Wave Interference and Diffraction

1. Wave Interference and Diffraction

2. Learning Goals • Explain the polarization of light waves by the following methods: • Selective Absorption • Reflection • Scattering

3. Learning Goals • Know what conditions are necessary for interference to be observed • Know how light changes phase when it is reflected from a surface • Understand the necessary conditions for interference to occur in thin films.

4. Polarization of Light Waves • Light waves are electromagnetic waves • E-field wave (electric) • B-field wave (magnetic) • Light waves are emitted by the vibration atoms or molecules of the light source • the result is an unpolarized light wave

5. Polarization of Light Waves • Three methods of polarizing light • Selective Absorption • Reflection • Scattering

6. Polarization of Light Waves • Selective Absorption • Polaroid Material ( chains of hydrocarbons ) • The direction perpendicular to the molecular chains is the Transmission Axes

7. Polarization of Light Waves • Polarization by Reflection • If incident Angle is 0o or 90o the reflected beam is unpolarized • If the incident angle is between 0o & 90o, the light is polarized to some extent • There exists an angle when the light will be completely polarized

8. Polarization of Light Waves • Polarization by Reflection • Common when sunlight is reflected by water, snow, or glass • If the surface is horizontal, the reflected light has a strong horizontal component. • The transmission axes is vertical for sunglasses

9. Polarization of Light Waves • Polarization by Scattering • Occurs when light is incident on a system of particles • The electrons in the medium can absorb and reradiate part of the light. • The absorption and reradiation of light by the medium is what causes sunlight to be polarized

10. Polarizing Light Intensity: If = ½ Ii

11. Malus’ Law If = Ii cos2θ

12. Concept Test 1 • Vertically polarized light with an initial intensity of 515 W/m2 passes through a polarizer oriented at an angle θ to the vertical. Find the final transmitted intensity of the light for (a) θ = 10º (b) θ = 90º

13. Concept Test 1 Discussion • Vertically polarized light with an initial intensity of 515 W/m2 passes through a polarizer oriented at an angle θ to the vertical. Find the final transmitted intensity of the light for (a) θ = 10º If=(515 W/m2)(cos 10º) = 499 W/m2 (b) θ = 90º If = 0 W/m2

14. Concept Test 2 Consider a set of three polarizers, as shown below. #1 has a vertical transmission axis, and #3 has a horizontal transmission axis. Taken together, #1 and #3 are crossed polarizers, from which no light would be expected to transmit. #2 is set at 45º and placed in between the other two polarizers. A beam of unpolarized light shines on #1 from the left. Is any light transmitted through the three polarizers or is transmission completely blocked? Be prepared to explain.

15. Polarizing Sunglasses

16. Scattering causes polarization • Molecules can naturally polarize light by scattering. • Scattering is a general physical process where a moving particle is forced to deviate from a straight trajectory by one or more paths due to localized non-uniformities in the medium through which they pass.

17. Effect of scattering on sunlight An observer looking at a right angle to the direction of the Sun will see more blue light than red. The observer looking directly toward the Sun will see more unpolarized light, appearing more red. The light is scattered by molecules in Earth’s atmosphere.

18. Reflection can also cause polarization Notice how the sky appears darker in the reflected surface of the lake. A smooth reflecting surface can also act as a polarizing filter.

19. Concept Test 3 • Why is the sky blue? Give your answer in one summary sentence.

20. Conditions for Light Wave Interference • We saw what occurred with the interference of sound waves (constructive or destructive, resulting in changes to wave amplitude) There is an analogy in light. • The sources must be coherent; that is, they must maintain a constant phase with respect to each other • The sources must have identical wavelengths (be monochromatic) • The superposition principle must apply

21. Interference in all EM waves • Constructive interference: path length difference = mλ where m = 1, 2, …. • Destructive interference: path length difference = (m+1/2) λ where m = 1, 2, ….

22. Two radio antennas transmitting the same signal • Midway between the 2 antennas, at point M, waves travel the same distance, hence they interfere constructively. At point P, the distance l2 is greater than the distance of l1 by one wavelength. So P is also a point of constructive interference. At point Q, the distance is different by ½ wavelength, so we observe destructive interference. A1 A2

23. Concept Test 4 • Two friends tune radios to the same f and pick up a signal transmitted simultaneously from a pair of antennas. Friend M receives a strong signal; Friend Q receives a very weak signal. Find the wavelength of the radio waves if d = 3.75 km, L = 14.0 km, and y = 1.88 km in the figure below. Assume Q is the first point of minimum signal as one moves away from M. (This means that the path-length difference at Q is half a wavelength.) A1 l1 d Q y M d l2 A2 L

24. Concept Test 4 Discussion • Strategy: Use Pythagorean Thm to find path length differences, then solve for λ. • Check for reality. Is this a reasonable λ for a radio wave?

25. Change of Phase due to Reflection • 180o phase change • Occurs when a ray is reflected from a medium that has a higher index of refraction • No phase change • Occurs when a ray is reflected from a medium with a lower index of refraction

26. Change of Phase due to Reflection 180o phase change No Phase Change N=1 N=1.33 t N=1

27. Learning Objectives • Explain how Young’s double slit experiment gave credibility to the wave theory of light • Know how Young’s double slit apparatus created areas of constructive and destructive interference • Explain why interference is observed from a single slit and diffraction grating.

28. Young’s Double-Slit Experiment • Double slit experiment of 1801 • Demonstrated constructive and destructive interference of light • Gave credibility to the wave theory of light

29. Huygen’s Principle Waves spread out when they pass through an opening in an obstruction. Every point on a wave front acts like a point source for new waves.

30. Young’s Double Slit Experiment Produces Interference Pattern

31. Young’s Double Slit Experiment • Interference occurs because of the path difference, d, the light waves travel • Constructive interference occurs when: d sin q = ml (m = 0, + 1, + 2, … ) Ybright = mLl/d (m = 0, + 1, + 2, … ) • Destructive interference occurs when: d sin q = (m + ½) l (m = 0, + 1, + 2, … ) Linear Distance from Central Bright Fringe: y = L tan θ

32. Reproducing Young’s Experiment http://phet.colorado.edu/en/simulation/wave-interference

33. Interference in Thin Films • Examine the following: • 180o phase change if n2>n1 • No phase change if n2< n1 • ln = l/n • ln is the wavelength of light in the medium • l is the wavelength of light in a vacuum • n is the index of refraction of the medium

34. Interference in Thin Films • Condition for constructive interference • 2nt = (m + ½)l • n is the index of refraction • t = the thickness of the film • Condition for destructive interference • 2nt = ml where m = 0, 1, 2, 3, …

35. Interference in Thin Films • Two factors that influence interference: • phase reversals on reflection • differences in travel distance • Only holds true when the medium above and below the film is the same • What happens if the medium above and below are not the same?

36. Interference in Thin Films • Newton’s Rings • Result from constructive and destructive interference • Used to test optical lenses • Supports the wave theory of light

37. Diffraction • Diffraction is the divergence of light from its initial line of travel and occurs when waves: • Pass through small openings • Travel around obstacles • By sharp edges

38. Single Slit Diffraction • Diffraction pattern consists of the following: • Central maximum • Secondary maximum • Minima • Each portion of the slit acts as a source of waves

39. Single Slit Diffraction

40. Single Slit Diffraction • Condition for destructive interference • sin qdark= m( l/a ) • m = +1, +2, +3, … • a is the slit width • Constructive interference • Occur halfway between the dark fringes • Central bright fringe is twice as wide as the weaker maximum

41. Diffraction Grating • Consists of many equally spaced parallel slits • Each slit produces a diffraction

42. Diffraction Grating • Condition for maximum constructive interference: • d sin qbright = ml ( m = 0, 1, 2, 3, … ) • m is called the order number of the diffraction pattern