Tài liệu Chapter XVIII Diffraction of light pdf

GENERAL PHYSICS III GENERAL PHYSICS III Optics & Quantum Physics Chapter XVIII Chapter XVIII Diffraction of light Diffraction of light §1. Fresnel diffraction §2. Fraunhofer diffraction §3. X-ray diffraction from crystals  “Diffraction” of light can be understood as any deviation of light rays from their geometrical propagation line (that is straight in a homogeneuos material) Point source Area of illuminaton Geometrical shadow Straight edge 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.  Diffraction of light can be considered as an argument for wave characteristics of light, like other wave processes (sound, etc.) §1. Fresnel diffraction: We consider separatly two kinds of diffraction:  Near-field diffraction → Fresnel diffraction  Far-field diffraction → Fraunhofer diffraction S Screen Screen with a circular aperture 1.1 Diffraction through a circular aperture: According to geometric optics, 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. 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  Huygen’s principle (1678): All points on wavefront are point 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. Wavefront at t=0 Wavefront at time t 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 ?  Fresnel’s complementary statement: Observation point Wave front  Fresnel states that for the vibration at P due to waves from the secondary source dS we have the following formula: dS where  a 0 & (t +   ) → the amplitude & phase of vibration of secondary sources at dS on the wave front S  K → a coefficient which depends on the angle   K decreases as  increases; K = 0 when  = / 2  The total vibration at P is 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 Calculate the area of the m-th zone: where S m is the area of the m-th spherical segment: (h m – the hight of segment) h m is defined from the following equation:  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: r m 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 ,,, … [...]... 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... Diffraction gratings are excellent tools for studying visible light because the slit spacing is on the order of the wavelength of the light (~ few tenths of microns) Visible light is a very small part of the spectrum of electromagnetic waves How can we study e-m waves with smaller wavelengths (e.g., Xrays with ~ 10-10 m)? We can’t use a standard diffraction grating to do this Why? Calculate for first... seen here     Y (cm)   This one does it all Increased spacing between maxima and constant diffraction 2.3 Diffraction Gratings: Diffraction gratings rely on N-slit interference.They simply consist  of a large number of evenly spaced parallel slits Recall that the intensity pattern produced by light of wavelength  passing through N slits with spacing d is given by:  2 25I1 20 IN N=5 I where:... d  1  Notes: Diffraction gratings can be made using special techonology In applications there are two types of diffraction gratings: ’transmission grating’ and ‘reflection grating’ One of common example of reflection gratings is a compact disk (CD) The rainbow-color reflection are due Reflection grating to the reflection-grating effect §3.X-Ray Diffraction from Crystals:    Diffraction gratings... 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 ? Screen with aperture Screen y y Suppose that the part of. .. for odd m - for even m (odd m) (even m) 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... Angular splitting of the Sodium doublet Consider the two closely spaced spectral (yellow) lines of sodium (Na),  = 589 nm and  = 589.6 nm, mentioned earlier If light 1 2 from a sodium lamp illuminates a diffraction grating with 4000 slits/cm, what is the angular separation of these two lines in the second-order (m=2) spectrum? Solving: First find the slit spacing d from the number of slits per centimeter... 2.1.1 Diffraction minima: First we define the location of minima in the screen a/2   a   sin   min 2 2   sin   min a P Incident Wave (wavelength  ) • At this angle the light from the top and the middle of the slit destructively interfere Below two these points we can find corresponding pairs from which two waves destructively interfere at P The second minimum is at an angle such that the light. .. I1 is roughly constant    L The position of the first principal maximum is given by sin   = /d (can’t assume small   Different colors  different angles !) Width of the principal maximum varies as 1/N  one can improve ability to resolve closely spaced spectral lines By examples we will see how effective are diffraction gratings at resolving light of different wavelengths (i.e separating closely-spaced... - occurs when the maximum of  overlaps with  2 1 2 the first diffraction minimum of  (  min =   /Nd) 1  min  IN = N2I1  sin  d  d 0 /Nd       d  min     min    d Nd sin   N = number of slits in grating   d “Rayleigh Criterion”  min 1    N Larger N Smaller  min  (Higher spectral resolution)  We can squeeze more resolution out of a given grating by working . III Optics & Quantum Physics Chapter XVIII Chapter XVIII Diffraction of light Diffraction of light §1. Fresnel diffraction §2. Fraunhofer diffraction §3. X-ray diffraction. Fresnel diffraction: We consider separatly two kinds of diffraction:  Near-field diffraction → Fresnel diffraction  Far-field diffraction → Fraunhofer diffraction S Screen Screen