27.E: Vision and Optical Instruments (Exercise)
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Conceptual Questions
27.1: The Wave Aspect of Light: Interference
1. What type of experimental evidence indicates that light is a wave?
2. Give an example of a wave characteristic of light that is easily observed outside the laboratory.
27.2: Huygens's Principle: Diffraction
3. How do wave effects depend on the size of the object with which the wave interacts? For example, why does sound bend around the corner of a building while light does not?
4. Under what conditions can light be modeled like a ray? Like a wave?
5. Go outside in the sunlight and observe your shadow. It has fuzzy edges even if you do not. Is this a diffraction effect? Explain.
6. Why does the wavelength of light decrease when it passes from vacuum into a medium? State which attributes change and which stay the same and, thus, require the wavelength to decrease.
7. Does Huygens’s principle apply to all types of waves?
27.3: Young’s Double Slit Experiment
8. Young’s double slit experiment breaks a single light beam into two sources. Would the same pattern be obtained for two independent sources of light, such as the headlights of a distant car? Explain.
9. Suppose you use the same double slit to perform Young’s double slit experiment in air and then repeat the experiment in water. Do the angles to the same parts of the interference pattern get larger or smaller? Does the color of the light change? Explain.
10. Is it possible to create a situation in which there is only destructive interference? Explain.
11. The figrue shows the central part of the interference pattern for a pure wavelength of red light projected onto a double slit. The pattern is actually a combination of single slit and double slit interference. Note that the bright spots are evenly spaced. Is this a double slit or single slit characteristic? Note that some of the bright spots are dim on either side of the center. Is this a single slit or double slit characteristic? Which is smaller, the slit width or the separation between slits? Explain your responses.
This double slit interference pattern also shows signs of single slit interference. (credit: PASCO)
27.4: Multiple Slit Diffraction
12. What is the advantage of a diffraction grating over a double slit in dispersing light into a spectrum?
13. What are the advantages of a diffraction grating over a prism in dispersing light for spectral analysis?
14. Can the lines in a diffraction grating be too close together to be useful as a spectroscopic tool for visible light? If so, what type of EM radiation would the grating be suitable for? Explain.
15. If a beam of white light passes through a diffraction grating with vertical lines, the light is dispersed into rainbow colors on the right and left. If a glass prism disperses white light to the right into a rainbow, how does the sequence of colors compare with that produced on the right by a diffraction grating?
16. Suppose pure-wavelength light falls on a diffraction grating. What happens to the interference pattern if the same light falls on a grating that has more lines per centimeter? What happens to the interference pattern if a longer-wavelength light falls on the same grating? Explain how these two effects are consistent in terms of the relationship of wavelength to the distance between slits.
17. Suppose a feather appears green but has no green pigment. Explain in terms of diffraction.
18. Suppose a feather appears green but has no green pigment. Explain in terms of diffraction.
27.5: Single Slit Diffraction
19. As the width of the slit producing a single-slit diffraction pattern is reduced, how will the diffraction pattern produced change?
27.6: Limits of Resolution: The Rayleigh Criterion
20. A beam of light always spreads out. Why can a beam not be created with parallel rays to prevent spreading? Why can lenses, mirrors, or apertures not be used to correct the spreading?
27.7: Thin Film Interference
21. What effect does increasing the wedge angle have on the spacing of interference fringes? If the wedge angle is too large, fringes are not observed. Why?
22. How is the difference in paths taken by two originally in-phase light waves related to whether they interfere constructively or destructively? How can this be affected by reflection? By refraction?
23. Is there a phase change in the light reflected from either surface of a contact lens floating on a person’s tear layer? The index of refraction of the lens is about 1.5, and its top surface is dry.
24. In placing a sample on a microscope slide, a glass cover is placed over a water drop on the glass slide. Light incident from above can reflect from the top and bottom of the glass cover and from the glass slide below the water drop. At which surfaces will there be a phase change in the reflected light?
25. Answer the above question if the fluid between the two pieces of crown glass is carbon disulfide.
26. While contemplating the food value of a slice of ham, you notice a rainbow of color reflected from its moist surface. Explain its origin.
27. An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a non-reflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?
28. A non-reflective coating like the one described in Example works ideally for a single wavelength and for perpendicular incidence. What happens for other wavelengths and other incident directions? Be specific.
29. Why is it much more difficult to see interference fringes for light reflected from a thick piece of glass than from a thin film? Would it be easier if monochromatic light were used?
27.8: Polarization
30. Under what circumstances is the phase of light changed by reflection? Is the phase related to polarization?
31. Can a sound wave in air be polarized? Explain.
32. No light passes through two perfect polarizing filters with perpendicular axes. However, if a third polarizing filter is placed between the original two, some light can pass. Why is this? Under what circumstances does most of the light pass?
33. Explain what happens to the energy carried by light that it is dimmed by passing it through two crossed polarizing filters.
34. When particles scattering light are much smaller than its wavelength, the amount of scattering is proportional to \(\displaystyle 1/λ^4\). Does this mean there is more scattering for small \(\displaystyle λ\) than large \(\displaystyle λ\)? How does this relate to the fact that the sky is blue?
35. Using the information given in the preceding question, explain why sunsets are red.
36. When light is reflected at Brewster’s angle from a smooth surface, it is \(\displaystyle 100%\) polarized parallel to the surface. Part of the light will be refracted into the surface. Describe how you would do an experiment to determine the polarization of the refracted light. What direction would you expect the polarization to have and would you expect it to be \(\displaystyle 100%\)?
27.9: *Extended Topic* Microscopy Enhanced by the Wave Characteristics of Light
37. Explain how microscopes can use wave optics to improve contrast and why this is important.
38. A bright white light under water is collimated and directed upon a prism. What range of colors does one see emerging?
Problems & Exercises
27.1: The Wave Aspect of Light: Interference
39. Show that when light passes from air to water, its wavelength decreases to 0.750 times its original value.
Solution
1/1.333 = 0.750
40. Find the range of visible wavelengths of light in crown glass.
41. What is the index of refraction of a material for which the wavelength of light is 0.671 times its value in a vacuum? Identify the likely substance.
Solution
1.49, Polystyrene
42. Analysis of an interference effect in a clear solid shows that the wavelength of light in the solid is 329 nm. Knowing this light comes from a He-Ne laser and has a wavelength of 633 nm in air, is the substance zircon or diamond?
43. What is the ratio of thicknesses of crown glass and water that would contain the same number of wavelengths of light?
Solution
0.877 glass to water
27.3: Young’s Double Slit Experiment
44. At what angle is the first-order maximum for 450-nm wavelength blue light falling on double slits separated by 0.0500 mm?
Solution
\(\displaystyle 0.516^{\circ}\)
45. Calculate the angle for the third-order maximum of 580-nm wavelength yellow light falling on double slits separated by 0.100 mm.
46. What is the separation between two slits for which 610-nm orange light has its first maximum at an angle of \(\displaystyle 30.0^{\circ}\)?
Solution
\(\displaystyle 1.22 \times 10^{-6} m\)
47. Find the distance between two slits that produces the first minimum for 410-nm violet light at an angle of \(\displaystyle 45.0^{\circ}\).
48. Calculate the wavelength of light that has its third minimum at an angle of \(\displaystyle 30.0^{\circ}\) when falling on double slits separated by \(\displaystyle 3.00 \mu m\) Explicitly, show how you follow the steps in "Problem-Solving Strategies for Wave Optics."
Solution
\(\displaystyle 600 nm\)
49. What is the wavelength of light falling on double slits separated by \(\displaystyle 2.00 \mu m\) if the third-order maximum is at an angle of \(\displaystyle 60.0^{\circ}\)?
50. At what angle is the fourth-order maximum for the situation in the first exercise?
Solution
\(\displaystyle 2.06^{\circ}\)
51. What is the highest-order maximum for 400-nm light falling on double slits separated by \(\displaystyle 25.0 \mu m\)?
52. Find the largest wavelength of light falling on double slits separated by \(\displaystyle 1.20 \mu m\) for which there is a first-order maximum. Is this in the visible part of the spectrum?
Solution
1200 nm (not visible)
53. What is the smallest separation between two slits that will produce a second-order maximum for 720-nm red light?
54. (a) What is the smallest separation between two slits that will produce a second-order maximum for any visible light?
(b) For all visible light?
Solution
(a) 760 nm
(b) 1520 nm
55. (a) If the first-order maximum for pure-wavelength light falling on a double slit is at an angle of \(\displaystyle 10.0^{\circ}\), at what angle is the second-order maximum?
(b) What is the angle of the first minimum?
(c) What is the highest-order maximum possible here?
56. The figure shows a double slit located a distance \(\displaystyle x\) from a screen, with the distance from the center of the screen given by \(\displaystyle y\). When the distance \(\displaystyle d\) between the slits is relatively large, there will be numerous bright spots, called fringes. Show that, for small angles (where \(\displaystyle \sin{\theta} \approx \theta\), with \(\displaystyle \theta\) in radians), the distance between fringes is given by \(\displaystyle \delta y = x \lambda / d\).
The distance between adjacent fringes is \(\displaystyle \delta y = x \lambda / d\), assuming the slit separation \(\displaystyle d\) is large compared with \(\displaystyle \lambda\).
Solution
For small angles \(\displaystyle \sin{\theta} - \tan{\theta} \approx \theta \left( in radians \right).\)
For two adjacent fringes we have, \[d \sin{\theta_{m}} = m \lambda \tag{27.4.5}\] and \[d \sin{\theta_{m+1}} = \left( m + 1 \right) \lambda \tag{27.4.6}\] Subtracting these equations gives \[d \left( \sin{\theta_{m+1}} - \sin{\theta_{m}}\right) = \left[ \left( m+1 \right) - m \right] \lambda \tag{27.4.7}\] \[d \left( \theta_{m+1} - \theta_{m} \right) = \lambda \tag{27.4.8}\] \[\tan{\theta_{m}} = \frac{y_{m}}{x} \approx \theta_{m} \rightarrow d \left( \frac{y_{m+1}}{x} - \frac{y_{m}}{x} \right) = \lambda \tag{27.4.9}\] \[d \frac{\delta y}{x} = \lambda \rightarrow \delta y = \frac{x \lambda}{d} \tag{27.4.10}\]
57. Using the result of the problem above, calculate the distance between fringes for 633-nm light falling on double slits separated by 0.0800 mm, located 3.00 m from a screen as in Figure 8.
58. Using the result of the problem two problems prior, find the wavelength of light that produces fringes 7.50 mm apart on a screen 2.00 m from double slits separated by 0.120 mm (see Figure 8).
Solution
450 nm
27.4: Multiple Slit Diffraction
59. A diffraction grating has 2000 lines per centimeter. At what angle will the first-order maximum be for 520-nm-wavelength green light?
Solution
\(\displaystyle 5.97^{\circ}\)
60. Find the angle for the third-order maximum for 580-nm-wavelength yellow light falling on a diffraction grating having 1500 lines per centimeter.
61. How many lines per centimeter are there on a diffraction grating that gives a first-order maximum for 470-nm blue light at an angle of \(\displaystyle 25.0^{\circ}\)?
Solution
\(\displaystyle 8.99 \times 10^{3}\)
62. What is the distance between lines on a diffraction grating that produces a second-order maximum for 760-nm red light at an angle of \(\displaystyle 60.0^{\circ}\)?
63. Calculate the wavelength of light that has its second-order maximum at \(\displaystyle 45.0^{\circ}\) when falling on a diffraction grating that has 5000 lines per centimeter.
Solution
707 nm
64. An electric current through hydrogen gas produces several distinct wavelengths of visible light. What are the wavelengths of the hydrogen spectrum, if they form first-order maxima at angles of \(\displaystyle 24.2^{\circ}\), \(\displaystyle 25.7^{\circ}\), \(\displaystyle 29.1^{\circ}\), and \(\displaystyle 41.0^{\circ}\) when projected on a diffraction grating having 10,000 lines per centimeter? Explicitly show how you follow the steps in "Problem-Solving Strategies for Wave Optics."
65. (a) What do the four angles in the above problem become if a 5000-line-per-centimeter diffraction grating is used?
(b) Using this grating, what would the angles be for the second-order maxima?
(c) Discuss the relationship between integral reductions in lines per centimeter and the new angles of various order maxima.
Solution
(a) \(\displaystyle 11.8^{\circ}\), \(\displaystyle 12.5^{\circ}\), \(\displaystyle 14.1^{\circ}\), \(\displaystyle 19.2^{\circ}\)
(b) \(\displaystyle 24.2^{\circ}\), \(\displaystyle 25.7^{\circ}\), \(\displaystyle 29.1^{\circ}\), \(\displaystyle 41.0^{\circ}\)
(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.
66. What is the maximum number of lines per centimeter a diffraction grating can have and produce a complete first-order spectrum for visible light?
67. The yellow light from a sodium vapor lamp seems to be of pure wavelength, but it produces two first-order maxima at \(\displaystyle 36.093^{\circ}\) and \(\displaystyle 36.129^{\circ}\) when projected on a 10,000 line per centimeter diffraction grating. What are the two wavelengths to an accuracy of 0.1 nm?
Solution
589.1 nm and 589.6 nm
68. What is the spacing between structures in a feather that acts as a reflection grating, given that they produce a first-order maximum for 525-nm light at a \(\displaystyle 30.0^{\circ}\) angle?
69. Structures on a bird feather act like a reflection grating having 8000 lines per centimeter. What is the angle of the first-order maximum for 600-nm light?
Solution
\(\displaystyle 28.7^{\circ}\)
70. An opal such as that shown in Figure 2 acts like a reflection grating with rows separated by about \(\displaystyle 8 \mu m\). If the opal is illuminated normally,
(a) at what angle will red light be seen and
(b) at what angle will blue light be seen?
71. At what angle does a diffraction grating produces a second-order maximum for light having a first-order maximum at \(\displaystyle 20.0^{\circ}\)?
Solution
\(\displaystyle 43.2^{\circ}\)
72. Show that a diffraction grating cannot produce a second-order maximum for a given wavelength of light unless the first-order maximum is at an angle less than \(\displaystyle 30.0^{\circ}\).
73. If a diffraction grating produces a first-order maximum for the shortest wavelength of visible light at \(\displaystyle 30.0^{\circ}\), at what angle will the first-order maximum be for the longest wavelength of visible light?
Solution
\(\displaystyle 90.0^{\circ}\)
74. (a) Find the maximum number of lines per centimeter a diffraction grating can have and produce a maximum for the smallest wavelength of visible light.
(b) Would such a grating be useful for ultraviolet spectra?
(c) For infrared spectra?
75. (a) Show that a 30,000-line-per-centimeter grating will not produce a maximum for visible light.
(b) What is the longest wavelength for which it does produce a first-order maximum?
(c) What is the greatest number of lines per centimeter a diffraction grating can have and produce a complete second-order spectrum for visible light?
Solution
(a) The longest wavelength is 333.3 nm, which is not visible.
(b) 333 nm (UV)
(c) \(\displaystyle 6.58 \times 10^{3} cm\)
76.  The brightest spot is the reflected beam at an angle equal to the angle of incidence. However, fringes are also observed. If the wall is 1.50 m from the CD, and the first fringe is 0.600 m from the central maximum, what is the spacing of grooves on the CD?
77. The analysis shown in the figure below also applies to diffraction gratings with lines separated by a distance \(\displaystyle d\). What is the distance between fringes produced by a diffraction grating having 125 lines per centimeter for 600-nm light, if the screen is 1.50 m away?
The distance between adjacent fringes is \(\displaystyle \delta y = x \lambda / d\), assuming the slit separation \(\displaystyle d\) is large compared with \(\displaystyle \lambda\).
Solution
\(\displaystyle 1.13 \times 10^{-2} m\)
78. Unreasonable Results:
Red light of wavelength of 700 nm falls on a double slit separated by 400 nm.
(a) At what angle is the first-order maximum in the diffraction pattern?
(b) What is unreasonable about this result?
(c) Which assumptions are unreasonable or inconsistent?
79. Unreasonable Results:
(a) What visible wavelength has its fourth-order maximum at an angle of \(\displaystyle 25.0^{\circ}\) 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?
Solution
(a) 42.3 nm
(b) Not a visible wavelength
(c) 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.
80. Construct Your Own Problem:
Consider a spectrometer based on a diffraction grating. Construct a problem in which you calculate the distance between two wavelengths of electromagnetic radiation in your spectrometer. Among the things to be considered are the wavelengths you wish to be able to distinguish, the number of lines per meter on the diffraction grating, and the distance from the grating to the screen or detector. Discuss the practicality of the device in terms of being able to discern between wavelengths of interest.
27.5: Single Slit Diffraction
81. (a) At what angle is the first minimum for 550-nm light falling on a single slit of width \(\displaystyle 1.00 \mu m\)?
(b) Will there be a second minimum?
Solution
(a) \(\displaystyle 33.4^{\circ}\)
(b) No
82. (a) Calculate the angle at which a \(\displaystyle 2.00 - \mu m\)-wide slit produces its first minimum for 410-nm violet light.
(b) Where is the first minimum for 700-nm red light?
83. (a) How wide is a single slit that produces its first minimum for 633-nm light at an angle of \(\displaystyle 28.0^{\circ}\)?
(b) At what angle will the second minimum be?
Solution
(a) \(\displaystyle 1.35 \times 10^{-6} m\)
(b) \(\displaystyle 69.9^{\circ}\)
84. (a) What is the width of a single slit that produces its first minimum at \(\displaystyle 60.0^{\circ}\) for 600-nm light?
(b) Find the wavelength of light that has its first minimum at \(\displaystyle 62.0^{\circ}\).
85. Find the wavelength of light that has its third minimum at an angle of \(\displaystyle 48.6^{\circ}\) when it falls on a single slit of width \(\displaystyle 3.00 \mu m\).
Solution
750 nm
86. Calculate the wavelength of light that produces its first minimum at an angle of \(\displaystyle 36.9^{\circ}\) when falling on a single slit of width \(\displaystyle 1.00 \mu m\).
87. (a) Sodium vapor light averaging 589 nm in wavelength falls on a single slit of width \(\displaystyle 7.50 \mu m\). At what angle does it produces its second minimum?
(b) What is the highest-order minimum produced?
Solution
(a) \(\displaystyle 9.04^{\circ}\)
(b) 12
88. (a) Find the angle of the third diffraction minimum for 633-nm light falling on a slit of width \(\displaystyle 20.0 \mu m\).
(b) What slit width would place this minimum at \(\displaystyle 85.0^{\circ}\)? Explicitly show how you follow the steps in "Problem-Solving Strategies fro Wave Optics."
89. (a) Find the angle between the first minima for the two sodium vapor lines, which have wavelengths of 589.1 and 589.6 nm, when they fall upon a single slit of width \(\displaystyle 2.00 \mu m\).
(b) What is the distance between these minima if the diffraction pattern falls on a screen 1.00 m from the slit?
(c) Discuss the ease or difficulty of measuring such a distance.
Solution
(a) \(\displaystyle 0.0150^{\circ}\)
(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.
90. (a) What is the minimum width of a single slit (in multiples of \(\displaystyle \lambda\)) that will produce a first minimum for a wavelength \(\displaystyle \lambda\)?
(b) What is its minimum width if it produces 50 minima?
(c) 1000 minima?
91. (a) If a single slit produces a first minimum at \(\displaystyle 14.5^{\circ}\), at what angle is the second-order minimum?
(b) What is the angle of the third-order minimum?
(c) Is there a fourth-order minimum?
(d) Use your answers to illustrate how the angular width of the central maximum is about twice the angular width of the next maximum (which is the angle between the first and second minima).
Solution
(a) \(\displaystyle 30.1^{\circ}\)
(b) \(\displaystyle 48.7^{\circ}\)
(c) No
(d) \(\displaystyle 2 \theta_{1} = \left(2\right)\left(14.5^{\circ}\right) = 29^{\circ}, \theta_{2} - \theta_{1} = 30.05^{\circ} - 14.5^{\circ} = 15.56^{\circ}.\) Thus, \(\displaystyle 29^{\circ} \approx \left(2\right)\left(15.56^{\circ}\right) = 31.1^{\circ}\).
92. A double slit produces a diffraction pattern that is a combination of single and double slit interference. Find the ratio of the width of the slits to the separation between them, if the first minimum of the single slit pattern falls on the fifth maximum of the double slit pattern. (This will greatly reduce the intensity of the fifth maximum.)
93. 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?
Solution
\(\displaystyle 23.6^{\circ}, 53.1^{\circ}\)
94. Integrated Concepts:
An aircraft maintenance technician walks past a tall hangar door that acts like a single slit for sound entering the hangar. Outside the door, on a line perpendicular to the opening in the door, a jet engine makes a 600-Hz sound. At what angle with the door will the technician observe the first minimum in sound intensity if the vertical opening is 0.800 m wide and the speed of sound is 340 m/s?
27.6: Limits of Resolution: The Rayleigh Criterion
95. The 300-m-diameter Arecibo radio telescope pictured in Figure 4 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?
Solution
(a) \(\displaystyle 1.63 \times 10^{-4} rad\)
(b) 326 ly
96. Assuming the angular resolution found for the Hubble Telescope in the "Calculating Diffraction Limits of the Hubble Space Telescope" example, what is the smallest detail that could be observed on the Moon?
97. Diffraction spreading for a flashlight is insignificant compared with other limitations in its optics, such as spherical aberrations in its mirror. To show this, calculate the minimum angular spreading of a flashlight beam that is originally 5.00 cm in diameter with an average wavelength of 600 nm.
Solution
\(\displaystyle 1.46 \times 10^{-5} rad\)
98. (a) What is the minimum angular spread of a 633-nm wavelength He-Ne laser beam that is originally 1.00 mm in diameter?
(b) If this laser is aimed at a mountain cliff 15.0 km away, how big will the illuminated spot be?
(c) How big a spot would be illuminated on the Moon, neglecting atmospheric effects? (This might be done to hit a corner reflector to measure the round-trip time and, hence, distance.) Explicitly show how you follow the steps in "Problem-Solving Strategies for Wave Optics."
99. 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?
(b) Neglecting atmospheric effects, what is the size of the spot this beam would make on the Moon, assuming a lunar distance of \(\displaystyle 3.84 \times 10^{8} m\)?
Solution
(a) \(\displaystyle 3.04 \times 10^{-7} rad\)
(b) diameter of 235 m
100. The limit to the eye’s acuity is actually related to diffraction by the pupil.
(a) What is the angle between two just-resolvable points of light for a 3.00-mm-diameter pupil, assuming an average wavelength of 550 nm?
(b) Take your result to be the practical limit for the eye. What is the greatest possible distance a car can be from you if you can resolve its two headlights, given they are 1.30 m apart?
(c) What is the distance between two just-resolvable points held at an arm’s length (0.800 m) from your eye?
(d) How does your answer to (c) compare to details you normally observe in everyday circumstances?
101. What is the minimum diameter mirror on a telescope that would allow you to see details as small as 5.00 km on the Moon some 384,000 km away? Assume an average wavelength of 550 nm for the light received.
Solution
5.15 cm
102.  If the eyes are separated by 6.5 cm and the diameter of your pupil is 5.0 mm, at what distance can you resolve the two eyes using light of wavelength 555 nm?
103. (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 \(\displaystyle 4.50 \times 10^{9} km\) 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?
Solution
(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.
104. The headlights of a car are 1.3 m apart. What is the maximum distance at which the eye can resolve these two headlights? Take the pupil diameter to be 0.40 cm.
105. When dots are placed on a page from a laser printer, they must be close enough so that you do not see the individual dots of ink. To do this, the separation of the dots must be less than Raleigh’s criterion. Take the pupil of the eye to be 3.0 mm and the distance from the paper to the eye of 35 cm; find the minimum separation of two dots such that they cannot be resolved. How many dots per inch (dpi) does this correspond to?
106. Unreasonable Results:
An amateur astronomer wants to build a telescope with a diffraction limit that will allow him to see if there are people on the moons of Jupiter.
a) What diameter mirror is needed to be able to see 1.00 m detail on a Jovian Moon at a distance of \(\displaystyle 7.50 \times 10^{8} km\) from Earth? The wavelength of light averages 600 nm.
(b) What is unreasonable about this result?
(c) Which assumptions are unreasonable or inconsistent?
107. Construct Your Own Problem:
Consider diffraction limits for an electromagnetic wave interacting with a circular object. Construct a problem in which you calculate the limit of angular resolution with a device, using this circular object (such as a lens, mirror, or antenna) to make observations. Also calculate the limit to spatial resolution (such as the size of features observable on the Moon) for observations at a specific distance from the device. Among the things to be considered are the wavelength of electromagnetic radiation used, the size of the circular object, and the distance to the system or phenomenon being observed.
27.7: Thin Film Interference
108. A soap bubble is 100 nm thick and illuminated by white light incident perpendicular to its surface. What wavelength and color of visible light is most constructively reflected, assuming the same index of refraction as water?
Solution
532 nm (green)
109. An oil slick on water is 120 nm thick and illuminated by white light incident perpendicular to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is 1.40?
110. Calculate the minimum thickness of an oil slick on water that appears blue when illuminated by white light perpendicular to its surface. Take the blue wavelength to be 470 nm and the index of refraction of oil to be 1.40.
Solution
83.9 nm
111. Find the minimum thickness of a soap bubble that appears red when illuminated by white light perpendicular to its surface. Take the wavelength to be 680 nm, and assume the same index of refraction as water.
112. A film of soapy water (\(\displaystyle n=1.33\)) on top of a plastic cutting board has a thickness of 233 nm. What color is most strongly reflected if it is illuminated perpendicular to its surface?
Solution
620 nm (orange)
113. What are the three smallest non-zero thicknesses of soapy water (\(\displaystyle n=1.33\)) on Plexiglas if it appears green (constructively reflecting 520-nm light) when illuminated perpendicularly by white light? Explicitly show how you follow the steps in Problem Solving Strategies for Wave Optics.
114 . Suppose you have a lens system that is to be used primarily for 700-nm red light. What is the second thinnest coating of fluorite (magnesium fluoride) that would be non-reflective for this wavelength?
Solution
380 nm
115. (a) As a soap bubble thins it becomes dark, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the bubble can be and appear dark at all visible wavelengths? Assume the same index of refraction as water.
(b) Discuss the fragility of the film considering the thickness found.
116. A film of oil on water will appear dark when it is very thin, because the path length difference becomes small compared with the wavelength of light and there is a phase shift at the top surface. If it becomes dark when the path length difference is less than one-fourth the wavelength, what is the thickest the oil can be and appear dark at all visible wavelengths? Oil has an index of refraction of 1.40.
Solution
33.9 nm
117. Figure shows two glass slides illuminated by pure-wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on a 0.100-mm-diameter hair at the other end, forming a wedge of air.
(a) How far apart are the dark bands, if the slides are 7.50 cm long and 589-nm light is used?
(b) Is there any difference if the slides are made from crown or flint glass? Explain.
118. Figure in Exercise 117 shows two 7.50-cm-long glass slides illuminated by pure 589-nm wavelength light incident perpendicularly. The top slide touches the bottom slide at one end and rests on some debris at the other end, forming a wedge of air. How thick is the debris, if the dark bands are 1.00 mm apart?
Solution
\(\displaystyle 4.42×10^{−5}m\)
119. Repeat Exercise, but take the light to be incident at a 45º angle.
120. Repeat Exercise, but take the light to be incident at a 45º angle.
Solution
The oil film will appear black, since the reflected light is not in the visible part of the spectrum.
121. Unreasonable Results
To save money on making military aircraft invisible to radar, an inventor decides to coat them with a non-reflective material having an index of refraction of 1.20, which is between that of air and the surface of the plane. This, he reasons, should be much cheaper than designing Stealth bombers.
(a) What thickness should the coating be to inhibit the reflection of 4.00-cm wavelength radar?
(b) What is unreasonable about this result?
(c) Which assumptions are unreasonable or inconsistent?
27.8: Polarization
122. What angle is needed between the direction of polarized light and the axis of a polarizing filter to cut its intensity in half?
Solution
45.0º
123. The angle between the axes of two polarizing filters is 45.0º . By how much does the second filter reduce the intensity of the light coming through the first?
124. If you have completely polarized light of intensity \(\displaystyle 150 W/m^2\), what will its intensity be after passing through a polarizing filter with its axis at an \(\displaystyle 89.0º\) angle to the light’s polarization direction?
Solution
\(\displaystyle 45.7mW/m^2\)
125. What angle would the axis of a polarizing filter need to make with the direction of polarized light of intensity \(\displaystyle 1.00kW/m^2\) to reduce the intensity to \(\displaystyle 10.0W/m^2\)?
126. At the end of Example, it was stated that the intensity of polarized light is reduced to 90.0% of its original value by passing through a polarizing filter with its axis at an angle of 18.4º to the direction of polarization. Verify this statement.
Solution
90.0%
127. Show that if you have three polarizing filters, with the second at an angle of 45º to the first and the third at an angle of 90.0º to the first, the intensity of light passed by the first will be reduced to 25.0% of its value. (This is in contrast to having only the first and third, which reduces the intensity to zero, so that placing the second between them increases the intensity of the transmitted light.)
128. Prove that, if \(\displaystyle I\) is the intensity of light transmitted by two polarizing filters with axes at an angle \(\displaystyle θ\) and \(\displaystyle I'\) is the intensity when the axes are at an angle \(\displaystyle 90.0º−θ\), then \(\displaystyle I+I'=I_0\), the original intensity. (Hint: Use the trigonometric identities \(\displaystyle cos(90.0º−θ)=sinθ \) and \(\displaystyle cos^2θ+sin^2θ=1\).)
Solution
\(\displaystyle I_0\)
129. At what angle will light reflected from diamond be completely polarized?
130. What is Brewster’s angle for light traveling in water that is reflected from crown glass?
Solution
\(\displaystyle 48.8º\)
131. A scuba diver sees light reflected from the water’s surface. At what angle will this light be completely polarized?
132. At what angle is light inside crown glass completely polarized when reflected from water, as in a fish tank?
Solution
\(\displaystyle 41.2º\)
133. Light reflected at \(\displaystyle 55.6º\) from a window is completely polarized. What is the window’s index of refraction and the likely substance of which it is made?
134. (a) Light reflected at \(\displaystyle 62.5º\) from a gemstone in a ring is completely polarized. Can the gem be a diamond?
(b) At what angle would the light be completely polarized if the gem was in water?
Solution
(a) 1.92, not diamond (Zircon)
(b) \(\displaystyle 55.2º\)
135. If \(\displaystyle θ_b\) is Brewster’s angle for light reflected from the top of an interface between two substances, and \(\displaystyle θ'b\) is Brewster’s angle for light reflected from below, prove that \(\displaystyle θ_b+θ'_b=90.0º\).
136. Integrated Concepts
If a polarizing filter reduces the intensity of polarized light to \(\displaystyle 50.0%\) of its original value, by how much are the electric and magnetic fields reduced?
Solution
\(\displaystyle B_2=0.707B_1\)
137. Integrated Concepts
Suppose you put on two pairs of Polaroid sunglasses with their axes at an angle of \(\displaystyle 15.0º\). How much longer will it take the light to deposit a given amount of energy in your eye compared with a single pair of sunglasses? Assume the lenses are clear except for their polarizing characteristics.
138. Integrated Concepts
(a) On a day when the intensity of sunlight is \(\displaystyle 1.00kW/m^2\), a circular lens 0.200 m in diameter focuses light onto water in a black beaker. Two polarizing sheets of plastic are placed in front of the lens with their axes at an angle of \(\displaystyle 20.0º\). Assuming the sunlight is unpolarized and the polarizers are \(\displaystyle 100%\) efficient, what is the initial rate of heating of the water in \(\displaystyle ºC/s\), assuming it is \(\displaystyle 80.0%\) 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.
Solution
(a) \(\displaystyle 2.07×10^{-2} °C/s\)
(b) Yes, the polarizing filters get hot because they absorb some of the lost energy from the sunlight.
Contributors and Attributions
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Paul Peter Urone (Professor Emeritus at California State University, Sacramento) and Roger Hinrichs (State University of New York, College at Oswego) with Contributing Authors: Kim Dirks (University of Auckland) and Manjula Sharma (University of Sydney). This work is licensed by OpenStax University Physics under a Creative Commons Attribution License (by 4.0) .


5.7: Gauss’ Law - Differential Form
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The integral form of Gauss’ Law is a calculation of enclosed charge \(Q_{encl}\) using the surrounding density of electric flux:
\[\oint_{\mathcal S} {\bf D}\cdot d{\bf s} = Q_{encl} \label{m0045_eGLIF} \]
where \({\bf D}\) is electric flux density and \({\mathcal S}\) is the enclosing surface. It is also sometimes necessary to do the inverse calculation (i.e., determine electric field associated with a charge distribution). This is sometimes possible using Equation \ref{m0045_eGLIF} if the symmetry of the problem permits; see examples in Section 5.5 and 5.6. If the problem does not exhibit the necessary symmetry, then it seems that one must fall back to the family of techniques presented in Section 5.4 requiring direct integration over the charge, which is derived from Coulomb’s Law.
However, even the Coulomb’s Law / direct integration approach has a limitation that is very important to recognize: It does not account for the presence of structures that may influence the electric field. For example, the electric field due to a charge in free space is different from the electric field due to the same charge located near a perfectly-conducting surface. In fact, these approaches do not account for the possibility of any spatial variation in material composition, which rules out their use in many engineering applications.
To address this broader scope of problems, we require an alternative form of Gauss’ Law that applies at individual points in space. That is, we require Gauss’ Law expressed in the form of a differential equation, as opposed to an integral equation. This facilitates the use of Gauss’ Law even in problems that do not exhibit sufficient symmetry and that involve material boundaries and spatial variations in material constitutive parameters. Given this differential equation and the boundary conditions imposed by structure and materials, we may then solve for the electric field in these more complicated scenarios. In this section, we derive the desired differential form of Gauss’ Law. Elsewhere (in particular, in Section 5.15) we use this equation as a tool to find electric fields in problems involving material boundaries.
There are in fact two methods to develop the desired differential equation. One method is via the definition of divergence, whereas the other is via the divergence theorem. Both methods are presented below because each provides a different bit of insight. Let’s explore the first method:
Derivation via the Definition of Divergence
Let the geometrical volume enclosed by \({\mathcal S}\) be \({\mathcal V}\), which has volume \(V\) (units of m\(^3\)). Dividing both sides of Equation \ref{m0045_eGLIF} by \(V\) and taking the limit as \(V \to 0\):
\[\lim_{V \to 0} \frac{ \oint_{\mathcal S}{ {\bf D} \cdot d{\bf s} } }{V} = \lim_{V \to 0} \frac{ Q_{encl} }{V} \nonumber \]
The quantity on the right hand side is the volume charge density \(\rho_v\) (units of C/m\(^3\)) at the point at which we converge after letting the volume go to zero. The left hand side is, by definition, the divergence of \({\bf D}\), indicated in mathematical notation as “\(\nabla \cdot {\bf D}\)” (Section 4.6). Thus, we have Gauss’ Law in differential form :
\[\boxed{ \nabla \cdot {\bf D} = \rho_v } \label{m0045_eGLDF} \]
To interpret this equation, recall that divergence is simply the flux (in this case, electric flux) per unit volume.
Gauss’ Law in differential form (Equation \ref{m0045_eGLDF}) says that the electric flux per unit volume originating from a point in space is equal to the volume charge density at that point.
Derivation via the Divergence Theorem
Equation \ref{m0045_eGLDF} may also be obtained from Equation \ref{m0045_eGLIF} using the Divergence Theorem, which in the present case may be written:
\[\int_{\mathcal V} \left( \nabla \cdot {\bf D} \right) dv = \oint_{\mathcal S} {\bf D}\cdot d{\bf s} \nonumber \]
From Equation \ref{m0045_eGLIF}, we see that the right hand side of the equation may be replaced with the enclosed charge:
\[\int_{\mathcal V} \left( \nabla \cdot {\bf D} \right) dv = Q_{encl} \nonumber \]
Furthermore, the enclosed charge can be expressed as an integration of the volume charge density \(\rho_v\) over \({\mathcal V}\): \[\int_{\mathcal V} \left( \nabla \cdot {\bf D} \right) dv = \int_{\mathcal V}\rho_v dv \nonumber \] The above relationship must hold regardless of the specific location or shape of \({\mathcal V}\). The only way this is possible is if the integrands are equal. Thus, \(\nabla \cdot {\bf D}=\rho_v\), and we have obtained Equation \ref{m0045_eGLDF}.
Example \(\PageIndex{1}\): Determining the charge density at a point, given the associated electric field
The electric field intensity in free space is \[{\bf E}({\bf r}) = \hat{\bf x}A x^2 + \hat{\bf y}B z + \hat{\bf z}C x^2 z \nonumber \] where \(A=3 \: ~\mbox{V/m}^3\), \(B=2 \: ~\mbox{V/m}^2\), and \(C=1 \: ~\mbox{V/m}^4\). What is the charge density at \({\bf r} = \hat{\bf x}2 - \hat{\bf y}2\) m?
Solution
First, we use \({\bf D} = \epsilon {\bf E}\) to get \({\bf D}\). Since the problem is in free space, \(\epsilon = \epsilon_0\). Thus we have that the volume charge density is
\[\begin{aligned} \rho _ { v } & = \nabla \cdot \mathbf { D } \\ & = \nabla \cdot \left( \epsilon _ { 0 } \mathbf { E } \right) = \epsilon _ { 0 } \nabla \cdot \mathbf { E } \\ & = \epsilon _ { 0 } \left[ \frac { \partial } { \partial x } \left( A x ^ { 2 } \right) + \frac { \partial } { \partial y } ( B z ) + \frac { \partial } { \partial z } \left( C x ^ { 2 } z \right) \right] \\ & = \epsilon _ { 0 } \left[ 2 A x + 0 + C x ^ { 2 } \right] \end{aligned} \nonumber \]
Now calculating the charge density at the specified location \({\bf r}\):
\[\begin{array} { l } { \epsilon _ { 0 } \left[ 2 \left( 3 \: \mathrm { V } / \mathrm { m } ^ { 3 } \right) ( 2 \: \mathrm { m } ) + 0 + \left( 1 \: \mathrm { V } / \mathrm { m } ^ { 4 } \right) ( 2 \: \mathrm { m } ) ^ { 2 } \right] } \\ { = \epsilon _ { 0 } ( 16 \: \mathrm { V } / \mathrm { m } ) } \\ { = 142 \: \mathrm { pC } / \mathrm { m } ^ { 3 } } \end{array} \nonumber \]
To obtain the electric field from the charge distribution in the presence of boundary conditions imposed by materials and structure, we must enforce the relevant boundary conditions. These boundary conditions are presented in Sections 5.17 and 5.18. Frequently, a simpler approach requiring only the boundary conditions on the electric potential (\(V(\bf{r})\)) is possible; this is presented in Section 5.15.
Furthermore, the reader should note the following. Gauss’ Law does not always necessarily fully constrain possible solutions for the electric field. For that, we might also need Kirchoff’s Voltage Law; see Section 5.11.
Before moving on, it is worth noting that Equation \ref{m0045_eGLDF} can be solved in the special case in which there are no boundary conditions to satisfy; i.e., for charge only, in a uniform and unbounded medium. In fact, no additional electromagnetic theory is required to do this. Here’s the solution:
\[{\bf D}({\bf r}) = \frac{1}{4\pi} \int_{\mathcal V} { \frac{{\bf r}-{\bf r}'}{\left|{\bf r}-{\bf r}'\right|^3}~\rho_v({\bf r}')~dv} \nonumber \]
which we recognize as one of the results obtained in Sections 5.4 (after dividing both sides by \(\epsilon\) to get \({\bf E}\)). It is reasonable to conclude that Gauss’ Law (in either integral or differential form) is fundamental, whereas Coulomb’s Law is merely a consequence of Gauss’ Law.


11: Homogeneous Tax Rates
After completing this chapter, you should be able to: (1) construct effective tax rates on defending and challenging investments; (2) describe how effective tax rates depend on investment types; and (3) find effective tax rates on defending and challenging investments by calculating tax adjust coefficients.
To achieve your learning goals, you should complete the following objectives:
- Learn how to compute a defender’s after-tax IRR by finding its tax adjustment coefficient.
- Learn how capital gains alter a defending investment’s tax adjustment coefficient.
- Learn how to compute the tax adjustment coefficient for a single period.
- Learn how to compare the average effective tax rate for an investment.
- Learn how to find the tax adjustment coefficient for loans and investments with increasing (decreasing) cash flow.
- Learn how taxes and tax adjustment coefficients influence investment rankings.
Introduction
According to Benjamin Franklin, taxes are one of the two constants in life (the other one is death). Taxes need to be included in present value (PV) models because only what the firm earns and keeps after paying taxes really matters. Therefore, proper construction of PV models requires that taxes be consistently applied to defending and challenging investments.
A single-equation PV model compares two investments: a challenger, described by time-dated cash flow, and a defender, usually described by its internal rate of return (IRR) of its time-dated cash flow. Homogeneous tax rates require that the same units are used when describing the cash flow of the challenger as the cash flow used when finding the IRR of the defender. And in particular, homogeneous tax rates require that when the cash flow of the challenger are adjusted for taxes, the IRR of the defender must also be adjusted for taxes. This chapter shows how to find homogeneous tax rates in PV models.
It is popular to adjust a defender’s before-tax IRR to an after-tax IRR by multiplying it by (1 – T ), where T is the average tax rate applied to the investment. This chapter points out that this method for adjusting the defender’s IRR for taxes is only appropriate in a few special cases. Finally, this chapter shows how to find appropriate tax rates for adjusting the defender’s IRR for taxes in a variety of tax environments.
After-tax PV Models for Defenders and Challengers
The goal is to find a defender’s after-tax IRR. The first step is to find the defender’s before-tax IRR. Recall that the defender’s before-tax IRR is that rate of return such that the NPV of the defender’s before-tax net cash flow is zero. The defender’s after-tax IRR discounts the defender’s after-tax cash flow so that its NPV remains equal to zero.
There is an easy test to determine if taxes have been properly introduced into a PV model. First consider the defender’s cash flow stream and its IRR. Then introduce taxes into the defender’s cash flow stream and its IRR. If taxes have been properly introduced, then the NPV of the defender’s after-tax cash flow stream discounted by its after-tax IRR is still zero.
To illustrate, consider a defending investment that earns constant net cash flow of R dollars per period in perpetuity. The maximum bid (minimum sell) price V 0 for this defending investment whose before-tax IRR is r . We describe this investment in Equation \ref{11.1}.
(11.1)
where r = R / V 0 is the before-tax IRR. Now suppose that we describe the defender’s cash flow and the IRR for this same investment on an after-tax basis. We do so by introducing the constant marginal income tax rate T into the model:
(11.2)
Income R and the before-tax IRR of the defender are both adjusted for taxes by multiplying by (1 – T ). We can be sure that r (1 – T ) is the after-tax IRR, since V 0 is the same in the before and after-tax models. However, we need to be aware that only in the special case of constant infinite income can we multiply r by (1 – T ) to obtain the after-tax IRR.
A General Approach for Finding the Defender’s After-tax IRR
A general approach for finding the defender’s after-tax IRR follows. Consider a defender that has a market-determined value of V 0 and earns a before-tax cash flow stream of R 1 , R 2 , …, and whose defender’s before-tax IRR is r . A before-tax PV model for this defender can be written as:
(11.3)
Now we express the PV model for the defender one period later as:
(11.4)
Note that the bracketed expression in Equation \ref{11.3} equals the right-hand side of Equation \ref{11.4}. This equality allows us to substitute V1 for the bracketed expression in Equation \ref{11.3}. After making the substitution in Equation \ref{11.3} we solve for capital gains equal to:
(11.5)
Using the expression in Equation \ref{11.5} we can solve for the defender’s before-tax IRR equal to:
(11.6)
We now want to find the after-tax IRR for the defender described in Equation \ref{11.6}. To do so, the IRR and the cash flow for the defender must be adjusted for taxes in such a way that V 0 in Equation \ref{11.3} is not changed (and the defender’s NPV is still zero). The defender’s cash flow is adjusted for taxes by multiplying them by (1 – T ). The defender’s before-tax IRR is adjusted for taxes by multiplying it by where , a tax adjustment coefficient, adjusts r to its after-tax equivalent such that V 0 is the same whether calculated on a before-tax or after-tax basis. Besides these tax adjustments, let T j equal other taxes applied to the defender’s cash flow that may include property taxes in the case of land, depreciation tax credits in the case of depreciable investments, and capital gains taxes when appropriate for assets earning capital gains. What all these taxes have in common is their functional dependence on the previous period’s asset value.
The after-tax PV model corresponding to Equation \ref{11.3} that leaves V 0 unchanged can be written as:
(11.7)
Similarly, V 1 can be expressed as:
(11.8)
Replacing the bracketed expression in Equation \ref{11.7} with V 1 , we find the after-tax IRR for the defender as:
(11.9)
Finally, substituting for r in Equation \ref{11.9} from the right-hand side of Equation \ref{11.6} and solving for , we obtain:
(11.10)
The value for in Equation \ref{11.10} adjusts the defender’s before-tax IRR to obtain homogeneity of measures between the defender’s after-tax IRR and the defender’s after-tax cash flow in Equation \ref{11.7}. When homogeneity of measures is maintained, the defender’s NPV is still zero whether calculated on a before-tax or after-tax basis. We now find some specific after-tax IRRs for various types of defenders.
Case 11.1. and . In this case, the defender earns neither capital gains nor suffers capital losses, in which case T j = 0 . This case is illustrated by an infinite constant cash flow series, described in equations 11.1 and 11.2. Because the cash flow is constant, capital gains are zero. Furthermore, the defender’s return in each period is equal to its cash receipts, which is fully taxed at income tax rate T . Therefore, the entire before-tax rate of return must be adjusted by (1 – T ). In Equation \ref{11.10}, substituting zero for capital gains and depreciation or capital gains tax results in:
(11.11)
Case 11.2. 0" title="Rendered by QuickLaTeX.com" height="18" width="103" style="vertical-align: -4px;"> and . In this case the investment earns capital gains that are not taxed, which lowers the defender’s effective tax rate. Thus, and . The greater the capital gains, the lower the effective tax rate in period t. We illustrate this type of model next.
Consider a defending investment whose before-tax cash flow grows geometrically at rate g . Then before-tax cash flow in period t , R t , equals: R 0 (1 + g ) t , and we write the investment’s IRR model as:
(11.12)
One period later, we can write:
(11.13)
and capital gains equal:
(11.14)
Then, substituting the right-hand side of Equation \ref{11.12} for V 0 in Equation \ref{11.14}, we obtain:
(11.15)
Now we are ready to find the tax-adjustment coefficient in this model. Substituting into Equation \ref{11.10} for capital gains and first period cash flow, we obtain:
(11.16)
Suppose that we naively introduced taxes into the growth model described in Equation \ref{11.12} and obtained:
(11.17)
In this case, an increase in T increases V 0 :
(11.18) 0. \end{equation*} " title="Rendered by QuickLaTeX.com">
Since increasing T increases V 0 , we know that r (1 – T ) cannot equal the after-tax IRR for the defending investment.
If, for example, r = 8%, g = 3%, then
(11.19)
Furthermore, for an investor in the 35% tax bracket whose cash flow is increasing at 3%, the effective tax rate would be .
Case 11.3. and . In this case the investment suffers capital losses for which no tax savings are allowed, which increases the defender’s effective tax rate. Thus, 1" title="Rendered by QuickLaTeX.com" height="13" width="40" style="vertical-align: -1px;"> and (1 - T)" title="Rendered by QuickLaTeX.com" height="18" width="145" style="vertical-align: -4px;">. The greater the capital losses, the higher is the effective tax rate in period t . We illustrate this type of model next.
Consider a defending investment whose before-tax cash flow declines geometrically at rate d > 0. Then before-tax cash flow in period t , R t , equals: , and we write the investment’s maximum bid price as:
(11.20)
One period later, we can write:
(11.21)
and capital losses equal:
(11.22)
Then, substituting the right-hand side of Equation \ref{11.20} for V 0 in Equation \ref{11.22} we obtain:
(11.23)
Now we are ready to find the tax-adjustment coefficient in this model. Substituting into Equation \ref{11.10} for capital losses and first period cash flow, we obtain:
(11.24)
If for example, r = 8%, d = 3%, then,
(11.25)
Furthermore, for an investor in the 35% tax bracket, the effective tax rate on cash flow described by a geometric growth pattern would be The conclusion of Case 11.3 is that suffering losses on investments which are not used to shield income from taxes increases the investor’s effective tax rate.
Case 11.4. 0 " title="Rendered by QuickLaTeX.com" height="18" width="103" style="vertical-align: -4px;"> and In this model, the defending investment earns capital gains but pays property tax on its previous period’s value times the property tax rate T p . However, property taxes are tax deductible, so we can write the after-tax IRR model for the defending land investment as:
(11.26)
We can sum the IRR model above by recognizing that it consists of two geometric sums. However, after summing, V 0 appears on both sides of the equation. Therefore, we solve for V 0 using Equation \ref{11.26} and obtain:
(11.27)
Capital gains and V 0 in this model are the same as in Case 11.2, which allows us to write:
(11.28)
Consider the effect of property taxes compared to the geometric model without property taxes. In Case 11.2, we assumed that r = 8% and g = 3%, so that . We continue with the assumption that the decision maker is in the 35% tax bracket and add the assumption that the property tax rate is 2%. We now substitute into Equation \ref{11.28} and find:
(11.29)
Case 11.5. 0" title="Rendered by QuickLaTeX.com" height="18" width="103" style="vertical-align: -4px;"> and A most fortuitous tax environment is an investment that appreciates but is considered to depreciate at rate d for tax purposes. In this case the investor is not required to pay taxes on the capital gains and is allowed a tax credit for depreciation that occurs only on the books. An example of such a model is the following:
(11.30)
Because the before-tax model is the geometric growth model, we can write:
(11.31)
To illustrate, suppose that d = 5%. Then, using numbers from our previous example, we substitute into Equation \ref{11.31} and find:
(11.32)
In other words, the tax rate for the investment is effectively zero.
Tax Adjustment Coefficients in Finite Models
So far we have obtained tax adjustment coefficient models for infinite time examples. As a result, the tax adjustment coefficients have been the same for each period. This approach has been convenient for exposition purposes. However, this result is not generally applicable. In practice, the tax adjustment coefficients vary by time period. We demonstrate using a finite time horizon model, a loan model in which the interest paid is tax deductible.
In many applications, the defender is a loan. When a loan is the defender, we immediately want to know if the interest paid is tax deductible. To keep our analysis simple, assume a constant-payment loan made for n periods at an interest rate of i percent. In which case, loan amount L 0 at interest rate i with annuity payments A can be written as:
(11.33)
And the firm’s before-tax IRR is i . The loan balance after one period can be written as:
(11.34)
Furthermore, capital loss for loans can be expressed as:
(11.35)
The after-tax IRR model can be expressed as:
(11.36)
With Equation \ref{11.36} in a familiar form, we can write as:
(11.37)
In words, the effective after-tax IRR for a loan whose interest is tax-deductible is (1 – T ).
Ranking Investments and Taxes
In this section, we acknowledge that the effective tax rates vary by the different types of investments being considered. Furthermore, if the effective tax rate for defenders and challengers differ, then before and after-tax investment rankings may be changed. To make this point, consider the following example. Assume the defender’s before-tax IRR is 8%, the income tax rate is 35%, the rate of growth in net cash flow is 3%, and the last year’s before-tax net cash flow was $100. The maximum bid price for this investment with geometric growing income is:
(11.38)
If the minimum sell price is $3,000, then the NPV for the investment is a negative $940 and the challenger is rejected.
Now suppose we introduce taxes into the model and recalculate the investment’s NPV. To simplify we assume an infinite life for the investment and assume for the defender and find:
(11.39)
Now the challenger’s NPV is positive. Introducing taxes into the NPV model has reversed the ranking: the challenger is now preferred to the defender. These results should alert financial managers to the importance of after-tax rankings.
Summary and Conclusions
Taxes are important when valuing and ranking investments because what really matters is the amount of your earnings you get to keep after paying taxes. It’s not the before-tax IRR that matters but the after-tax IRR that counts because that’s the rate you really earned on your defending investment.
In this chapter, we have introduced a method for finding a defender’s after-tax IRR. The method required that when a defender’s cash flow adjusted for taxes and was discounted by its after-tax IRR its change in NPV was zero. Then we used the equality to find the after-tax discount rate IRR. The implied tax rate in the after-tax IRR was compared to the firm’s average marginal income tax-rate.
While the usual practice is to multiply the defender’s before-tax IRR by (1 – T ) to obtain the defender’s after-tax IRR, this chapter demonstrated that this cannot be generally relied on to obtain the defender’s properly adjusted after-tax IRR. Indeed, we demonstrated that capital gains that are not taxed can lower the effective tax rate. Capital losses that do not create tax shields can increase the effective tax rate. Property tax paid on land and other real property increases the effective tax rate paid. Allowing investors to write off an investment’s depreciation lowers the effective tax rate.
Finally, this chapter has employed simplifying assumptions, such as large values for n and average depreciation and growth rates, to find effective after-tax discount rates that have nice, closed-form solutions. Usually, this is not the case. It is often more difficult to find effective after-tax discount rates, and often these are not closed-form solutions.
Questions
- Describe the appropriate test for determining whether or not taxes have been properly introduced into the defender’s IRR.
- When finding after-tax IRRs for defenders, we solve for and claim that is the defender’s effective tax rate. Interpret the meaning of .
- Under what condition is the defender’s effective tax rate equal to its income tax rate T or that ?
- Assume that a defending investment’s price is V 0 , that d is the defender’s book value depreciation rate, that R is the constant stream of income earned by the defender, that T is the income tax rate, and that r is the defender’s before-tax IRR. Also assume that the defender is allowed to write off book value depreciation even though its income stream is constant. We write such a model as:
(Q11.1)
In this model, the depreciation is constant for 100/d periods beyond which the discounted cash flow is small enough to be ignored. The before-tax IRR model can be written as:
(Q11.2)
Solve for in the after-tax IRR model by substituting for V 0 the right hand side of the second model R / r . Next, solve for a numerical value for assuming that d = 3% and r = 8%.
- Suppose that you have found the NPV for a challenger. What impact will an increase in the defender’s effective tax rate have on the challenger’s IRR? Please explain your answers.
- Calculate the effective tax rate for a depreciating challenger where d is the depreciation rate, T is the income tax rate, r is the defender’s before-tax IRR, and V 0 is the price of defender. The IRR model is written below. (Hint: find by setting the right hand side of the equation below to the equivalent model without taxes ( T = 0) and solve for .
(Q11.3)
- Discuss the following. Suppose that cash flow was constant ( R t = R 0 ), yet tax laws allowed the owner of the durable to claim depreciation at rate d . Without solving for , can you deduce whether it would be equal to, less than, or greater than one? Defend your answer.
- Assume you are investing in land and that you borrow money to finance its purchase. Also assume that you pay property taxes on the land. In other words, land is the challenger and the loan is the defender. Which investment has the higher effective tax rate under three scenarios: g = 0, g > 0, and g < 0? What do you need to know about the property tax rate T p to answer this question?
- In this chapter, five different investment tax scenarios were described. Please provide an example of each type of investment and how the tax scenario might change its rankings compared to before-tax rankings.


Learning Objectives
By the end of this section, you will be able to:
- Describe the obstacles that sperm must overcome to reach an oocyte
- Explain capacitation and its importance in fertilization
- Summarize the events that occur as a sperm fertilizes an oocyte
Fertilization occurs when a sperm and an oocyte (egg) combine and their nuclei fuse. Because each of these reproductive cells is a haploid cell containing half of the genetic material needed to form a human being, their combination forms a diploid cell. This new single cell, called a zygote , contains all of the genetic material needed to form a human—half from each biological parent.
Transit of Sperm
Fertilization is a numbers game. During ejaculation, hundreds of millions of sperm (spermatozoa) are released into the vagina. Almost immediately, millions of these sperm are overcome by the acidity of the vagina (approximately pH 3.8), and millions more may be blocked from entering the uterus by thick cervical mucus. Of those that do enter, thousands are destroyed by phagocytic uterine leukocytes. Thus, the race into the uterine tubes, which is the most typical site for sperm to encounter the oocyte, is reduced to a few thousand contenders. Their journey—thought to be facilitated by uterine contractions—usually takes from 30 minutes to 2 hours. If the sperm do not encounter an oocyte immediately, they can survive in the uterine tubes for another 3–5 days. Thus, fertilization can still occur if intercourse takes place a few days before ovulation. In comparison, an oocyte can survive independently for only approximately 24 hours following ovulation. Intercourse more than a day after ovulation will therefore usually not result in fertilization.
During the journey, fluids in the female reproductive tract prepare the sperm for fertilization through a process called capacitation , or priming. The fluids improve the motility of the spermatozoa. They also deplete cholesterol molecules embedded in the membrane of the head of the sperm, thinning the membrane in such a way that will help facilitate the release of the lysosomal (digestive) enzymes needed for the sperm to penetrate the oocyte’s exterior once contact is made. Sperm must undergo the process of capacitation in order to have the “capacity” to fertilize an oocyte. If they reach the oocyte before capacitation is complete, they will be unable to penetrate the oocyte’s thick outer layer of cells.
Contact Between Sperm and Oocyte
Upon ovulation, the oocyte released by the ovary is swept into—and along—the uterine tube. Fertilization must occur in the distal uterine tube because an unfertilized oocyte cannot survive the 72-hour journey to the uterus. As you will recall from your study of the oogenesis, this oocyte (specifically a secondary oocyte) is surrounded by two protective layers. The corona radiata is an outer layer of follicular (granulosa) cells that form around a developing oocyte in the ovary and remain with it upon ovulation. The underlying zona pellucida (pellucid = “transparent”) is a transparent, but thick, glycoprotein membrane that surrounds the cell’s plasma membrane.
As it is swept along the distal uterine tube, the oocyte encounters the surviving capacitated sperm, which stream toward it in response to chemical attractants released by the cells of the corona radiata. To reach the oocyte itself, the sperm must penetrate the two protective layers. The sperm first burrow through the cells of the corona radiata. Then, upon contact with the zona pellucida, the sperm bind to receptors in the zona pellucida. This initiates a process called the acrosomal reaction in which the enzyme-filled “cap” of the sperm, called the acrosome , releases its stored digestive enzymes. These enzymes clear a path through the zona pellucida that allows sperm to reach the oocyte. Finally, a single sperm makes contact with sperm-binding receptors on the oocyte’s plasma membrane (Figure \(\PageIndex{1}\)). The plasma membrane of that sperm then fuses with the oocyte’s plasma membrane, and the head and mid-piece of the “winning” sperm enter the oocyte interior.
How do sperm penetrate the corona radiata? Some sperm undergo a spontaneous acrosomal reaction, which is an acrosomal reaction not triggered by contact with the zona pellucida. The digestive enzymes released by this reaction digest the extracellular matrix of the corona radiata. As you can see, the first sperm to reach the oocyte is never the one to fertilize it. Rather, hundreds of sperm cells must undergo the acrosomal reaction, each helping to degrade the corona radiata and zona pellucida until a path is created to allow one sperm to contact and fuse with the plasma membrane of the oocyte. If you consider the loss of millions of sperm between entry into the vagina and degradation of the zona pellucida, you can understand why a low sperm count can cause male infertility.
When the first sperm fuses with the oocyte, the oocyte deploys two mechanisms to prevent polyspermy , which is penetration by more than one sperm. This is critical because if more than one sperm were to fertilize the oocyte, the resulting zygote would be a triploid organism with three sets of chromosomes. This is incompatible with life.
The first mechanism is the fast block, which involves a near instantaneous change in sodium ion permeability upon binding of the first sperm, depolarizing the oocyte plasma membrane and preventing the fusion of additional sperm cells. The fast block sets in almost immediately and lasts for about a minute, during which time an influx of calcium ions following sperm penetration triggers the second mechanism, the slow block. In this process, referred to as the cortical reaction , cortical granules sitting immediately below the oocyte plasma membrane fuse with the membrane and release zonal inhibiting proteins and mucopolysaccharides into the space between the plasma membrane and the zona pellucida. Zonal inhibiting proteins cause the release of any other attached sperm and destroy the oocyte’s sperm receptors, thus preventing any more sperm from binding. The mucopolysaccharides then coat the nascent zygote in an impenetrable barrier that, together with hardened zona pellucida, is called a fertilization membrane .
The Zygote
Recall that at the point of fertilization, the oocyte has not yet completed meiosis; all secondary oocytes remain arrested in metaphase of meiosis II until fertilization. Only upon fertilization does the oocyte complete meiosis. The unneeded complement of genetic material that results is stored in a second polar body that is eventually ejected. At this moment, the oocyte has become an ovum, the female haploid gamete. The two haploid nuclei derived from the sperm and oocyte and contained within the egg are referred to as pronuclei. They decondense, expand, and replicate their DNA in preparation for mitosis. The pronuclei then migrate toward each other, their nuclear envelopes disintegrate, and the male- and female-derived genetic material intermingles. This step completes the process of fertilization and results in a single-celled diploid zygote with all the genetic instructions it needs to develop into a human.
Most of the time, a person releases a single egg during an ovulation cycle. However, in approximately 1 percent of ovulation cycles, two eggs are released and both are fertilized. Two zygotes form, implant, and develop, resulting in the birth of dizygotic (or fraternal) twins. Because dizygotic twins develop from two eggs fertilized by two sperm, they are no more identical than siblings born at different times.
Much less commonly, a zygote can divide into two separate offspring during early development. This results in the birth of monozygotic (or identical) twins. Although the zygote can split as early as the two-cell stage, splitting occurs most commonly during the early blastocyst stage, with roughly 70–100 cells present. These two scenarios are distinct from each other, in that the twin embryos that separated at the two-cell stage will have individual placentas, whereas twin embryos that form from separation at the blastocyst stage will share a placenta and a chorionic cavity.
Everyday Connection
In Vitro Fertilization
IVF, which stands for in vitro fertilization, is an assisted reproductive technology. In vitro, which in Latin translates to “in glass,” refers to a procedure that takes place outside of the body. There are many different indications for IVF. The procedure is used frequently in supporting LGBTQ people, people using surrogates/gestational carriers, and also people who have difficulties reproducing. For example, a person may produce healthy eggs, but the eggs cannot reach the uterus because the uterine tubes are blocked or otherwise compromised. A person may have a low sperm count, low sperm motility, sperm with an unusually high percentage of morphological abnormalities, or sperm that are incapable of penetrating the zona pellucida of an egg.
A typical IVF procedure begins with egg collection. A normal ovulation cycle produces only one oocyte, but the number can be boosted significantly (to 10–20 oocytes) by administering a short course of gonadotropins. The course begins with follicle-stimulating hormone (FSH) analogs, which support the development of multiple follicles, and ends with a luteinizing hormone (LH) analog that triggers ovulation. Right before the ova would be released from the ovary, they are harvested using ultrasound-guided oocyte retrieval. In this procedure, ultrasound allows a physician to visualize mature follicles. The ova are aspirated (sucked out) using a syringe.
In parallel, sperm are obtained from the male partner or from a sperm bank. The sperm are prepared by washing to remove seminal fluid because seminal fluid contains a peptide, FPP (or, fertilization promoting peptide), that—in high concentrations—prevents capacitation of the sperm. The sperm sample is also concentrated, to increase the sperm count per milliliter.
Next, the eggs and sperm are mixed in a petri dish. The ideal ratio is 75,000 sperm to one egg. If there are severe problems with the sperm—for example, the count is exceedingly low, or the sperm are completely nonmotile, or incapable of binding to or penetrating the zona pellucida—a sperm can be injected into an egg. This is called intracytoplasmic sperm injection (ICSI).
The embryos are then incubated until they either reach the eight-cell stage or the blastocyst stage. In the United States, fertilized eggs are typically cultured to the blastocyst stage because this results in a higher pregnancy rate. Finally, the embryos are transferred to a woman’s uterus using a plastic catheter (tube). Figure 28.3 illustrates the steps involved in IVF.
IVF is a relatively new and still evolving technology, and until recently it was necessary to transfer multiple embryos to achieve a good chance of a pregnancy. Today, however, transferred embryos are much more likely to implant successfully, so countries that regulate the IVF industry cap the number of embryos that can be transferred per cycle at two. This reduces the risk of multiple-birth pregnancies.
The rate of success for IVF is correlated with the age of the gestational carrier (which may be the child's mother or a surrogate). More than 40 percent of females under 35 succeed in giving birth following IVF, but the rate drops to a little over 10 percent in females over 40.
Interactive Link
Go to this site to view resources covering various aspects of fertilization, including movies and animations showing sperm structure and motility, ovulation, and fertilization.


2.4.2: pH, Buffers, Acids, and Bases
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Learning Objectives
- Explain the composition of buffer solutions and how they maintain a steady pH
Self-Ionization of Water
Hydrogen ions are spontaneously generated in pure water by the dissociation (ionization) of a small percentage of water molecules into equal numbers of hydrogen (H + ) ions and hydroxide (OH – ) ions. The hydroxide ions remain in solution because of their hydrogen bonds with other water molecules; the hydrogen ions, consisting of naked protons, are immediately attracted to un-ionized water molecules and form hydronium ions (H 3 0 + ). By convention, scientists refer to hydrogen ions and their concentration as if they were free in this state in liquid water.
2H2O⇋H3O++OH−2H2O⇋H3O++OH−
The concentration of hydrogen ions dissociating from pure water is 1 × 10 -7 moles H + ions per liter of water. The pH is calculated as the negative of the base 10 logarithm of this concentration:
pH = -log[H + ]
The negative log of 1 × 10 -7 is equal to 7.0, which is also known as neutral pH. Human cells and blood each maintain near-neutral pH.
pH Scale
The pH of a solution indicates its acidity or basicity (alkalinity). The pH scale is an inverse logarithm that ranges from 0 to 14: anything below 7.0 (ranging from 0.0 to 6.9) is acidic, and anything above 7.0 (from 7.1 to 14.0) is basic (or alkaline ). Extremes in pH in either direction from 7.0 are usually considered inhospitable to life. The pH in cells (6.8) and the blood (7.4) are both very close to neutral, whereas the environment in the stomach is highly acidic, with a pH of 1 to 2.
Non-neutral pH readings result from dissolving acids or bases in water. Using the negative logarithm to generate positive integers, high concentrations of hydrogen ions yield a low pH, and low concentrations a high pH.
An acid is a substance that increases the concentration of hydrogen ions (H + ) in a solution, usually by dissociating one of its hydrogen atoms. A base provides either hydroxide ions (OH – ) or other negatively-charged ions that react with hydrogen ions in solution, thereby reducing the concentration of H + and raising the pH.
Strong Acids and Strong Bases
The stronger the acid, the more readily it donates H + . For example, hydrochloric acid (HCl) is highly acidic and completely dissociates into hydrogen and chloride ions, whereas the acids in tomato juice or vinegar do not completely dissociate and are considered weak acids; conversely, strong bases readily donate OH – and/or react with hydrogen ions. Sodium hydroxide (NaOH) and many household cleaners are highly basic and give up OH – rapidly when placed in water; the OH – ions react with H + in solution, creating new water molecules and lowering the amount of free H + in the system, thereby raising the overall pH. An example of a weak basic solution is seawater, which has a pH near 8.0, close enough to neutral that well-adapted marine organisms thrive in this alkaline environment.
Buffers
How can organisms whose bodies require a near-neutral pH ingest acidic and basic substances (a human drinking orange juice, for example) and survive? Buffers are the key. Buffers usually consist of a weak acid and its conjugate base; this enables them to readily absorb excess H + or OH – , keeping the system’s pH within a narrow range.
Maintaining a constant blood pH is critical to a person’s well-being. The buffer that maintains the pH of human blood involves carbonic acid (H 2 CO 3 ), bicarbonate ion (HCO 3 – ), and carbon dioxide (CO 2 ). When bicarbonate ions combine with free hydrogen ions and become carbonic acid, hydrogen ions are removed, moderating pH changes. Similarly, excess carbonic acid can be converted into carbon dioxide gas and exhaled through the lungs; this prevents too many free hydrogen ions from building up in the blood and dangerously reducing its pH; likewise, if too much OH – is introduced into the system, carbonic acid will combine with it to create bicarbonate, lowering the pH. Without this buffer system, the body’s pH would fluctuate enough to jeopardize survival.
Antacids, which combat excess stomach acid, are another example of buffers. Many over-the-counter medications work similarly to blood buffers, often with at least one ion (usually carbonate) capable of absorbing hydrogen and moderating pH, bringing relief to those that suffer “heartburn” from stomach acid after eating.
Key Points
- A basic solution will have a pH above 7.0, while an acidic solution will have a pH below 7.0.
- Buffers are solutions that contain a weak acid and its a conjugate base; as such, they can absorb excess H + ions or OH – ions, thereby maintaining an overall steady pH in the solution.
- pH is equal to the negative logarithm of the concentration of H + ions in solution: pH = – log[H + ].
Key Terms
- alkaline : having a pH greater than 7; basic
- acidic : having a pH less than 7
- buffer : a solution composed of a weak acid and its conjugate base that can be used to stabilize the pH of a solution


1.8: The Human-Environment Relationship
The process of spatial diffusion can be profoundly affected by the physical terrain, such as is the case with a mountain range. Because migration and transportation over mountain ranges can be limited, diffusion can be slowed or even stopped by these physical barriers. This example is but one instance of the relationship between humans and their environment. The environment can significantly affect human activities, and vice versa, humans can shape and changethe Earth’s surface and its atmosphere. Two major perspectives on the humanenvironment relationship in the field of geography are environmental determinism, which has been largely rejected, and possibilism. Environmental determinism is the idea that the natural or physical environment shapes and creates cultures; in other words, the environment essentially dictates culture. For example, environmental determinists in the 1920s thought that people who lived in the tropics were slothful and backward because finding food in the tropics was thought to be rather easy. In contrast, Europeans, who lived in “stimulating” climates with a sharp change in seasons were more industrious and inventive. The racist undertones of this sort of perspective is clear, but modern geographers still recognize the definitive impact of the environment on societies, as can be seen, for example, in the theorized demise of the Anasazi people in the American Southwest because of extended drought ( Figure 1.14 ).
The Anasazi, who are believed to have inhabited in the Southwestern U.S. from 100 B.C. to 1300 A.D., were ill-equipped to deal with drought, compared to those who inhabit the modern-day states of New Mexico and Arizona. In the view of possibilism , people can adapt to their environmental conditions, despite the limitations they might pose, and if a society has better technology, the people are better able to adapt and develop their culture in a number of possible ways. The possibilities are greater, hence, the term possibilism. One excellent example of possibilism is found in Dubai, in the United Arab Emirates. Although snow skiing in the Middle East may seem preposterous, plans are in place to build the longest indoor ski slope in the world in this city, where one ski slope already exists ( Figure 1.15 ). Even technological advanced societies, however, can still be tremendously affected by the environment and have little or no control over the power of nature. The devastating impact of hurricanes in the United States, tsunamis in Japan, and fire in the United States are but a few examples ( Figures 1.16 and 1.17 ).


5.4: Active Transport
By the end of this section, you will be able to:
- Understand how electrochemical gradients affect ions
- Distinguish between primary active transport and secondary active transport
Active transport mechanisms require the use of the cell’s energy, usually in the form of adenosine triphosphate (ATP). If a substance must move into the cell against its concentration gradient—that is, if the concentration of the substance inside the cell is greater than its concentration in the extracellular fluid (and vice versa)—the cell must use energy to move the substance. Some active transport mechanisms move small-molecular weight materials, such as ions, through the membrane. Other mechanisms transport much larger molecules.
Electrochemical Gradient
We have discussed simple concentration gradients—differential concentrations of a substance across a space or a membrane—but in living systems, gradients are more complex. Because ions move into and out of cells and because cells contain proteins that do not move across the membrane and are mostly negatively charged, there is also an electrical gradient, a difference of charge, across the plasma membrane. The interior of living cells is electrically negative with respect to the extracellular fluid in which they are bathed, and at the same time, cells have higher concentrations of potassium (K + ) and lower concentrations of sodium (Na + ) than does the extracellular fluid. So in a living cell, the concentration gradient of Na + tends to drive it into the cell, and the electrical gradient of Na + (a positive ion) also tends to drive it inward to the negatively charged interior. The situation is more complex, however, for other elements such as potassium. The electrical gradient of K + , a positive ion, also tends to drive it into the cell, but the concentration gradient of K + tends to drive K + out of the cell (Figure \(\PageIndex{1}\)). The combined gradient of concentration and electrical charge that affects an ion is called its electrochemical gradient .
Moving Against a Gradient
To move substances against a concentration or electrochemical gradient, the cell must use energy. This energy is harvested from ATP generated through the cell’s metabolism. Active transport mechanisms, collectively called pumps , work against electrochemical gradients. Small substances constantly pass through plasma membranes. Active transport maintains concentrations of ions and other substances needed by living cells in the face of these passive movements. Much of a cell’s supply of metabolic energy may be spent maintaining these processes. (Most of a red blood cell’s metabolic energy is used to maintain the imbalance between exterior and interior sodium and potassium levels required by the cell.) Because active transport mechanisms depend on a cell’s metabolism for energy, they are sensitive to many metabolic poisons that interfere with the supply of ATP.
Two mechanisms exist for the transport of small-molecular weight material and small molecules. Primary active transport moves ions across a membrane and creates a difference in charge across that membrane, which is directly dependent on ATP. Secondary active transport describes the movement of material that is due to the electrochemical gradient established by primary active transport that does not directly require ATP.
Carrier Proteins for Active Transport
An important membrane adaption for active transport is the presence of specific carrier proteins or pumps to facilitate movement: there are three types of these proteins or transporters (Figure \(\PageIndex{2}\)). A uniporter carries one specific ion or molecule. A symporter carries two different ions or molecules, both in the same direction. An antiporter also carries two different ions or molecules, but in different directions. All of these transporters can also transport small, uncharged organic molecules like glucose. These three types of carrier proteins are also found in facilitated diffusion, but they do not require ATP to work in that process. Some examples of pumps for active transport are Na + -K + ATPase, which carries sodium and potassium ions, and H + -K + ATPase, which carries hydrogen and potassium ions. Both of these are antiporter carrier proteins. Two other carrier proteins are Ca 2+ ATPase and H + ATPase, which carry only calcium and only hydrogen ions, respectively. Both are pumps.
Primary Active Transport
The primary active transport that functions with the active transport of sodium and potassium allows secondary active transport to occur. The second transport method is still considered active because it depends on the use of energy as does primary transport (Figure \(\PageIndex{3}\)).
One of the most important pumps in animals cells is the sodium-potassium pump (Na + -K + ATPase), which maintains the electrochemical gradient (and the correct concentrations of Na + and K + ) in living cells. The sodium-potassium pump moves K + into the cell while moving Na + out at the same time, at a ratio of three Na + for every two K + ions moved in. The Na + -K + ATPase exists in two forms, depending on its orientation to the interior or exterior of the cell and its affinity for either sodium or potassium ions. The process consists of the following six steps.
- With the enzyme oriented towards the interior of the cell, the carrier has a high affinity for sodium ions. Three ions bind to the protein.
- ATP is hydrolyzed by the protein carrier and a low-energy phosphate group attaches to it.
- As a result, the carrier changes shape and re-orients itself towards the exterior of the membrane. The protein’s affinity for sodium decreases and the three sodium ions leave the carrier.
- The shape change increases the carrier’s affinity for potassium ions, and two such ions attach to the protein. Subsequently, the low-energy phosphate group detaches from the carrier.
- With the phosphate group removed and potassium ions attached, the carrier protein repositions itself towards the interior of the cell.
- The carrier protein, in its new configuration, has a decreased affinity for potassium, and the two ions are released into the cytoplasm. The protein now has a higher affinity for sodium ions, and the process starts again.
Several things have happened as a result of this process. At this point, there are more sodium ions outside of the cell than inside and more potassium ions inside than out. For every three ions of sodium that move out, two ions of potassium move in. This results in the interior being slightly more negative relative to the exterior. This difference in charge is important in creating the conditions necessary for the secondary process. The sodium-potassium pump is, therefore, an electrogenic pump (a pump that creates a charge imbalance), creating an electrical imbalance across the membrane and contributing to the membrane potential.
Secondary Active Transport (Co-transport)
Secondary active transport brings sodium ions, and possibly other compounds, into the cell. As sodium ion concentrations build outside of the plasma membrane because of the action of the primary active transport process, an electrochemical gradient is created. If a channel protein exists and is open, the sodium ions will be pulled through the membrane. This movement is used to transport other substances that can attach themselves to the transport protein through the membrane (Figure \(\PageIndex{4}\)). Many amino acids, as well as glucose, enter a cell this way. This secondary process is also used to store high-energy hydrogen ions in the mitochondria of plant and animal cells for the production of ATP. The potential energy that accumulates in the stored hydrogen ions is translated into kinetic energy as the ions surge through the channel protein ATP synthase, and that energy is used to convert ADP into ATP.
Summary
The combined gradient that affects an ion includes its concentration gradient and its electrical gradient. A positive ion, for example, might tend to diffuse into a new area, down its concentration gradient, but if it is diffusing into an area of net positive charge, its diffusion will be hampered by its electrical gradient. When dealing with ions in aqueous solutions, a combination of the electrochemical and concentration gradients, rather than just the concentration gradient alone, must be considered. Living cells need certain substances that exist inside the cell in concentrations greater than they exist in the extracellular space. Moving substances up their electrochemical gradients requires energy from the cell. Active transport uses energy stored in ATP to fuel this transport. Active transport of small molecular-sized materials uses integral proteins in the cell membrane to move the materials: These proteins are analogous to pumps. Some pumps, which carry out primary active transport, couple directly with ATP to drive their action. In co-transport (or secondary active transport), energy from primary transport can be used to move another substance into the cell and up its concentration gradient.
Review Questions
Active transport must function continuously because __________.
- plasma membranes wear out
- not all membranes are amphiphilic
- facilitated transport opposes active transport
- diffusion is constantly moving solutes in opposite directions
- Answer
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Answer:D
How does the sodium-potassium pump make the interior of the cell negatively charged?
- by expelling anions
- by pulling in anions
- by expelling more cations than are taken in
- by taking in and expelling an equal number of cations
- Answer
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Answer:C
What is the combination of an electrical gradient and a concentration gradient called?
- potential gradient
- electrical potential
- concentration potential
- electrochemical gradient
- Answer
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Answer:D
Free Response
Where does the cell get energy for active transport processes?
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Answer: The cell harvests energy from ATP produced by its own metabolism to power active transport processes, such as the activity of pumps.
How does the sodium-potassium pump contribute to the net negative charge of the interior of the cell?
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Answer: The sodium-potassium pump forces out three (positive) Na + ions for every two (positive) K + ions it pumps in, thus the cell loses a positive charge at every cycle of the pump.
Study Figure \(\PageIndex{1}\). Injection of a potassium solution into a person’s blood is lethal; this is used in capital punishment and euthanasia. Why do you think a potassium solution injection is lethal?
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Answer: Cells typically have a high concentration of potassium in the cytoplasm and are bathed in a high concentration of sodium. Injection of potassium dissipates this electrochemical gradient. In heart muscle, the sodium/potassium potential is responsible for transmitting the signal that causes the muscle to contract. When this potential is dissipated, the signal can’t be transmitted, and the heart stops beating. Potassium injections are also used to stop the heart from beating during surgery.
Study Figure \(\PageIndex{4}\). If the pH outside the cell decreases, would you expect the amount of amino acids transported into the cell to increase or decrease?
- Answer
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Answer: A decrease in pH means an increase in positively charged H + ions, and an increase in the electrical gradient across the membrane. The transport of amino acids into the cell will increase.
Glossary
- active transport
- method of transporting material that requires energy
- antiporter
- transporter that carries two ions or small molecules in different directions
- electrochemical gradient
- gradient produced by the combined forces of an electrical gradient and a chemical gradient
- electrogenic pump
- pump that creates a charge imbalance
- primary active transport
- active transport that moves ions or small molecules across a membrane and may create a difference in charge across that membrane
- pump
- active transport mechanism that works against electrochemical gradients
- secondary active transport
- movement of material that is due to the electrochemical gradient established by primary active transport
- symporter
- transporter that carries two different ions or small molecules, both in the same direction
- transporter
- specific carrier proteins or pumps that facilitate movement
- uniporter
- transporter that carries one specific ion or molecule
Contributors and Attributions
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Connie Rye (East Mississippi Community College), Robert Wise (University of Wisconsin, Oshkosh), Vladimir Jurukovski (Suffolk County Community College), Jean DeSaix (University of North Carolina at Chapel Hill), Jung Choi (Georgia Institute of Technology), Yael Avissar (Rhode Island College) among other contributing authors. Original content by OpenStax (CC BY 4.0; Download for free at<EMAIL_ADDRESS>).