280 likes | 308 Views
Explore diffraction and interference phenomena in waves with topics such as single-slit diffraction, intensity distribution, and the Young's Double Slit Experiment. Learn about phasors, constructive and destructive interference, and the geometric interpretation behind diffraction patterns. Discover the relationship between wave behavior and particles.
E N D
PHY 102: Waves & Quanta Topic 7 Diffraction John Cockburn (j.cockburn@... Room E15)
Interference re-cap • Phasors • Single slit diffraction • Intensity distribution for single slit
Electromagnetic Waves Where E0and B0 are related by: E0 = cB0 INTENSITY of an EM wave E02 NB. we will see later that EM radiation sometimes behaves like a stream of particles (Photons) rather than a wave………………
Interference First, consider case for sound waves, emitted by 2 loudspeakers: Path difference =nλ Constructive Interference Path difference =(n+1/2)λ Destructive Interference (n = any integer, m = odd integer)
Young’s Double Slit Experiment • Demonstrates wave nature of light • Each slit S1 and S2 acts as a separate source of coherent light (like the loudspeakers for sound waves)
Young’s Double Slit Experiment Constructive interference: Destructive interference:
Young’s Double Slit Experiment Y-position of bright fringe on screen: ym = Rtanm Small , ie r1, r2 ≈ R, so tan ≈ sin So, get bright fringe when: (small only)
Young’s Double Slit Experiment: Intensity Distribution For some general point P, the 2 arriving waves will have a path difference which is some fraction of a wavelength. This corresponds to a difference in the phases of the electric field oscillations arriving at P:
Young’s Double Slit Experiment: Intensity Distribution Total Electric field at point P: Trig. Identity: With = (t + ), = t, get:
So, ETOT has an “oscillating” amplitude: Since intensity is proportional to amplitude squared: Or, since I0E02, and proportionality constant the same in both cases:
Young’s Double Slit Experiment: Intensity Distribution
Light 2-slit intensity distribution: “phasor” treatment • Remember from Lecture 1, harmonic oscillation with amplitude A and angular frequency can be represented as projection on x or y axis of a rotating vector (phasor) of magnitude (length) A rotating about origin. • We can use this concept to add oscillations with the same frequency, but different phase constant by “freezing” this rotation in time and treating the 2 oscillations as fixed vectors…… • So called “phasor method”
2-slit intensity distribution: “phasor” treatment Use phasor diagram to do the addition E1 + E2 Using cosine rule:
Single Slit Diffraction “geometrical” picture breaks down when slit width becomes comparable with wavelength
Single Slit Diffraction observed for all types of wave motion eg water waves in ripple tank
Single Slit Diffraction • Explain/analyse by treating the single slit as a linear array of coherent point sources that interfere with one another (Huygen’s principle)……………………. All “straight ahead” wavelets in phase → central bright maximum Fraunhofer (“far-field”) case Destructive interference of light from sources within slit for certain angles
Single Slit Diffraction • From diagram, can see that for slit of width A, we will get destructive interference (dark band on screen) at angles which satisfy…..: Choice of a/2 and a/4 in diagram is entirely arbitrary, so in general we have a dark band whenever; (m=±1, ±2, ±3………..)
Position of dark fringes in single-slit diffraction If, like the 2-slit treatment we assume small angles, sin ≈ tan =ymin/R, then Positions of intensity MINIMA of diffraction pattern on screen, measured from central position. Very similar to expression derived for 2-slit experiment: But remember, in this case ym are positions of MAXIMA In interference pattern
Intensity distribution Width of central maximum • We can define the width of the central maximum to be the distance between the m = +1 minimum and the m=-1 minimum: Ie, the narrower the slit, the more the diffraction pattern “spreads out” image of diffraction pattern
Single-slit diffraction: intensity distribution To calculate this, we treat the slit as a continuous array of infinitesimal sources: Can be done algebraically, but more nicely with phasors………………..
Single-slit diffraction: intensity distribution E0 is E-field amplitude at central maximum = total phase difference for “wavelets” from top and bottom of slit
Single-slit diffraction: intensity distribution How is related to our slit/screen setup? Path difference between light rays from top and bottom of slit is From earlier (2-slit)