Vågrörelselära och optik Kapitel 35 - Interferens 1 Vågrörelselära och optik Kurslitteratur: University Physics by Young & Friedman Harmonisk oscillator: Kapitel 14.1 14.4 Mekaniska vågor: Kapitel 15.1 15.8 Ljud och hörande: Kapitel 16.1 16.9 Elektromagnetiska vågor: Kapitel 32.1 & 32.3 & 32.4 Ljusets natur: Kapitel 33.1 33.4 & 33.7 Stråloptik: Kapitel 34.1 34.8 Interferens: Kapitel 35.1 35.5 Diffraktion: Kapitel 36.1-36.5 & 36.7 2
Vågrörelselära och optik kap 14 kap 14+15 kap 15 kap 15+16 kap 16 kap 16+32 kap 32+33 kap 33 kap 34 kap 34 kap 34+35 kap 35 kap 36 kap 36 3 What is interference? 4
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: Wave overlap in space Coherent sources: Same frequency (or wavelength) and constant phase relationship (not necessarily in phase). 7 Contructive interference Destructive interference 8
Contructive interference Antinodal curves = Contructive interference Destructive interference 9 Black: Amplitude = zero Red: Amplitude > 0 Blue: Amplitude < 0 http://www.opensourcephysics.org/items/detail.cfm?id=9989 10
Contructive Destructive 11 Geometry: m=-1 m=0 m=1 m=2 m=3 R y y Contructive interference: 12
Constructive interference Destructive interference Longer λ Longer d 13 Problem solving 14
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Intensity of the light after interference 17 Power per unit area: Intensity = the average value of S The average of cos 2 (x) = 1/2 How to eliminate μ 0 ε 0 μ 0 = 1 / c 2 μ 0 = 1 / ε 0 c 2 Intensity of an electromagnetic wave: I 18
Intensity of an electromagnetic wave: where E max is the amplitude of the electric field Strategy: Calculate the amplitude of the electric field after the superposition of two interfering waves. Use phasors to calculate this new E max. Put the new E max into the formula: Derive a relationship between intensity and d, y and R. 19 Phasors 20
Phasors for a cos- and sin-function 21 Phasor for a harmonic oscillation 22
By adding phasors as vectors one can obtain the combined wave from two waves with the same frequency that are out of phase 23 By adding phasors as vectors one can obtain the combined wave from two waves with different frequency that are out of phase 24
same frequency different phase different frequency different phase 25 Combining two electric fields by using phasors to get the amplitude 26
Combine two electric fields with 1. The same amplitude E 2. The same frequency ω 3. Different phase - φ 27 Trigonometry b c a 28
Find this amplitude (E p ) E p E π φ E Step 1 Step 2 E p2 = 2E 2 (1+cos(φ)) 29 Step3 E p2 = 2E 2 (1+cos(φ)) E p2 = 2E 2 (1 + 2cos 2 (φ/2) - 1) E p2 = 4E 2 cos 2 (φ/2) 30
Amplitude of the waves after interference: Intensity of the waves after interference: The intensity of light (I) is proportional to the square of the amplitude of the electric field (E p ) where is the maximum intensity. 31 Relationship between intensity and d, y and R 32
E p 2E π 2π φ Conclusions: Constructive interference occur when the phase difference is 2π Destructive interference occur when the phase difference is π 33 Path difference A path difference of one wavelength corresponds to a phase difference of 2π 34
Introduce y in the formula θ y small θ 35 m=-1 Intensity: m=0 m=1 m=2 m=3 y 36
Summary Contructive interference: Intensity: 37 Problem solving 38
y = 700 tan(4.0 o ) = 48.9 m R=700m d = 10.0 m θ = 4.0 o 39 Thin-film interference 40
41 Different colours have different wavelengths so some will interfer constructively and other destructively. 42
Reflected Amplitude Incoming Amplitude Reflections for θ = 0 Positive if n a > n b Negative if n b > n a No phase shift Phase shift with π 43 If we have one reflection with phase shift we get the following: Constructive reflections: Destructive reflections: Phase shift with π This is opposite to what we normally have without a phase shift. 44
Problem solving 45 Destructive reflections: x t L h 46
Newton s rings 47 48
Newton s rings can be used to study the surface of lenses to a very high precision. Between each dark ring the distance t has changed with one half wavelength. Destructive reflections: 49 Non-reflecting coating 50
Non-reflecting film Film thickness: λ film /4 Film refractive index: n film < n glass λ air Destructive interference = no reflections λ film λ = ν / f n > 1 λ 0 = c / f n = 1 n = c / ν = λ 0 / λ The wavelength in the film has to be a quarter of the film thickness. This is not the same wavelength as that of the incoming light but it can be easily calculated with: λ film = λ air / n film 51 Problem solving 52
λ film = λ air / n film Film thickness = λ/4 = 400 / 100 = 100 nm 53 The Michelson Interferometer 54
The Michelson Interferometer y The compensator plate compensates for this The observer will see an interference pattern with rings. The fringes in the pattern will move when the mirror is moved. The number of fringes (m) can be used to calculate y or λ 55 56