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Transcript of © Houghton Mifflin Harcourt Publishing Company Preview Objectives Combining Light Waves...
© Houghton Mifflin Harcourt Publishing Company
Preview
• Objectives
• Combining Light Waves
• Demonstrating Interference
• Sample Problem
Chapter 15 Section 1 Interference
© Houghton Mifflin Harcourt Publishing Company
Section 1 InterferenceChapter 15
Objectives
• Describe how light waves interfere with each other to produce bright and dark fringes.
• Identify the conditions required for interference to occur.
• Predict the location of interference fringes using the equation for double-slit interference.
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Section 1 InterferenceChapter 15
Combining Light Waves
• Interference takes place only between waves with the same wavelength. A light source that has a single wavelength is called monochromatic.
• In constructive interference, component waves combine to form a resultant wave with the same wavelength but with an amplitude that is greater than the either of the individual component waves.
• In the case of destructive interference, the resultant amplitude is less than the amplitude of the larger component wave.
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Chapter 15
Interference Between Transverse Waves
Section 1 Interference
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• Waves must have a constant phase difference for interference to be observed.
• Coherence is the correlation between the phases of two or more waves.– Sources of light for which the phase difference is
constant are said to be coherent.– Sources of light for which the phase difference is
not constant are said to be incoherent.
Section 1 InterferenceChapter 15
Combining Light Waves, continued
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Click below to watch the Visual Concept.
Visual Concept
Chapter 15 Section 1 Interference
Combining Light Waves
© Houghton Mifflin Harcourt Publishing Company
Section 1 InterferenceChapter 15
Demonstrating Interference
• Interference can be demonstrated by passing light through two narrow parallel slits.
• If monochromatic light is used, the light from the two slits produces a series of bright and dark parallel bands, or fringes, on a viewing screen.
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Chapter 15
Conditions for Interference of Light Waves
Section 1 Interference
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Section 1 InterferenceChapter 15
Demonstrating Interference, continued
• The location of interference fringes can be predicted.
• The path difference is the difference in the distance traveled by two beams when they are scattered in the same direction from different points.
• The path difference equals dsin.
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Click below to watch the Visual Concept.
Visual Concept
Chapter 15 Section 1 Interference
Interference Arising from Two Slits
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Section 1 InterferenceChapter 15
Demonstrating Interference, continued
• The number assigned to interference fringes with respect to the central bright fringe is called the order number. The order number is represented by the symbol m.
• The central bright fringe at q = 0 (m = 0) is called the zeroth-order maximum, or the central maximum.
• The first maximum on either side of the central maximum (m = 1) is called the first-order maximum.
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Section 1 InterferenceChapter 15
Demonstrating Interference, continued
• Equation for constructive interferenced sin = ±m m = 0, 1, 2, 3, …
The path difference between two waves = an integer multiple of the wavelength
• Equation for destructive interferenced sin = ±(m + 1/2) m = 0, 1, 2, 3, …
The path difference between two waves = an odd number of half wavelength
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Section 1 InterferenceChapter 15
Sample Problem
Interference
The distance between the two slits is 0.030 mm. The second-order bright fringe (m = 2) is measured on a viewing screen at an angle of 2.15º from the central maximum. Determine the wavelength of the light.
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Section 1 InterferenceChapter 15
Sample Problem, continued
Interference1. DefineGiven: d = 3.0 10–5 m
m = 2= 2.15º
Unknown: = ?
Diagram:
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Section 1 InterferenceChapter 15
Sample Problem, continued
Interference2. Plan
Choose an equation or situation: Use the equation for constructive interference.
d sin = m
Rearrange the equation to isolate the unknown:
d sinm
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Section 1 InterferenceChapter 15
Sample Problem, continued
Interference3. Calculate
Substitute the values into the equation and solve:
–5
–7 2
2
3.0 10 m sin2.15º
2
5.6 10 m 5.6 10 nm
5.6 10 nm
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Section 1 InterferenceChapter 15
Sample Problem, continued
Interference4. Evaluate
This wavelength of light is in the visible spectrum. The wavelength corresponds to light of a yellow-green color.
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Preview
• Objectives
• The Bending of Light Waves
• Diffraction Gratings
• Sample Problem
• Diffraction and Instrument Resolution
Chapter 15 Section 2 Diffraction
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Section 2 DiffractionChapter 15
Objectives
• Describe how light waves bend around obstacles and produce bright and dark fringes.
• Calculate the positions of fringes for a diffraction grating.
• Describe how diffraction determines an optical instrument’s ability to resolve images.
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Section 2 DiffractionChapter 15
The Bending of Light Waves
• Diffraction is a change in the direction of a wave when the wave encounters an obstacle, an opening, or an edge.
• Light waves form a diffraction pattern by passing around an obstacle or bending through a slit and interfering with each other.
• Wavelets (as in Huygens’ principle) in a wave front interfere with each other.
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Chapter 15
Destructive Interference in Single-Slit Diffraction
Section 2 Diffraction
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Section 2 DiffractionChapter 15
The Bending of Light Waves, continued
• In a diffraction pattern, the central maximum is twice as wide as the secondary maxima.
• Light diffracted by an obstacle also produces a pattern.
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Section 2 DiffractionChapter 15
Diffraction Gratings
• A diffraction grating uses diffraction and interference to disperse light into its component colors.
• The position of a maximum depends on the separation of the slits in the grating, d, the order of the maximum m,, and the wavelength of the light, .
d sin = ±m m = 0, 1, 2, 3, …
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Chapter 15
Constructive Interference by a Diffraction Grating
Section 2 Diffraction
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Section 2 DiffractionChapter 15
Sample Problem
Diffraction Gratings
Monochromatic light from a helium-neon laser ( = 632.8 nm) shines at a right angle to the surface of a diffraction grating that contains 150 500 lines/m. Find the angles at which one would observe the first-order and second-order maxima.
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Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
1. Define
Given: = 632.8 nm = 6.328 10–7 m
m = 1 and 2
Unknown: = ? 2 = ?
d 1
150 500lines
m
1
150 500m
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Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
1. Define, continued
Diagram:
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Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
2. Plan
Choose an equation or situation: Use the equation for a diffraction grating.
d sin = ±mRearrange the equation to isolate the unknown:
–1sinm
d
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Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
3. Calculate
Substitute the values into the equation and solve:
For the first-order maximum, m = 1:
–7–1 –1
1
1
6.328 10 msin sin
1m
150 500
5.465º
d
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Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
3. Calculate, continued
For m = 2:
–12
–7
–12
2
2sin
2 6.328 10 msin
1m
150 500
10.98º
d
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Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
4. Evaluate
The second-order maximum is spread slightly more than twice as far from the center as the first-order maximum. This diffraction grating does not have high dispersion, and it can produce spectral lines up to the tenth-order maxima (where sin = 0.9524).
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Click below to watch the Visual Concept.
Visual Concept
Chapter 15 Section 2 Diffraction
Function of a Spectrometer
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Section 2 DiffractionChapter 15
Diffraction and Instrument Resolution
• The ability of an optical system to distinguish between closely spaced objects is limited by the wave nature of light.
• Resolving power is the ability of an optical instrument to form separate images of two objects that are close together.
• Resolution depends on wavelength and aperture width. For a circular aperture of diameter D:
1.22
D
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Chapter 15
Resolution of Two Light Sources
Section 2 Diffraction
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Preview
• Objectives
• Lasers and Coherence
• Applications of Lasers
Chapter 15 Section 3 Lasers
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Section 3 LasersChapter 15
Objectives
• Describe the properties of laser light.
• Explain how laser light has particular advantages in certain applications.
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Section 3 LasersChapter 15
Lasers and Coherence
• A laser is a device that produces coherent light at a single wavelength.
• The word laser is an acronym of “light amplification by stimulated emission of radiation.”
• Lasers transform other forms of energy into coherent light.
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Click below to watch the Visual Concept.
Visual Concept
Chapter 15 Section 3 Lasers
Comparing Incoherent and Coherent Light
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Click below to watch the Visual Concept.
Visual Concept
Chapter 15 Section 3 Lasers
Laser
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Section 3 LasersChapter 15
Applications of Lasers
• Lasers are used to measure distances with great precision.
• Compact disc and DVD players use lasers to read digital data on these discs.
• Lasers have many applications in medicine.– Eye surgery– Tumor removal– Scar removal
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Chapter 15
Components of a Compact Disc Player
Section 3 Lasers
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Chapter 15
Incoherent and Coherent Light
Section 3 Lasers