Interference of Light

Transcription

1 Interference of Light

2 Review: Principle of Superposition When two or more waves interact they interfere. Wave interference is governed by the principle of superposition. The superposition principle says that when two, or more waves pass through the same point, the displacement is the arithmetic sum of the individual displacements. In the figure below, (a) exhibits destructive interference and (b) exhibits constructive interference.

3 These figures show the sum of two waves. In (a) they add constructively; in (b) they add destructively; and in (c) they add partially destructively. Contructive interference Destructive interference

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5 Diffraction The amount of diffraction depends on the size of the obstacle compared to the wavelength. If the obstacle is much smaller than the wavelength, the wave is barely affected (a). If the object is comparable to, or larger than, the wavelength, diffraction is much more significant (b, c, d).

6 When waves encounter an obstacle, they bend around it, leaving a shadow region. This is called diffraction.

7 Huygens Principle and Diffraction Huygens principle: Every point on a wave front acts as a point source; the wavefront as it develops is tangent to their envelope. Chapter 24 of text

8 Huygens principle is consistent with diffraction:

9 Young s Double-Slit Experiment If light is a wave, interference effects will be seen, where one part of wavefront can interact with another part. One way to study this is to do a double-slit experiment:

10 If light is a wave, there should be an interference pattern.

11 The interference occurs because each point on the screen is not the same distance from both slits. Depending on the path length difference, the wave can interfere constructively (bright spot) or destructively (dark spot).

12 We can use geometry to find the conditions for constructive and destructive interference: (24-2a) (24-2b)

13 Between the maxima and the minima, the interference varies smoothly.

14 Example: A parallel beam of light from a He-Ne laser, with a wavelength 656 nm, falls on two very narrow slits mm apart. How far apart are the fringes in the center of the pattern on a screen 3.6 m away?

15 Solution:

16 Since the position of the maxima (except the central one) depends on wavelength, the firstand higher-order fringes contain a spectrum of colors.

17 The Visible Spectrum and Dispersion Wavelengths of visible light: 400 nm to 750 nm Shorter wavelengths are ultraviolet; longer are infrared

18 This variation in refractive index is why a prism will split visible light into a rainbow of colors.

19 Actual rainbows are created by dispersion in tiny drops of water.

20 Diffraction Grating A diffraction grating consists of a large number of equally spaced narrow slits or lines. A transmission grating has slits, while a reflection grating has lines that reflect light. The more lines or slits there are, the narrower the peaks.

21 The maxima of the diffraction pattern are defined by. Which happens to be identical to the relationship in Young s Experiment.

22 Example: How many lines per centimeter must a grating have if there is to be no second-order spectrum for any visible wavelength?

23 Solution: Because the angle increases with wavelength, to miss a complete order we use the smallest wavelength.

24 Diffraction by a Single Slit or Disk Light will also diffract around a single slit or obstacle.

25 The resulting pattern of light and dark stripes is called a diffraction pattern. This pattern arises because different points along a slit create wavelets that interfere with each other just as a double slit would. D is the width of the single slit or disk.

26 The minima (destructive interference) of the single-slit diffraction pattern occur when: Do not confuse this relationship to that of Young s experiment and diffraction gratings. D is the slit width, not the distance between sources. Below is the condition for maxima (constructive interference). Dsinθ = m λ m = 1,2,3 m=0 is not a solution. See graph on previous page.

27 Example: Answer: (B)The spacing of the minima doesn t depend on the distance between the source and slit. It does depend on the width of the slit and the distance from the slit to the screen( and wavelength). There would be more minima (therefore more bright fringes) within a given angle for larger a larger slit width according to the relationship m = Dsinθ λ

28 The Spectrometer and Spectroscopy A spectrometer makes accurate measurements of wavelengths using a diffraction grating or prism.

29 The wavelength can be determined to high accuracy by measuring the angle at which the light is diffracted. Atoms and molecules can be identified when they are in a thin gas through their characteristic emission lines.

30 Michelson Interferometer The Michelson interferometer is centered around a beam splitter, which transmits about half the light hitting it and reflects the rest. It can be a very sensitive measure of length. It was most famously used in his experiment to detect an ether.

31 Thin Film Interference If there is a very thin film of material a few wavelengths thick light will reflect from both the bottom and the top of the layer, causing interference. This can be seen in soap bubbles and oil slicks, for example

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34 The wavelength of the light will be different in the oil and the air. At the boundaries there will be a transmitted and reflected wave. The reflected wave at A and the reflected wave at B will interfere. What is the condition for constructive interference for these reflected waves?

35 The path difference (PD) is just the exta distance one light ray travels. If the light is close to normally incident PD would equal twice the thickness of the material. Therefore: PD=2x If the two light rays are in phase, the condition for constructive interference would be: What is the condition for PD=2x=mλ destructive interference?

36 There is an important characteristic of reflected light before we can conclusively state the condition for constructive Interference. It turns out that when a light wave reflects of a second medium with a higher index of refraction compared to the first medium, will undergo a half wavelength shift(phase shift). If one of the rays undergoes a phase shift then the conditions for constructive interference, and destructive interference, are interchanged.

37 If both rays undergo a phase change, there is no relative shift, so the normal conditions for constructive and destructive interference apply. The other important point is that the wavelength stated in the path difference is the wavelength of light in the given material (not necessarily air).

38 The conditions for constructive and destructive interference can be stated as: Constructive Interference PD = 2x = mλ n Destructive Interference λ n = λ air n PD = 2x = (m )λ n * Remember possibility of relative phase change

39 A similar effect takes place when a shallowly curved piece of glass is placed on a flat one. When viewed from above, concentric circles appear that are called Newton s rings.

40 One can also create a thin film of air by creating a wedge-shaped gap between two pieces of glass.

41 Problem Solving: Interference 1. Interference occurs when two or more waves arrive simultaneously at the same point in space. 2. Constructive interference occurs when the waves are in phase. 3. Destructive interference occurs when the waves are out of phase. 4. An extra half-wavelength shift occurs when light reflects from a medium with higher refractive index.

42 Example: A soap bubble (n= 1.35) appears green (λ=540 nm) at its point nearest the viewer. What is its minimum thickness of the soap film? Solution: There is a relative phase change. Therefore the condition for constructive interference is: air soap film air 2x = (m )λ n, λ n = λ air n The minimum thickness occurs when m=0. solving for x gives: x= 100 nm

43 Example: What is the minimum thickness of an optical coating of MgF2 whose index of refraction is n=1.38 and is designed to eliminate wavelengths created at 550 nm when incident normally on glass for which n=1.50? Solution: There is no relative phase change, therefore the condition for destructive interference is: 2x = (m )λ n, λ n = λ air n Solving for x gives, x= 99.6 nm n=1.38 n=1.5