Vågrörelselära och optik

Transcription

1 Vågrörelselära och optik Kapitel 36 - Diffraktion 1 Vågrörelselära och optik Kurslitteratur: University Physics by Young & Friedman Harmonisk oscillator: Kapitel Mekaniska vågor: Kapitel Ljud och hörande: Kapitel Elektromagnetiska vågor: Kapitel 32.1 & 32.3 & 32.4 Ljusets natur: Kapitel & 33.7 Stråloptik: Kapitel Interferens: Kapitel Diffraktion: Kapitel &

2 Vågrörelselära och optik kap 14 kap kap 15 kap kap 16 kap kap kap 33 kap 34 kap 34 kap kap 35 kap 36 kap 36 3 What is diffraction? 4

3 5 6

4 Interference: Double slit experiment : single slit experiment 7 Fresnel diffraction or near-field diffraction. Frauenhofer diffraction or far-field diffraction. The lines to the screen are assumed to be parallel 8

5 Huygen s principle Each point in a wavefront is regarded as a new source of secondary wavelets. All the combined circles (wavelets) from all the points add up to create the new wavefronts. 9 10

6 Interference from many points in a slit d sin(θ) = mλ - destructive interference m = 1, 2, 3... d: width of the slit 11 For every point in the top half of the slit there is a corresponding point in the bottom half. 12

7 Geometry: Destructive Interference: Small angles: 13 Bright bands: Dark bands: 14

8 15 Problem solving 16

9 17 Intensity for single slit diffraction 18

10 The intensity of light (I) is proportional to the square of the amplitude of the total electric field (E p ) So what is E p? 19 Assume many small phasors with a total length E 0 are giving the total electric field strength (E p ) β is the phase difference between a ray at the top and bottom of the slit. 20

11 Step 1 Step 2 Find x! Find r! 180 o -β 180 o - β + 90 o + X + 90 o = 360 o X = β 21 Step 1 & 2 r = E 0 / β Step 3 Triangle: 22

12 Intensity But what is β? 23 r 2 r 1 A path difference of one wavelength corresponds to a phase difference of 2π Path difference: r 2 r 1 = a sin(θ) r 2 -r 1 is the path difference between a ray at the top and bottom of the slit. 24

13 25 Summary where 26

14 Destructive interference: Intensity is minimum Constructive interference: Intensity is maximum β = 6π β = 4π β = 2π β = 0 β = -2π β = -4π β = -6π This gives again: Gives maximum (and minimum) But the resulting equation has no analytical solution. The peaks are close but not exactly at β = 3π, 5π, 7π If the width of the slit is equal or smaller than λ then only one broad maximum is observed. A broader slit makes a narrower centre peak. 28

15 Intensity: 29 Problem solving 30

16 λ = 633 nm x = 6.00 m a = 0.24 mm y = 3.00 mm 5 x 10-4 = sin(θ) = β = 66 rad θ = 7.0 o 32

17 Two broad slits 33 In the analysis of interference from two slits it was assumed that they were very narrow. What if they are broad? Two narrow slits: One broad slit: Two broad slits: where 34

18 Two narrow slits: One broad slit: Two broad slits:

19 Multiple slits 37 The path difference between adjacent slits that gives maximum intensity with many slits is always: 2 slits 8 slits N-2 small peaks 38

20 N = 2 N = 8 N = 16 N-1 minima Principal maxima: 39 N = 8 mimima for φ = k 2π / N φ = 0 φ = 2π where k = 1, 2...N-1 40

21 41 In diffraction grating one uses devices with thousands of slits or reflecting surfaces. This gives very narrow principal maximum that can be used to determine the wavelength of light. Transmission grating Reflection grating 42

22 Problem solving 43 Grating: 1000 slits per mm 1st order maximum at 24 o What is λ? with d = 1 mm / 1000 slits = 10-6 m θ = 24 o λ = d sin(θ) = 10-6 sin(24 o ) = x 10-6 = 407 nm 44

23 Spectrometers 45 Spectrometer for astronomy Light incident on a grating is dispursed into a spectrum. The angles of deviations of the maxima are measured to calculate the wave length. 46

24 The ESO Very Large Telescope (VLT) in Chile The XSHOOTER spectrometer in the VLT ESO: European Southern Observatory 47 Light from the VLT 48

25 Chromatic resolving power: The minimum wavelength difference (Δλ) that can be distinguished by a spectrograph. R is higher for many slits and higher orders! 49 Pinhole diffraction 50

26 51 limits the angular resolution of optical intruments. 52

27 Rayleigh s criterion: Two point objects can be resolved by an optical system if their angular separation is larger than θ 1 where The limit for two objects to be resolved is when the center of one diffraction pattern is in the first minimum of the other. D 53 Resolution or resolving power: Minimum separation of two objects that can be resolved by an instrument. Higher resolution for larger optical aperture and shorter wavelenght. 54

28 DVD players 55 The principle of a DVD player: Destructive interference 56

29 57 58