrical shadow, and there are dark and light fringers in the area of illumination.

Chủ đề:

Nội dung Text: Chapter XVIII Diffraction of light

  1. GENERAL PHYSICS III Optics & Quantum Physics
  2. Chapter XVIII Diffraction of light §1. Fresnel diffraction §2. Fraunhofer diffraction §3. X-ray diffraction from crystals
  3. “Diffraction” of light can be understood as any deviation of light rays from their geometrical propagation line (that is straight in a homogeneuos material) Area of illuminaton An example (shown in the picture): The edge of shadow is never Point source perfectly sharp. Some light appears in the geometrical shadow, and there are dark and light fringers in the area of Straight Geometrical edge illumination. shadow  Diffraction of light can be considered as an argument for wave characteristics of light, like other wave processes (sound, etc.)
  4. We consider separatly two kinds of diffraction: Near-field diffraction → Fresnel diffraction Far-field diffraction → Fraunhofer diffraction §1. Fresnel diffraction: 1.1 Diffraction through a circular aperture: According to geometric optics, S the image of a circular aperture in the screen must be a light circular dick with a perfectly sharp edge. But it’s not so. Screen with a Screen circular aperture
  5. a b For analyzing this phenomenon, we use the Huyghen-Fresnel principle. 1.2 Huygen-Fresnel principle: For analyzing properties of wave processes, Huygen introduced the concept of wave front, and the rule how to draw a wave front from known sharp of at some former time
  6.  Huygen’s principle (1678): Wavefront at Wavefront at All points on wavefront are point t=0 time t sources for spherical secondary wavelets with speed, frequency that equal to initial wave. The wave front at a later time is the envelope of these wavelets. Basing on Huygen’s principle one can interpret diffraction as interference of light from secondary sources. For example, every point of circular aperture becomes a secondary source, and what we see in the screen is the interference of secondary sources. But this interpretation is only qualitively, for a quantitive analyze, we need more. It has not been known from Huygen’s principle: how can determine the amplitude and the phase of secondary waves ?
  7.  Fresnel’s complementary statement:  Fresnel states that for the vibration at P due to waves from the secondary dS source dS we have the following formula: Observation point where Wave front  0 & ( + a t  ) → the amplitude &  phase of vibration of secondary sources at dS on the wave front S  → a coefficient which depends on the angle K decreases K  as  increases; K = 0 when =  2 / The total vibration at P is
  8. 1.3 Analysis of diffraction through circular aperture: Having Huygen-Fresnel principle we tend to analyze the phenomenon of diffraction through a circular aperture. 1.3.1. The Frsenel method to devide a spherical wave front into adjacent zones (Fresnel zones): 1-th zone 2-nd zone 3-rd zone 4-th zone
  9. Calculate the area of the m-th zone: where Sm is the area of the m-th spherical segment: (hm – the hight of segment) hm is defined from the following equation: 
  10. Remark: The area of a zone does not depend on m. It means that the areas of zones are approximatly the same (for values of m that are not large). We have also the formula for the radius of the m-zone: rm is proportional to From the Fresnel formula for dξand all the described properties of Fresnel zones we can lead to the following formula for the amplitudes of vibrations at P which are sent from Fresnel spherical zones : from the 1-st zone ,,, … from the m-th zone
  11. Further, the phases of vibrations from two adjacent zones have the phase difference  we can write for the total amplitude of  → vibrations at P: It equals a half of amplitude due to the 1-st zone ! 1.3.2 Come back to the experiment of diffraction through an aperture: → What happens if there is a screen with circular aperture in the light propagation line ?
  12. Screen with aperture y y Screen Suppose that the part of wave front based on the aperture incorporates m zones: + for odd m - for even m (odd m) (even m)
  13. Conclusions of diffraction picture on the screen: Depending on the size of aperture, the number of open zones m is odd or even: • If m is odd → at the center point P there is a light spot • If m is even → at the center point P there is a dark spot Besides the center point P, at other points in the screen, the light intensity has maxima or minima, depending on the distance from the center point P. Owing to the symmetry, light & dark fringers on the screen are circles centered at P.
  14. a) b) c) Comparison the Fresnel zone pictures according to a) the axis SP; b) the axis SP’; c) the axis SP” To understand why the light intensity is different at P, P’ & P” we describe the Fresnel zones for three cases, and make a comparison. Here we suppose that the aperture exposes 3 zones.
  15. §2. Fraunhofer diffraction: We have studied an example of Fresnel diffraction. In Fresnel (or near-field) diffraction, we deal with spherical wave (wave front is spherical). Now we consider Fraunhofer (or far-field) diffraction that concerns plane waves. An example is diffraction of a beam of prallel rays on a single slit, or multiple slits. 2.1 Single-slit Diffraction: Consider a plane light wave that comes to a slit. Suppose that slit is very long (or its length is much larger the width) 2.1.1 Diffraction minima: First we define the location of minima in the screen
  16. a   sin   P  2 min 2 a/2 Incident Wave  (wavelength ) y   sin   min a • At this angle the light from the top and the middle of the slit a  destructively interfere. Below two O these points we can find corresponding pairs from which two L waves destructively interfere at P.  The second minimum is at an angle such that the light from the top and a point at a/4 destructively interfere: Location of nth-minimum: a  n   sin   sin  n  a/4 4 min,2 2 min, (n =±1, ±2,…) a 2   sin   min,2 a
  17. n • For small → n-th minimum corresponds to   , ( n  1, ,...) n  2 a and the corresponding y- coordinates in the screen are n yn L , ( n  1, ,...)  2 a  a a L • Note that for n = 0, that is  = 0, it is the point O in the screen,  y Light from entire slit arrives at O in phase. Here we have a bright fringer with the width 2 (twice in width than other bright fringers). /a 2.1.2 Intensity in the single-slit pattern: By using the phasor diagram method, we can have a detail analysis
  18.  To analyze diffraction, we treat it as  = a sina a Screen interference of light from strips (far away)  Model the single slit as M strips with  spacing between the strips of a/M. a  There are phase differences between waves coming from adjacent strips P  The phase difference between first and last source is given by L L >> a implies rays are     (asin   a  =     parallel. • Taking M = 10, we draw the phasor diagrams A0 Vibration vectors at O AP A0  The amplitude at O (in phase) and at P
  19.  As we let M   the polygon becomes , the arc of a circle.    The radius of the circle is determined A by the relation between angle and R Ao arc length:  Ao/R. = 2  a  Trigonometry: A/2 = R sin(/2) With R = Ao/, A = (2Ao/ /2) sin( sin( / 2) A A0 / 2 Intensity is related to amplitude: I = A2. So, here’s the final answer:   a sin  /   )  2 2  / 2)  sin( sin ( I I 0   / 2  I 0  ( sin  /     a     )  Remember:   (   sin  a 2 a   2   
  20. I • The graph of the intensity versus I0 sinis shown in picture • Intensity maxima: Consider the obtained equation as a function I = I ( we can find ) -2a –a   a 2a   sin maxima by solving dI/d 0. = This is a transcendental equation that has to be solved numerically. Here is the results:
Đồng bộ tài khoản