Upload
josoborned
View
3.961
Download
1
Tags:
Embed Size (px)
DESCRIPTION
Citation preview
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
• To View the presentation as a slideshow with effects select “View” on the menu bar and click on “Slide Show.”
• To advance through the presentation, click the right-arrow key or the space bar.
• From the resources slide, click on any resource to see a presentation for that resource.
• From the Chapter menu screen click on any lesson to go directly to that lesson’s presentation.
• You may exit the slide show at any time by pressing the Esc key.
How to Use This Presentation
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter Presentation
Transparencies Sample Problems
Visual Concepts
Standardized Test Prep
ResourcesResources
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Interference and DiffractionChapter 15
Table of Contents
Section 1 Interference
Section 2 Diffraction
Section 3 Lasers
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Interference Between Transverse Waves
Section 1 Interference
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Section 1 InterferenceChapter 15
Combining Light Waves, continued
• 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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Incoherent and Coherent Light
Section 3 Lasers
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Combining Light Waves
Section 1 Interference
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Conditions for Interference of Light Waves
Section 1 Interference
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Interference Arising from Two Slits
Section 1 Interference
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Section 1 InterferenceChapter 15
Sample Problem, continued
Interference1. DefineGiven: d = 3.0 10–5 m
m = 2= 2.15º
Unknown: = ?
Diagram:
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Destructive Interference in Single-Slit Diffraction
Section 2 Diffraction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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, …
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Constructive Interference by a Diffraction Grating
Section 2 Diffraction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Section 2 DiffractionChapter 15
Sample Problem, continued
Diffraction Gratings
1. Define, continued
Diagram:
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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).
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Function of a Spectrometer
Section 2 Diffraction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Resolution of Two Light Sources
Section 2 Diffraction
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Section 3 LasersChapter 15
Objectives
• Describe the properties of laser light.
• Explain how laser light has particular advantages in certain applications.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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.
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Comparing Incoherent and Coherent Light
Section 3 Lasers
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Laser
Section 3 Lasers
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
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
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Chapter 15
Components of a Compact Disc Player
Section 3 Lasers
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice
1. In the equations for interference, what does the term d represent?
A. the distance from the midpoint between the two slits to the viewing screen
B. the distance between the two slits through which a light wave passes
C. the distance between two bright interference fringes
D. the distance between two dark interference fringes
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
1. In the equations for interference, what does the term d represent?
A. the distance from the midpoint between the two slits to the viewing screen
B. the distance between the two slits through which a light wave passes
C. the distance between two bright interference fringes
D. the distance between two dark interference fringes
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference?
F. The waves must be in phase at all times.
G. The waves must be 90º out of phase at all times.
H. The waves must be 180º out of phase at all times.
J. The waves must be 270º out of phase at all times.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
2. Which of the following must be true for two waves with identical amplitudes and wavelengths to undergo complete destructive interference?
F. The waves must be in phase at all times.
G. The waves must be 90º out of phase at all times.
H. The waves must be 180º out of phase at all times.
J. The waves must be 270º out of phase at all times.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern?
A. dsin = /2
B. dsin = 3/2
C. dsin = 5/2
D. dsin = 3
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
3. Which equation correctly describes the condition for observing the third dark fringe in an interference pattern?
A. dsin = /2
B. dsin = 3/2
C. dsin = 5/2
D. dsin = 3
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
4. Why is the diffraction of sound easier to observe than the diffraction of visible light?
F. Sound waves are easier to detect than visible light waves.
G. Sound waves have longer wavelengths than visible light waves and so bend more around barriers.
H. Sound waves are longitudinal waves, which diffract more than transverse waves.
J. Sound waves have greater amplitude than visible light waves.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
4. Why is the diffraction of sound easier to observe than the diffraction of visible light?
F. Sound waves are easier to detect than visible light waves.
G. Sound waves have longer wavelengths than visible light waves and so bend more around barriers.
H. Sound waves are longitudinal waves, which diffract more than transverse waves.
J. Sound waves have greater amplitude than visible light waves.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
5. Monochromatic infrared waves with a wavelength of 750 nm pass through two narrow slits. If the slits are 25 µm apart, at what angle will the fourth order bright fringe appear on a viewing screen?
A. 4.3º
B. 6.0º
C. 6.9º
D. 7.8º
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
5. Monochromatic infrared waves with a wavelength of 750 nm pass through two narrow slits. If the slits are 25 µm apart, at what angle will the fourth order bright fringe appear on a viewing screen?
A. 4.3º
B. 6.0º
C. 6.9º
D. 7.8º
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
6. Monochromatic light with a wavelength of 640 nm passes through a diffraction grating that has 5.0 104 lines/m. A bright line on a screen appears at an angle of 11.1º from the central bright fringe.What is the order of this bright line?
F. m = 2
G. m = 4
H. m = 6
J. m = 8
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
6. Monochromatic light with a wavelength of 640 nm passes through a diffraction grating that has 5.0 104 lines/m. A bright line on a screen appears at an angle of 11.1º from the central bright fringe.What is the order of this bright line?
F. m = 2
G. m = 4
H. m = 6
J. m = 8
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
7. For observing the same object, how many times better is the resolution of the telescope shown on the left in the figure below than that of the telescope shown on the right?
A. 4
B. 2
C. 1/2
D. 1/4
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
7. For observing the same object, how many times better is the resolution of the telescope shown on the left in the figure below than that of the telescope shown on the right?
A. 4
B. 2
C. 1/2
D. 1/4
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
8. What steps should you employ to design a telescope with a high degree of resolution?
F. Widen the aperture, or design the telescope to detect light of short wavelength.
G. Narrow the aperture, or design the telescope to detect light of short wavelength.
H. Widen the aperture, or design the telescope to detect light of long wavelength.
J. Narrow the aperture, or design the telescope to detect light of long wavelength.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
8. What steps should you employ to design a telescope with a high degree of resolution?
F. Widen the aperture, or design the telescope to detect light of short wavelength.
G. Narrow the aperture, or design the telescope to detect light of short wavelength.
H. Widen the aperture, or design the telescope to detect light of long wavelength.
J. Narrow the aperture, or design the telescope to detect light of long wavelength.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
9. What is the property of a laser called that causes coherent light to be emitted?
A. population inversion
B. light amplification
C. monochromaticity
D. stimulated emission
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
9. What is the property of a laser called that causes coherent light to be emitted?
A. population inversion
B. light amplification
C. monochromaticity
D. stimulated emission
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
10. Which of the following is not an essential component of a laser?
F. a partially transparent mirror
G. a fully reflecting mirror
H. a converging lens
J. an active medium
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Multiple Choice, continued
10. Which of the following is not an essential component of a laser?
F. a partially transparent mirror
G. a fully reflecting mirror
H. a converging lens
J. an active medium
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Short Response
11. Why is laser light useful for the purposes of making astronomical measurements and surveying?
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Short Response, continued
11. Why is laser light useful for the purposes of making astronomical measurements and surveying?
Answer: The beam does not spread out much or lose intensity over long distances.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Short Response, continued
12. A diffraction grating used in a spectrometer causes the third-order maximum of blue light with a wavelength of 490 nm to form at an angle of 6.33º from the central maximum (m = 0). What is the separation between the lines of the grating?
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Short Response, continued
12. A diffraction grating used in a spectrometer causes the third-order maximum of blue light with a wavelength of 490 nm to form at an angle of 6.33º from the central maximum (m = 0). What is the separation between the lines of the grating?
Answer: 7.5 104 lines/m = 750 lines/cm
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Short Response, continued
13. Telescopes that orbit Earth provide better images of distant objects because orbiting telescopes are more able to operate near their theoretical resolution than telescopes on Earth. The orbiting telescopes needed to provide high resolution in the visible part of the spectrum are much larger than the orbiting telescopes that provide similar images in the ultraviolet and X-ray portion of the spectrum. Explain why the sizes must vary.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Short Response, continued
13. (See previous slide for question.)
Answer: The resolving power of a telescope depends on the ratio of the wavelength to the diameter of the aperture. Telescopes using longer wavelength radiation (visible light) must be larger than those using shorter wavelengths (ultraviolet, X ray) to achieve the same resolving power.
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response
14. Radio signals often reflect from objects and recombine at a distance. Suppose you are moving in a direction perpendicular to a radio signal source and its reflected signal. How would interference between these two signals sound on a radio receiver?
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
14. (See previous slide for question.)
Answer: The interference pattern for radio signals would “appear” on a radio receiver as an alternating increase in signal intensity followed by a loss of intensity (heard as static or “white noise”).
Standardized Test PrepChapter 15
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
Base your answers to questions 15–17 on the information below. In each problem, show all of your work.
A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.
Standardized Test PrepChapter 15
15. What is the wave-length of the light?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
Base your answers to questions 15–17 on the information below. In each problem, show all of your work.
A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.
Standardized Test PrepChapter 15
15. What is the wave-length of the light?
Answer: 589 nm
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
Base your answers to questions 15–17 on the information below. In each problem, show all of your work.
A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.
Standardized Test PrepChapter 15
16. At what angle would the third-order (m = 3) bright fringe appear?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
Base your answers to questions 15–17 on the information below. In each problem, show all of your work.
A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.
Standardized Test PrepChapter 15
16. At what angle would the third-order (m = 3) bright fringe appear?
Answer: 6.77º
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
Base your answers to questions 15–17 on the information below. In each problem, show all of your work.
A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.
Standardized Test PrepChapter 15
17. At what angle would the third-order (m = 3) dark fringe appear?
Copyright © by Holt, Rinehart and Winston. All rights reserved.
ResourcesChapter menu
Extended Response, continued
Base your answers to questions 15–17 on the information below. In each problem, show all of your work.
A double-slit apparatus for demonstrating interference is constructed so that the slits are separated by 15.0 µm. A first-order fringe for constructive interference appears at an angle of 2.25° from the zeroth-order (central) fringe.
Standardized Test PrepChapter 15
17. At what angle would the third-order (m = 3) dark fringe appear?
Answer: 7.90º