College Physics B - PHY2054C

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1 Young College - PHY2054C Wave Optics: 10/29/2014 My Office Hours: Tuesday 10:00 AM - Noon 206 Keen Building

2 Outline Young Young 4 5

3 Assume a thin soap film rests on a flat glass surface. Young

4 Young Assume a thin soap film rests on a flat glass surface. The upper surface of the soap film is similar to the beam splitter in the interferometer: It reflects part of the incoming light and allows the rest to be transmitted into the soap layer after refraction at the air-soap interface. The transmitted ray is partially reflected at the bottom surface. The two outgoing rays meet the conditions for interference: 1 Travel through different regions 2 Recombination 3 Coherence

5 Young Assume a thin soap film rests on a flat glass surface. Index of refraction of the film also needs to be accounted for: The wavelength changes as the light wave travels from a vacuum into the film: λ film = v f = c/n film f = λ vac n film λ air n film Number of extra wavelengths: N = 2d λ film = 2d λ/n film

6 Young Frequency of Wave at Interface When a light wave passes from one medium to another, the waves must stay in phase at the interface. The frequency must be the same on both sides of the interface.

7 Young Frequency of Wave at Interface 1 There is a phase change whenever the index of refraction on the incident side is less than the index of refraction of the opposite side (wave is inverted). 2 If the index of refraction is larger on the incident side the reflected ray in not inverted and there is no phase change.

8 Young Phase Changes in a Thin Film The total phase change in a thin film must be accounted for: The phase difference due to the extra distance traveled by the ray. Any phase change due to reflection. For a soap film on glass: n air < n film < n glass

9 Young Phase Changes in a Thin Film There are phase changes for both reflections at the soap film interfaces: B The reflections at both the top and bottom surfaces undergo a 180 phase change: 1 If the number of extra cycles, N, is an integer, there is constructive interference: 2d = mλ/n film 2 If the number of extra cycles is a half-integer, there is destructive interference: 2d = (m + 1/2)λ/n film

10 Young Phase Changes in a Thin Film Assume the soap bubble is surrounded by air. C There is a phase change at the top of the bubble. There is no phase change at the bottom of the bubble: 1 If the number of extra cycles, N, is a half-integer, there is constructive interference: 2d = (m + 1/2)λ/n film 2 If the number of extra cycles is an integer, there is destructive interference: 2d = mλ/n film

11 Young Phase Changes in a Thin Film Assume the soap bubble is surrounded by air. C There is a phase change at the top of the bubble. There is no phase change at the bottom of the bubble: 1 If the number of extra cycles, N, is a half-integer, there is constructive interference: 2d = (m + 1/2)λ/n film 2 If the number of extra cycles is an integer, there is destructive interference: 2d = mλ/n film

12 Outline Young Young 4 5

13 Young Light Through a Single Slit Light passes through a slit or opening and illuminates a screen: As the width of the slit becomes closer to the wavelength of the light, the intensity pattern on the screen and additional maxima become noticeable.

14 Young All types of waves undergo single-slit diffraction: Water waves have a wavelength easily visible. The outgoing wave front is approximately spherical. is the bending or spreading of a wave when it passes through an opening. Wave :

15 Young Huygens s Principle Christiaan Huygens (14 April July 1695) Huygens s Principle: All points on a wave front can be thought of as new sources of spherical waves.

16 Outline Young Young 4 5

17 Young Light passes through two very narrow slits: When the two slits are both very narrow, each slit acts as a simple point source of new waves. The outgoing waves from each slit are like simple spherical waves. The double slit experiment showed conclusively that light is a wave. Why? X X X X X X X?

18 Young Question 1 What will you observe on the screen in an experiment using light incident on two slits? A Nothing, the screen will be dark. B A complicated pattern with numerous bright lines. C Two bright lines corresponding to the number of slits. D No changes compared to a single slit. X X X X X X X?

19 Young Question 1 What will you observe on the screen in an experiment using light incident on two slits? A Nothing, the screen will be dark. B A complicated pattern with numerous bright lines. C Two bright lines corresponding to the number of slits. D No changes compared to a single slit.

20 Young Light passes through two very narrow slits: When the two slits are both very narrow, each slit acts as a simple point source of new waves. The outgoing waves from each slit are like simple spherical waves. The double slit experiment showed conclusively that light is a wave (observation of interference).

21 Young Question 2 What will you observe on the screen in an experiment using a beam of particles incident on two slits? A Nothing, the screen will be dark. B A complicated pattern with numerous bright lines. C Two bright lines corresponding to the number of slits. D No changes compared to a single slit. X X X X X X X?

22 Young Young s was first carried out by T. Young around 1800: Light is incident onto two slits and after passing through them strikes a screen. satisfies the general requirements for interference: 1 The interfering waves travel through different regions of space as they travel through different slits. 2 The waves come together at a common point on the screen where they interfere. 3 The waves are coherent because they come from the same source. will determine how the intensity of light on the screen varies with position. Quantum Wave : phet.colorado.edu/en/simulation/quantum-wave-interference

23 Young Determine the path length between each slit and the screen. Assume W is very large. If the slits are separated by a distance d, then the difference in length between the paths of the two rays is: L = d sin θ Bright fringe: d sin θ = m λ m = 0, ± 1, ± 2,... Dark fringe: d sin θ = (m )λ m = 0, ± 1, ± 2,...

24 Young Intensity Pattern The angle θ varies as you move along the screen: Each bright fringe corresponds to a different value of m. Negative values of m indicate that the path to those points on the screen from the lower slit is shorter than the path from the upper slit.

25 For m = 1: sin θ = λ/d Since the angle is very small: Example Young tan θ = θ and sin θ = θ and θ = λ/d Between m = 0 and m = 1: h = W tan θ = W λ/d

26 For m = 1: sin θ = λ/d Since the angle is very small: Example Young tan θ = θ and sin θ = θ and θ = λ/d Between m = 0 and m = 1: h = W tan θ = W λ/d

27 For m = 1: sin θ = λ/d Since the angle is very small: Example Young tan θ = θ and sin θ = θ and θ = λ/d Between m = 0 and m = 1: h = W tan θ = W λ/d Suppose the slits are d = 0.1 mm apart and the screen is located W = 50 cm from the slits. For light with a wavelength of 630 nm: h = W λ d = (0.5 m) m m = 3.2 mm This fringe spacing is large enough to be seen easily by the naked eye. This gives also a way to measure the wavelength of light.

28 Young Question 3 A Young s double-slit experiment is performed in air and then the apparatus is submerged in water. What happens to the fringe separation (d sin θ = m λ), and what can be used to explain the change, if any? A The separation stays the as it is the same experiment independent of the medium. B The separation decreases because the frequency of the light decreases in the water. C The separation increases because the wavelength of the light increases in the water. D The separation decreases because the wavelength of the light decreases in the water.

29 Young Question 3 A Young s double-slit experiment is performed in air and then the apparatus is submerged in water. What happens to the fringe separation (d sin θ = m λ), and what can be used to explain the change, if any? A The separation stays the as it is the same experiment independent of the medium. B The separation decreases because the frequency of the light decreases in the water. C The separation increases because the wavelength of the light increases in the water. D The separation decreases because the wavelength of the light decreases in the water.

30 Outline Young Young 4 5

31 Young Slits may be narrow enough to exhibit diffraction but not so narrow that they can be treated as a single point source of waves:

32 Young Assume the single slit has a width, w. Light is diffracted as it passes through the slit and then propagates to the screen:

33 Young Assume the single slit has a width, w. Light is diffracted as it passes through the slit and then propagates to the screen:

34 Young Assume the single slit has a width, w. Light is diffracted as it passes through the slit and then propagates to the screen. All points across the slit act as wave sources and interfere at the screen:

35 Young Fringe Locations If one point in each part of the slit satisfies the conditions for destructive interference (dark fringes), the waves from all similar sets of points will also interfere destructively: w sin θ = ± m λ, with m = 1, 2, 3,...

36 Young Fringe Locations If one point in each part of the slit satisfies the conditions for destructive interference (dark fringes), the waves from all similar sets of points will also interfere destructively: w sin θ = ± m λ, with m = 1, 2, 3,... Different from double slits!

37 Young Fringe Locations There is no simple formula for the angles at which the bright fringes occur: There is a central bright fringe with other bright fringes that are lower in intensity Central Maximum About 20 times more intense.

38 Young : Central Maximum The width of the central bright fringe is approximately the angular separation of the first dark fringes on either side: The full angular width of the central bright fringe is 2λ/w. If the slit is much wider than the light s wavelength, the light beam essentially passes straight through the slit with almost no diffraction.

39 Young The complete interference pattern that is produced by two slits is combination of A the double-slit pattern and B the single-slit pattern. C Intensity pattern for slits that are not extremely narrow. A full calculation of the intensity pattern is very complicated. Summary

40 Outline Young Young 4 5

41 Young An arrangement of many slits is called a diffraction grating. Assumptions: 1 The slits are narrow. 2 The screen is very far away.

42 Young An arrangement of many slits is called a diffraction grating. Assumptions: 1 The slits are narrow. 2 The screen is very far away. Since the screen is so far away, the rays striking the screen are approximately parallel making an angle θ with the horizontal axis: L = d sin θ = m λ Bright fringes: m = 0, ± 1, ± 2,...

43 Young The condition for bright fringes from a diffraction grating is identical to the condition for constructive inteference from a double slit: Overall intensity pattern depends on the number of slits. The larger the number of slits, the narrower the peaks.

44 Young A diffraction grating will produce an intensity pattern on the screen for each color: The different colors will have different angles and different places on the screen. gratings are widely used to analyze the colors in a beam of light.

45 Young and CDs Light reflected from the arcs in a CD acts as sources of Huygens waves: The reflected waves exhibit constructive interference at certain angles. Light reflected from a CD has the colors separated.

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