Every problem in the chapter beside the book’s answer, a chapter at a time as you scroll.
Chapter 27: Wave Optics
60 problems · 5 figures27-1 §27.1 source ↗
Answer
27-3 §27.1 source ↗
Answer
27-5 §27.1 source ↗
Answer
27-6 §27.3 source ↗
Answer
27-8 §27.3 source ↗
Answer
27-10 §27.3 source ↗
Answer
27-12 §27.3 source ↗
Answer
27-14 §27.3 source ↗
Answer
27-16 §27.3 source ↗
Answer
(a) 760 nm
(b) 1520 nm
27-18 §27.3 source ↗
Answer
For small angles .
For two adjacent fringes we have,
and
Subtracting these equations gives
27-20 §27.3 source ↗
Answer
27-21 §27.4 source ↗
Answer
27-23 §27.4 source ↗
Answer
27-25 §27.4 source ↗
Answer
27-27 §27.4 source ↗
Answer
(a)
(b)
(c) Decreasing the number of lines per centimeter by a factor of x means that the angle for the x‐order maximum is the same as the original angle for the first-order maximum.
27-29 §27.4 source ↗
Answer
27-31 §27.4 source ↗
Answer
27-33 §27.4 source ↗
Answer
27-35 §27.4 source ↗
Answer
27-37 §27.4 source ↗
Answer
(a) The longest wavelength is 333.3 nm, which is not visible.
(b) 333 nm (UV)
(c)
27-39 §27.4 source ↗
Answer
27-41 §27.4 source ↗
Unreasonable Results
(a) What visible wavelength has its fourth-order maximum at an angle of when projected on a 25,000-line-per-centimeter diffraction grating? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
Answer
(a) 42.3 nm
(b) Not a visible wavelength
The number of slits in this diffraction grating is too large. Etching in integrated circuits can be done to a resolution of 50 nm, so slit separations of 400 nm are at the limit of what we can do today. This line spacing is too small to produce diffraction of light.
27-43 §27.5 source ↗
Answer
(a)
(b) No
27-45 §27.5 source ↗
Answer
(a)
(b)
27-47 §27.5 source ↗
Answer
27-49 §27.5 source ↗
Answer
(a)
(b) 12
27-51 §27.5 source ↗
Answer
(a)
(b) 0.262 mm
(c) This distance is not easily measured by human eye, but under a microscope or magnifying glass it is quite easily measurable.
27-53 §27.5 source ↗
Answer
(a)
(b)
(c) No
(d) . Thus, .
27-55 §27.5 source ↗
Integrated Concepts
A water break at the entrance to a harbor consists of a rock barrier with a 50.0-m-wide opening. Ocean waves of 20.0-m wavelength approach the opening straight on. At what angle to the incident direction are the boats inside the harbor most protected against wave action?
Answer
27-57 §27.6 source ↗
The 300-m-diameter Arecibo radio telescope pictured in Figure 27.28 detects radio waves with a 4.00 cm average wavelength.
(a) What is the angle between two just-resolvable point sources for this telescope?
(b) How close together could these point sources be at the 2 million light year distance of the Andromeda galaxy?
Answer
(a)
(b) 326 ly
27-59 §27.6 source ↗
Answer
27-61 §27.6 source ↗
A telescope can be used to enlarge the diameter of a laser beam and limit diffraction spreading. The laser beam is sent through the telescope in opposite the normal direction and can then be projected onto a satellite or the Moon.
(a) If this is done with the Mount Wilson telescope, producing a 2.54-m-diameter beam of 633-nm light, what is the minimum angular spread of the beam (the angle between the beam's maximum and its first zero)?
(b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the Moon, assuming a lunar distance of ?
Answer
(a)
(b) Diameter of
27-63 §27.6 source ↗
Answer
27-65 §27.6 source ↗
(a) The planet Pluto and its Moon Charon are separated by 19,600 km. Neglecting atmospheric effects, should the 5.08-m-diameter Mount Palomar telescope be able to resolve these bodies when they are from Earth? Assume an average wavelength of 550 nm.
(b) In actuality, it is just barely possible to discern that Pluto and Charon are separate bodies using an Earth-based telescope. What are the reasons for this?
Answer
(a) Yes. Should easily be able to discern.
(b) The fact that it is just barely possible to discern that these are separate bodies indicates the severity of atmospheric aberrations.
27-70 §27.7 source ↗
Answer
27-72 §27.7 source ↗
Answer
27-74 §27.7 source ↗
Answer
27-76 §27.7 source ↗
| Medium | n |
|---|---|
| Gases at , 1 atm | |
| Air | 1.000293 |
| Carbon dioxide | 1.00045 |
| Hydrogen | 1.000139 |
| Oxygen | 1.000271 |
| Liquids at | |
| Benzene | 1.501 |
| Carbon disulfide | 1.628 |
| Carbon tetrachloride | 1.461 |
| Ethanol | 1.361 |
| Glycerine | 1.473 |
| Water, fresh | 1.333 |
| Solids at | |
| Diamond | 2.419 |
| Fluorite | 1.434 |
| Glass, crown | 1.52 |
| Glass, flint | 1.66 |
| Ice at | 1.309 |
| Magnesium Fluoride | 1.38 |
| Polystyrene | 1.49 |
| Plexiglas | 1.51 |
| Quartz, crystalline | 1.544 |
| Quartz, fused | 1.458 |
| Sodium chloride | 1.544 |
| Zircon | 1.923 |
Answer
27-78 §27.7 source ↗
Answer
27-80 §27.7 source ↗
Answer
27-82 §27.7 source ↗
Answer
27-84 §27.8 source ↗
Answer
27-86 §27.8 source ↗
Answer
27-88 §27.8 source ↗
Answer
27-90 §27.8 source ↗
Answer
27-92 §27.8 source ↗
Answer
27-94 §27.8 source ↗
Answer
27-96 §27.8 source ↗
Answer
(a) 1.92, not diamond (Zircon)
(b)
27-98 §27.8 source ↗
Integrated Concepts
If a polarizing filter reduces the intensity of polarized light to of its original value, by how much are the electric and magnetic fields reduced?
Answer
27-100 §27.8 source ↗
Integrated Concepts
(a) On a day when the intensity of sunlight is , a circular lens 0.200 m in diameter focuses light onto water in a black aluminum beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle of Assuming the sunlight is unpolarized and the polarizers are efficient, what is the initial rate of heating of the water in , assuming it is absorbed? The aluminum beaker has a mass of 30.0 grams and contains 250 grams of water. (b) Do the polarizing filters get hot? Explain.
Answer
(a) °C/s
(b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.
27-101 §27.8 source ↗
Answer
(a)
(b)
(c)
(d)
(e) No.
AP Test Prep 27.1 §27.2 source ↗
- a.It can be used to explain the particle behavior of waves.
- b.It states that each point on a wavefront can be considered a new wave source.
- c.It can be used to find the velocity of a wave.
- d.All of the above.
Answer
AP Test Prep 27.3 §27.3 source ↗
Superposition of which of the following light waves may produce interference fringes? Select two answers.
Wave1 = A1sin(2ωt)
Wave2 = A2sin(4ωt)
Wave3 = A3sin(2ωt + θ)
Wave4 = A4sin(4ωt + θ).
- a.Wave1 and Wave2
- b.Wave2 and Wave4
- c.Wave3 and Wave1
- d.Wave4 and Wave3
Answer
AP Test Prep 27.5 §27.3 source ↗
- a.The intensity of the bright fringe
- b.The intensity of the dark fringe
- c.The difference in the intensities of consecutive bright and dark fringes
- d.None of the above
Answer
AP Test Prep 27.7 §27.3 source ↗
- a.2a
- b.4a
- c.8a2
- d.16a2
Answer
AP Test Prep 27.9 §27.4 source ↗
- a.The central maximum will be white but the higher-order maxima will disperse into a rainbow of colors.
- b.The central maximum and higher-order maxima will be of equal widths.
- c.The lower wavelength components of light will have less diffraction compared to higher wavelength components for all maxima except the central one.
- d.None of the above.
Answer
AP Test Prep 27.11 §27.5 source ↗
- a.0.6 μm
- b.1 μm
- c.1.4 μm
- d.1.7 μm
Answer
AP Test Prep 27.13 §27.6 source ↗
- a.W increases as S increases.
- b.W decreases as S increases.
- c.W can increase or decrease as S decreases.
- d.W can neither increase nor decrease as S decreases.
Answer
AP Test Prep 27.15 §27.7 source ↗
- a.Light reflecting from a medium having an index of refraction less than that of the medium in which it is traveling.
- b.Light reflecting from a medium having an index of refraction greater than that of the medium in which it is traveling.
- c.Light changing its wavelength and speed after reflection.
- d.Light reflecting from the top and bottom surfaces of a film.
Answer
AP Test Prep 27.17 §27.8 source ↗
- a.It is parallel to the direction of propagation and perpendicular to the direction of the electric field.
- b.It is perpendicular to the direction of propagation and parallel to the direction of the electric field.
- c.It is parallel to the directions of propagation and the electric field.
- d.It is perpendicular to the directions of propagation and the electric field.
Answer
Loading chapter 28…
Problems and answers are adapted from College Physics for AP® Courses 2e by Gregg Wolfe, Erika Gasper, John Stoke, Julie Kretchman, David Anderson and others, OpenStax, and are used under the CC BY-NC-SA 4.0 licence. Access for free at https://openstax.org/books/college-physics-ap-courses-2e/pages/1-connection-for-ap-r-courses. Illustrations: Copyright Rice University, OpenStax, under CC BY-NC-SA 4.0 license. This adaptation is released under the same licence, is non-commercial, and is neither affiliated with nor endorsed by OpenStax or Rice University.