Mastering Physics Solutions Chapter 28 Physical Optics: Interference and Diffraction
Chapter 28 Physical Optics: Interference and Diffraction Q.1CQ
When two light waves interfere destructively, what happens to their energy?
Solution:
When two waves interfere destructively at one place, then at some other place, these waves interfere constructively. The energy at the point of destructive interference at one place is always balanced by that at constructive interference. In destructive interference, the net energy of the resultant wave is less than the sum of energies of two individual waves, which interfere destructively to give destructive interference.
In constructive interference, the net energy of the resultant wave is more than the sum of energies of two individual waves which interfere constructively to give constructive interference. Thus, when two waves interfere destructively at one place, then the energy of individual waves at that place goes to the point where these waves constructively interfere. Thus, at a place of destructive interference, the energy is nearly zero, and at a place of constructive interference, energy is more than the sum of energies of individual waves. Hence, energy is redistributed from a place of destructive interference to a place of constructive interference.
Chapter 28 Physical Optics: Interference and Diffraction Q.1P
Two sources emit waves that are coherent, in phase, and have wavelengths of 26.0 m. Do the waves interfere constructively or destructively at an observation point 78.0 m from one source and 143 m from the other source?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.2CQ
What happens to the two-slit interference pattern if the separation between the slits is less than the wavelength of light?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.2P
Repeat Problem 1 for observation points that are
(a) 91.0 m and 221 m and
(b) 44.0 m and 135 m from the two sources.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.3CQ
If a radio station broadcasts its signal through two different antennas simultaneously, does this guarantee that the signal you receive will be stronger than from a single antenna? Explain.
Solution:
When a radio station broadcasts its signal through two different antennas may interfere destructively or constructively. Then the net intensity will be minimum or maximum. So the signals receiving from single antenna may be stronger than the signals receiving from the two antennas.
Chapter 28 Physical Optics: Interference and Diffraction Q.3P
Two sources emit waves that are in phase with each other. What is the longest wavelength that will give constructive interference at an observation point 161 m from one source and 295 m from the other source?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.4CQ
How would you expect the interference pattern of a two-slit experiment to change if white light is used instead of monochromatic light?
Solution:
In a two-slit interference pattern when a white light is used instead of monochromatic light then always the location of bright and dark fringes depends on the wavelength of light.
So each bright fringe formed on the screen splits up into seven different colors called ‘VIGB YOR’ (or) it also known as “rainbow” effect.
Chapter 28 Physical Optics: Interference and Diffraction Q.4P
A person driving at 17 m/s crosses the line connectingtwo radio transmitters at right angles, as shown in Figure 28–31. The transmitters emit identical signals in phase with each other, which the driver receives on the car radio. When the car is at
point A, the radio picks up a maximum net signal.
(a) What is the longest possible wavelength of the radio waves?
(b) How long after the car passes point A does the radio experience a minimum in the net signal? Assume that the wavelength has the value found in part (a).
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.5CQ
Suppose a sheet of glass is placed in front of one of the slits in a two-slit experiment. If the thickness of the glass is such that the light reaching the two slits is 180° out of phase, how does this affect the interference pattern?
Solution:
If a sheet of glass is placed in front of one of the slits and the thickness of the glass is such that the light reaching the two slits is 180° out of phase. Then here in this case the interference pattern will be reversed.
Changing the phase of the light from one slit by 180°, results in destructive interference (a dark fringe) at the center of the pattern. Thus the entire fringe pattern will be reversed.
Chapter 28 Physical Optics: Interference and Diffraction Q.5P
Two students in a dorm room listen to a pure tone produced by two loudspeakers that are in phase. Students A and B in Figure 28–32 hear a maximum sound. What is the lowest possible frequency of the loudspeakers?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.6CQ
Describe the changes that would be observed in the two-slit interference pattern if the entire experiment were to be submerged in water.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.6P
If the loudspeakers in Problem 5 are 180° out of phase, determine whether a 185-Hz tone heard at location B is a maximum or a minimum.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.7CQ
Explain why the central spot in Newton’s rings is dark.
Solution:
When light incident from rarer to denser medium then there exist a phase difference of . When the light refracted from denser medium to rarer medium there will be no phase
change.
Chapter 28 Physical Optics: Interference and Diffraction Q.7P
A microphone is located on the line connecting two speakers that are 0.845 m apart and oscillating in phase. The microphone is 2.55 m from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone’s location?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.8CQ
Two identical sheets of glass are coated with films of different materials but equal thickness. The colors seen in reflected light from the two films are different. Give a reason that can account for this observation.
Solution:
The color (wavelength) that reflects strongly depends on the index of refraction of the material of the film. As the two films have different index of refractive, so the interference in the thin films happens at different wavelengths. The phase change in reflection from the film – glass interface will be different for the two films. This results in different colors appearing in the reflected light.
Chapter 28 Physical Optics: Interference and Diffraction Q.8P
A microphone is located on the line connecting two speakers that are 0.845 m apart and oscillating 180° out of phase. The microphone is 2.25 m from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone’s location?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.9CQ
Spy cameras use lenses with very large apertures. Why are large apertures advantageous in such applications?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.9P
Moe, Larry, and Curly stand in a line with a spacing of 1.00 m. Larry is 3.00 m in front of a pair of stereo speakers 0.800 m apart, as shown in Figure 28–33. The speakers produce a single-frequency tone, vibrating in phase with each other. What are the two lowest frequencies that allow Larry to hear a loud tone while Moe and Curly hear very little?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.10CQ
A cat’s eye has a pupil that is elongated in the vertical direction. How does the resolution of a cat’s eye differ in the horizontal and vertical directions?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.10P
IP In Figure 28–33 the two speakers emit sound that is 180° out of phase and of a single frequency, f.
(a) Does Larry hear a
sound intensity that is a maximum or a minimum? Does your answer depend on the frequency of the sound? Explain.
(b) Find the lowest two frequencies that produce amaximum sound intensity at the positions of Moe and Curly.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction
Which portion of the soap film in the accompanying photograph is thinnest? Explain.
Solution:
The top of the film in the photograph appears black, so the soap film is thinnest at the top.
Light reflected from the front surface of the film has undergone a phase change of because the light is reflected at the surface backed by denser medium. The light that reflects from the back surface of the film does not undergo any phase change because the light is reflected at the surface backed by rarer medium. Therefore light from front & back surfaces of the film will undergo destructive interference as the path length between the surfaces goes to zero. This is why the top of the film where the film is thinnest, appears black in the photo.
Chapter 28 Physical Optics: Interference and Diffraction Q.11P
IP Suppose the car radio in Problem 4 picks up a minimum net signal at point A.
(a) What is the largest possible value for the wavelength of the radio waves? (b) If the radio transmitters use a wavelength that is half the value found in part (a), will the car radio pick up a net signal at point A that is a maximum or a minimum? Explain.
(c) What is the second largest wavelength that will result in a minimum signal at point A?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.12CQ
The color of an iridescent object, like a butterfly wing or a feather, appears to be different when viewed from different directions. The color of a painted surface appears the saine from all viewing angles. Explain the difference.
Solution:
In iridescent objects. the colors are reflected through the process of thin film of interferenca There will be two reflections one from the top and one from the bottom of the thin film that then
constructively interfere The conditions for interference depends on path length and path length depends on the angle from which one views the film
A pointed object however reflects light of a given color. There is only one reflection at the surface of the points. So no matter what angle you view it from, it always reflects the same color It is the interference process for the thin film that causes the iridescence.
Chapter 28 Physical Optics: Interference and Diffraction Q.12P
CE Consider a two-slit interference pattern, with monochromatic light of wavelength λ. What is the path difference ∆ℓ for
(a) the fourth bright fringe and
(b) the third dark fringe above the centra] bright fringe? Give your answers in terms of the wavelength of the light.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.13P
CE
(a) Does the path-length difference ∆ℓ increase or decrease as you move from one bright fringe of a two-slit experiment to the next bright fringe farther out?
(b) What is ∆ℓ in terms of the wavelength λ of the light?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.14P
CEPredict/Explain A two-slit experiment with red light produces a set of bright fringes.
(a) Will the spacing between the fringes increase, decrease, or stay the same if the color of the light is changed to blue?
(b) Choose the best explanation from among the following:
I. The spacing between the fringes will increase because blue light has a greater frequency than red light.
II. The fringe spacing decreases because blue light has a shorter wavelength than red light.
III. Only the wave property of light is important in producing the fringes, not the color of the light. Therefore the spacing stays the same.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.15P
CE Atwo-slit experiment with blue light produces a set of bright fringes. Will the spacing between the fringes increase, decrease, or stay the same if
(a) the separation of the slits is decreased, or
(b) the experiment is immersed in water?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.16P
Laser light with a wavelength λ = 670 nm illuminates a pair of slits at normal incidence. What slit separation will produce first-order maxima at angles of ±35° from the incident direction?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.17P
Monochromatic light passes through two slits separated by a distance of 0.0334 mm. If the angle to the third maximum above the central fringe is 3.21°, what is the wavelength of the light?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.18P
In Young’s two-slit experiment, the first dark fringe above the central bright fringe occurs at an angle of 0.31°. What is the ratio of the siit separation, d, to the wavelength of the light,λ?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.19P
IP A two-slit experiment with slits separated by 48.0 × 10−5 m produces a second-order maximum at an angle of 0.0990°.
(a) Find the wavelength of the light used in this experiment.
(b) If the slit separation is increased but the second-order maximum stays at the same angle, does the wavelength increase, decrease, or stay the same? Explain.
(c) Calculate the wavelength for a slit separation of 68.0 × 10−5 m.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.20P
A two-slit pattern is viewed on a screen 1.00 m from the slits. If the two third-order minima are 22.0 cm apart, what is the width (in cm) of the centralbright fringe?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.21P
Light from a He-Ne laser (λ = 632.8 nm) strikes apair of slits at normal incidence, forming a double-slit interference pattern on a screen located 1.40 m from the slits. Figure 28–34
shows the interference pattern observed on the screen. What is the slit separation?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.22P
Light with a wavelength of 546 nm passes through two slits and forms an interference pattern on a screen 8.75 m away. If the linear distance on the screen from the central fringe to the first bright fringe above it is 5.36 cm, what is the separadon of the slits?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.23P
A set of parallel slits for optical interference can be made by holding two razor blades together (carefully!) and scratching a pair of lines on a glass microscope slide that has been painted black. When monochromatic light strikes these slits at normal incidence, an interference pattern is formed on a distant screen. The thickness of each razor blade used to make the slits is 0.230 mm, and the screen is 2.50 m from the slits, if the center-to-center separation of the fringes is 7.15 mm, what is the wavelength of the light?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.24P
IP Suppose the interference pattern shown in Figure 28–34 is produced by monochromatic light passing through two slits, with a separation of 135 μm, and onto a screen 1.20 m away.
(a) What is the wavelength of the light?
(b) If the frequency of this light is increased, will the bright spots of the pattern move closer together or farther apart? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.25P
A physics instructor wants to produce a double-slit interference pattern large enough for her class to see. For the size of the room, she decides that the distance between successive bright fringes on the screen should be at least 2.50 cm. If the slits have a separation d = 0.0220 mm, what is the minimum distance from the slits to the screen when 632.8-nm light from a He-Ne laser is used?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.26P
IP When green light (λ = 505 nm) passes through a pair of double slits, the interference pattern shown in Figure 28–35
(a) is observed. When light of a different color passes throughthe same pair of slits, the pattern shown in Figure 28–35
(b) is observed.
(a) Is the wavelength of the second color Longer or shorter than 505 nm? Explain.
(b) Find the wavelength of the second color. (Assume that the angles involved are small enough to set sin θ ≈ tan θ.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.27P
IP The interference pattern shown in Figure 28–35 (a) is produced by green light with a wavelength of λ = 505 nm passing through two slits with a separation of 127 μ m. After passing through the slits, the light forms a pattern of bright and dark spots on a screen located 1.25 m from the slits.
(a) What is the distance between the two vertical, dashed lilies in Figure 28–35 (a)?
(b) If it is desired to produce a more rightly packed interference pattern, like the one shown in Figure 28–35 (b), should the frequency of the light be increased or decreased? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.28P
CE Figure 28–36 shows four different cases where light of wavelength λ reflects from both the top and the bottom of a thin film of thickness d. The indices of refraction of the film and the media above and below it are indicated in the figure. For which of the cases will the two reflected rays undergo constructive interference if
(a) d = λ/4 or
(b) d = λ/2?
Solution:
Here, as shown in figure, it is clear that,
In figure (1), light travels from high refractive index to low refractive index in both situations. In this case, both waves are inverted, so they end up having no shift with respect to one another.
In figure (2), light travels from low refractive index to high refractive index in both cases. In this case, there is phase shift. In this case, neither of the waves is inverted, so they have also no shift with respect to one another.
In figure (3), light travels from low refractive index to high refractive index and then from high refractive index to low refractive index. In this case, when the light is refracted from low refractive wavelength to high refractive wavelength, then the waves is inverted, which is equivalent to half the wavelength shift and secondly when the light is refracted from the high refractive wavelength to low refractive wavelength, then it experiences no shift. This gives the effect that the waves are shifted by half a wavelength.
In figure (4), light travels from low refractive index to high refractive index and then from high refractive index to low refractive index. In this case, when the light is refracted from low refractive wavelength to high refractive wavelength, then the waves is inverted, which is equivalent to half the wavelength shift and secondly when the light is refracted from the high refractive wavelength to low refractive wavelength, then it experiences no shift. This gives the effect that the waves are shifted by half a wavelength.
This concludes:
For the figure (1) there is no phase shift.
Chapter 28 Physical Optics: Interference and Diffraction Q.29P
CE The oil film floadng on water in the accompanying photo appears dark near the edges, where it is thinnest. Is the index of refraction of the oil greater than or less than that of the water? Explain.
Solution:
The light reflected from the top surface of the oil has its phase changed by half wavelength, since the refractive index of oil is greater than the refractive index of air. At the boundary between oil and water the reflected wave experiences no phase change if the refractive index of oil is greater than that of water. This must be the case, because where the oil is thinnest the phase difference between the light reflected from the top and bottom of the oil is just the phase difference due to the reflection – the phase difference due to the path length difference goes to zero. With a half wavelength phase change between light reflected from the top and bottom of the oil, light will undergo destructive interference and the film will appear dark. Therefore, a dark appearance near the edges implies oil has the greater refractive index than water.
Chapter 28 Physical Optics: Interference and Diffraction Q.30P
A soap bubble with walls 401 nm thick floats in air. If this bubble is illuminated perpendicularly with sunlight, what wavelength (and color) will be absent in the reflected light? Assume that the index of refraction of the soap film is 1.33. (Refer to Example 25-3 for the connection between wavelength and color.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.31P
A soap film (n = 1.33) is 825 nm thick. White light strikes the film at normal incidence. What visible wavelengths will be constructively reflected if the film is surrounded by air on both sides? (Refer to Example 25-3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.32P
White light is incident on a soap film (n = 1.30) in air. The reflected light looks bluish because the red light (λ = 670 nm) is absent in the reflection. What is the minimum thickness of the soap film?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.33P
A 742-nm-thick soap film (nfilm = 1.33) rests on a glass plate (nglass = 1.52). White light strikes the film at normal incidence. What visible wavelengths will be constructively reflected from the film? (Refer to Example 25-3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.34P
An oil film (n = 1.38) floats on awater puddle. You notice that green light (λ = 521 nm) is absent in the reflection. What is the minimum thickness of the oil film?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.35P
A radio broadcast antenna is 36.00 km from your house. Suppose an airplane is flying 2.230 km above the line connecting the broadcast antenna and your radio, and that waves reflected from the airplane travel 88.00 wavelengths farther than waves that travel directly from the antenna to your house.
(a) Do you observe constructive or destructive interference
between the direct and reflected waves? (Hint: Does a phase change occur when the waves are reflected?)
(b) The situation just described occurs when the plane is above a point on the ground that is two-thirds of the way from the antenna to your house. What is the wavelength of the radio waves?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.36P
IP Newton’s Rings Monochromatic light with λ = 648 nm shines down on a plano-convex lens lying on a piece of plate glass, as shown in Figure 28–37. When viewed from above, one sees a set of concentric dark and bright fringes, referred to as Newton’s rings (See Figure 28–11 for a photo of Newton’s rings.).
(a) If the radius of the twelfth dark ring from the center is measured to be 1.56 cm, what is the radius of curvature, R, of the lens?
(b) If light with a longer wavelength is used with this system, will the radius of the twelfth dark ring be greater than or less than 1.56 cm? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.37P
Light is incident from above on two plates of glass, separated on both ends by small wires of diameter
d = 0.600 μ m. Considering only interference between light reflected from the bottom surface of the upper plate and light reflected from the upper surface of the lower plate, state whether the following wavelengths give constructive or destructive interference:
(a) λ = 600.0 nm;
(b) λ = 800.0 nm;
(c) λ = 343.0 nm.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.38P
(a) What is the minimum soap-film thickness (n = 1.33) in air that will produce constructive interference in reflection for red (λ = 652 nm) tight?
(b) Which visible wavelengths will destructively interfere when reflected from this film? (Refer to Example 25-3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.39P
IP A thin layer of magnesium fluoride (n = 1.38) is used to coat a flint-glass lens (n = 1.61).
(a) What thickness should the magnesium fluoride film have if the reflection of 565-nm light is to be suppressed? Assume that the light is incident at right angles to the film.
(b) If it is desired to suppress the reflection of light with a higher frequency, should the coating of magnesium fluoride be made thinner or thicker? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.40P
White light is incident normally on a thin soap film (n = 1.33) suspended in air.
(a) What are the two minimum thicknesses that will constructively reflect yellow (λ = 590 nm) light?
(b) What arc the two minimum thicknesses that will destructively reflect yellow (λ = 590 nm) light?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.41P
A thin coating (t = 340.0 nm, n = 1.480) is placed on a glass lens. Which visible (400 nm < λ < 700 nm) wavelength(s) will be absent in the reflected beam if
(a) the glass has an index of refraction n = 1.350, and
(b) the glass has an index of refraction n = 1.675?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.42P
Two glass plates are separated by fine wires with diameters d1 = 0.0500 mm and d2 = 0.0520 mm, as indicated in Figure 28–38. The wires are parallel and separated by a distance of 7.00 cm. If monochromatic light with λ = 589 nm is incident
from above, what is the distance (in cm) between adjacent dark bands in the reflected light? (Consider interference only between light reflected from the bottom surface of the upper plate and light reflected from the upper surface of the lower plate.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.43P
CE A single-slit diffraction pattern is formed on a distant screen. Assuming the angles involved are small, by what factor will the width of the central bright spot on the screen change if
(a) the wavelength is doubled,
(b) the slit width is doubled, or
(c) the distance from the slit to the screen is doubled?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.44P
What width single slit will produce first-order diffraction minima at angles of ±23° from the central maximum with 690-nm light?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.45P
Diffraction also occurs with sound waves. Consider 1300-Hz sound waves diffracted by a door that is 84 cm wide. What is the angle between the two first-order diffraction minima?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.46P
Green light (λ = 546 nm) strikes a single slit atnormal incidence. What width slit will produce a central maximum that is 2.50 cm wide on a screen 1.60 m from the slit?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.47P
Light with a wavelength of 676 nm passes through a slit 7.64 μ mwide and falls on a screen 1.85 m away. Find the linear distance on the screen from the central bright fringe to the first bright fringe above it.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.48P
Repeat Problem 47, only this time find the distance on the screen from the central bright fringe to the third dark fringe above it.
IP A single slit is illuminated with 610-nm light, and the resulting diffraction pattern is viewed on a screen 2.3 m away.
(a) If the lineardistance between the first and second dark fringes of the pattern is 12 cm, what is the width of the slit?
(b) If the slit is made wider, will the distance between the first and second dark fringes increase or decrease? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.49P
IP A single slit is illuminated with 610-nm light, and the resulting diffraction pattern is viewed on a screen 2.3 m away.
(a) If the lineardistance between the first and second dark fringes of the pattern is 12 cm, what is the width of the slit?
(b) If the slit is made wider, will the distance between the first and second dark fringes increase or decrease? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.50P
How many dark fringes will be produced on either side of the central maximum if green light (λ = 553 nm) is incident on a slit that is 8.00 μ m wide?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.51P
IP The diffraction pattern shown in Figure 28–39 is produced by passing He-Ne laser light (λ = 632.8 nm) through a single slit and viewing the pattern an a screen 1.50 m behind the slit.
(a) What is the width of the slit?
(b) If monochromatic yellow light with a wavelength oi 591 nm is used with this slit instead, will the distance indicated in Figure 28–39 be greater than or less than 15.2 cm? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.52P
A screen is placed 1.00 m behind a single slit. The central maximum in the resulting diffraction pattern on the screen is 1.60 cm wide—that is, the two first-order diffraction minima
are separated by 1.60 cm. What is the distance between the two second-order minima?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.53P
CE Predict/Explain
(a) In principle, do your eyes have greater resolution on a dark cloudy day or on a bright sunny day?
(b) Choose the best explanation from among the following:
I. Your eyes have greater resolution on a cloudy day because your pupils are open wider to allow more light to enter the eye.
II. Your eyes have greater resolution on a sunny day because the bright light causes your pupil to narrow down to a smaller opening.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.54P
CE Is resolution greater with blue light or red light, all other factors being equal? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.55P
Two point sources of light are separated by 5.5 cm. As viewed through a 12-μm-diameter pinhole, what is the maximum distance from which they can be resolved
(a) if red light (λ = 690 nm) is used, or
(b) if violet light (λ = 420 nm) is used?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.56P
A spy camera is said to be able to read the numbers on a car’s license plate. Tf the numbers on the plate are 5.0 cm apart, and the spy satellite is at an altitude of 160 km, what must be the diameter of the camera’s aperture? (Assume light with a wavelength of 550 nm.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.57P
Splitting Binary Stars As seen from Earth, the red dwarfs Krüger 60A and Krüger 60B form a binary star system with an angular separation of 2.5 arc seconds. What is the smallest diameter telescope that could theoretically resolve these stars using550-nm light? (Note: 1 arc sec = 1/3600°)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.58P
Find the minimum aperture diameter of a camera that can resolve detail on the ground the size of a person (2.0 m) from an SR-71 Blackbird airplane flying at an altitude of 27 km. (As-sume light with a wavelength of 450 nm.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.59P
The Resolution of Hubble The Hubble Space Telescope (HST) orbits Earth at an altitude of 613 km. It has an objective mirror that is 2.4 m in diameter. If the HST were to look down on Earth’s surface (rather than up at the stars), what is the minimum separation of two objects that could be resolved using 550-nm light? [Note: The HST is used only for astronomical work, but a (classified) number of similar telescopes are in orbit for spy purposes.]
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.60P
A lens that is “optically perfect” is still limited by diffraction effects. Suppose a lens has a diameter of 120 mm and a focal length of 640 mm.
(a) Find the angular width (that is, the angle from the bottom to the top) of the central maximum in the diffraction pattern formed by this lens when illuminated with 540-nm light.
(b) What is the linear width (diameter) of the central maximum at the focal distance of the lens?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.61P
The resolution of a telescope is ultimately limited by the diameter of its objective lens or mirror. A typical amateur astronomer’s telescope may have a 6.0-in.-diameter mirror.
(a) What is the minimum angular separation (in arc seconds) of two stars that can be resolved with a 6.0-in. scope? (Take λ to be atthe center of the visible spectrum, about 550 run, and see Froblem 57 for the definition of an arc second.)
(b) What is the minimum distance (in km) between two points on the Moon’s surface that can be resolved by a 6.0-in. scope? (Note: The average distance from Earth to the Moon is 384,400 km.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.62P
Early cameras were little more than a box with a pinhole on the side opposite the film.
(a) What angular resolution would you expect from a pinhole with a 0.50-mm diameter?
(b) What
is the greatest distance from the camera at which two point objects 15 cm apart can be resolved? (Assume light with a wavelength of 520 nm.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.63P
A grating has 787 lines per centimeter. Find the angles of the first three principal maxima above the central fringe when this grating is illuminated with 655-nm light.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.64P
Suppose you want to produce a diffraction pattern with X-rays whose wavelength is 0.030 nm. If you use a diffraction grating, what separation between lines is needed to generate a pattern with the first maximum at an angle of 14°? (For comparison, a typical atom is a few tenths of a nanometerindiameter.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.65P
A diffraction grating has 2200 lines/cm. What is the angle between the first-order maxima for red light (λ = 680 nm) and blue light (λ =410 nm)?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.66P
A diffractiongrating with 345 lines/mm is 1.00 m in front of a screen. What is the wavelength of light whose first-order maxima will be 16.4 cm from the central maximum on the screen?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.67P
The yellow light from a helium discharge tube has a wavelength of 587.5 nm. When this light illuminates a certain diffraction grating it produces a first-order principal maximum at an angle of 1.250°. Calculate the number of lines per centimeter on the grating.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.68P
IP The second-order maximum produced by a diffraction grating with 560 lines per centimeter is at an angle of 3.1°.
(a) What is the wavelength of the light that illuminates the grating?
(b) If a grating with a larger number of lines per centimeter is used with this light, is the angle of the second-order maximum greater than or less than 3.1°? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.69P
White light strikes a gratingwith 7600 lines/cm at normal incidence. How many complete visible spectra will be formed on either side of the central maximum? (Refer to Example 25-3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.70P
White light strikes a diffraction grating (890 lines/mm) at normal incidence. What is the highest-order visible maximum that is formed? (Refer to Example 25-3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.71P
White light strikes a diffraction grating (760 lines/mm) at normal incidence. What is the longest wavelength that forms a second-order maximum?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.72P
A light source emits two distinct wavelengths [λ1 = 430 nm (violet); λ2 = 630 nm (orange)]. The light strikes a diffraction grating with 450 lines/mm at normal incidence. Identify the colors of the first eight interference maxima on either side of the central maximum.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.73P
A laser emits two wavelengths (λ1 = 420 nm; λ2 = 630 nm). When these two wavelengths strike a grating with 450 lines/mm, they produce maxima (in different orders) that coincide.
(a) What is the order (m)of each of the two overlapping lines?
(b) At what angle does this overlap occur?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.74P
IP When blue light with a wavelength of 465 nm illuminates a diffraction grating, it produce’s a first-order principal maximum but no second-order maximum.
(a) Explain the absence of higher-order principal maxima.
(b) What is the maximum spacing between lines on this grating?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.75P
Monochromatic light strikes a diffraction grating at normal incidence before illuminating a screen 2.10 m away. If the first-order maxima are separated by 1.53 m on the screen, what is the distance between the two second-order maxima?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.76P
A diffraction grating with a slit separation d is illuminated bya beam of monochromatic light of wavelength λ. The diffracted beam is observed at an angle ϕ relative to the incident direction. If the plane of the grating bisects the angle between the incident and diffracted beams, show that the mth maximum will be observed at anangle that satisfies the relation mλ = 2d sin(ϕ/2), with m = 0, ±1, ±2,….
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.77GP
CE Monochromatic light with a wavelength λ passes through a single slit of width W and forms a diffraction pattern of alternating bright and dark fringes, (a) if the width of the slit is decreased, do the dark fringes move outward or inward? Explain. (b) What width is necessary for the first dark fringe to move outward to infinity? Give your answer in terms of λ.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.78GP
CE Predict/Explain (a) If a thin liquid film floating on water has an index of refraction less than that of water, will the film appear bright or dark in reflected light as its thickness goes to zero? (b) Choose the best explanation from among the following:
I. The film will appear bright because as the thickness of the film goes to zero the phase difference for reflected rays goes to zero.
II. The film will appear dark because there is a phase change at both interfaces, and this will cause destructive interference of the reflected rays.
Solution:
(b)
Therefore the best explanation is I.
Chapter 28 Physical Optics: Interference and Diffraction Q.79GP
CE If the index of refraction of an eye could be magically reduced, would the eye’s resolution increase or decrease? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.80GP
CE In orderto increase the resolution of a camera, should its f-number be increased or decreased? Explain.
Solution:
The resolution of camera increases, when we increase the diameter of aperture of the camera. But f-number is inversely proportional to diameter. Therefore in order increase the resolution of a camera the f-numbers must be decreased.
Chapter 28 Physical Optics: Interference and Diffraction Q.81GP
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.82GP
When reading the printout from a laser printer, you are actually looking at an array of tiny dots. If the pupil of your eye is 4.3 mm in diameter when reading a page held 28 cm from your eye, what is the minimum separation of adjacent dots that can be resolved? (Assume light with a wavelength of 540 nm, and use 1.36 as the index of refraction for the interior of the eye.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.83GP
The headlights of a pickup truck are 1.32 m apart. What is the greatest distance at which these headlights can be resolved as separate points of light on a photograph taken with a camera whose aperture lias a diameter of 12.5 mm? (Take λ = 555 rim.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.84GP
Antireflection Coating A glass lens (nglass = 1.52) has an antireflcction coating of MgF2 (n = 1.38). (a) For 517-nm light, what minimum thickness of MgF2 will cause the reflected rays R2 and R4 in Figure 28–40 to interfere destmctively, assuming normal incidence? (b) Interference will also occur between the forward-moving rays R1 and R3 in Figure 28–40. What minimum thickness of MgF2 will cause these two rays to interfere constructively?
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.85GP
IP White light reflected at normal incidence from a soap bubble (n = 1.33) in air produces an interference maximum at λ = 575 nm but no interference minima in the visible spectrum.
(a) Explain the absence of interferenceminima in the visible.
(b) What are the possible thicknesses of the soap film? (Refer to Example 25–3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.86GP
A thin film of oil (n = 1.30) floats on water (n − 1.33). When sunlight is incident at right angles to this film, the only colors that are enhanced by reflection are blue (458 nm) and red (687 nm). Estimate the thickness of the oil film.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.87GP
The yellow light of sodium, with wavelengths of 588.99 nm and 589.59 nm, is normally incident on a grating with 494 lines/cm. Find the linear distance between the first-order maxima for these two wavelengths on a screen 2.55 m from the grating.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.88GP
IP A thin soap film (n = 1.33) suspended in air has a uniform thickness. When white light strikes the film at normal incidence, violet light (λv = 420 nm) is constructively reflected. (a) If we would like green light (λG = 560 nm) to be constructively reflected, instead, should the film’s thickness be increased or decreased? (b) Find the new thickness of the film. (Assume the film has the minimum thickness that can produce these reflections.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.89GP
IP A thin film of oil (n = 1.40) floats on water (n = 1.33). When sunlight is incident at right angles to this film, the only colors that are absent from the reflected light are blue (458 nm) and red (687 nm). Estimate the thickness of the oil film.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.90GP
IP Sodium light, with a wavelength of λ = 589 nm, shines downward onto the system shown in Figure 28–37. When viewed from above, you see a series of concentric circles known as Newton’s rings, (a) Do you expect a bright or a dark spot at the center of the pattern? Explain, (b) If the radius of curvature of the plano-convex lens is R = 26.1 m, what is the radius of the tenth-largest dark ring? (Only rings of nonzero radius will be counted as “rings.”)
Solution:
a) In Newton’s ring arrangement the thickness of air film at the point of contact is zero, i.e., there is no path difference between the rays reflected from upper and lower surfaces of air film. Light reflected from the lower surface of the lens has no phase change. Light reflected from the top surface of the flat glass piece undergoes half – wavelength phase change. As a result they are half a wavelength out of phase resulting in destructive interference. Hence, the center of the pattern is a dark spot.
Chapter 28 Physical Optics: Interference and Diffraction Q.91GP
IP Figure 28–39 shows a single-slit diffraction pattern formed by light passing througha slit of width W = 11.2 μm and illuminating a screen 0.855 m behind the slit, (a) What is the wavelength of the light? (b) If the width of the slit is decreased, will the distance indicated in Figure 28–39 be greater than or less than 15.2 cm? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.92GP
B10Entoptic Halos Images produced by structures within the eye (like lens fibers or cell fragments) are referred to as entoptic images. These images can sometimes take the form of “halos” around a bright light seen against a dark background. The halo in such a case is actually the bright outer rings of a circular diffraction pattern, like Figure 28–21, with the central bright spot not visible because it overlaps the direct image of tine light. Find the diameter of the eye structure that causes a circular diffraction pattern with the first dark ring at an angle of 3.7° when viewed with monochromatic light of wavelength 630 nm. (Typical eye structures of this type have diameters on the orderof 10 μm. Also, the index of refraction of the vitreous humor is 1.336.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.93GP
White light is incident on a soap film (n = 1.33, thickness = 800.0 nm) suspended in air. If the incident light makes a 45° angle with the normal to the film, what visible wavelength(s) will be constructively reflected? (Refer to Example 25-3 for the range of visible wavelengths.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.94GP
A system like that shown in Figure consists of N slits, each transmitting light of intensity I0. The light from each slit has the same phase and the same wavelength. The net intensity I observed at an angle θ due to all N slits is
In this expression, ϕ = (2πd/λ) sin θ, where λ is the wavelength of the light, (a) Show that the intensity in the limit θ→0 is I = N2I0. This is the maximum intensity of the interference pattern, (b) Show that the first points of zero intensity on cither side of θ = 0 occur atϕ = 2π/N and ϕ = −2π/N. (c) Does the central maximum (θ = 0) of this pattern become narrower or broader as the number of slits is increased? Explain.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.95GP
Two plates of glass are separated on both ends by small wires of diameter d. Derive an expression for the condition for constructive interference when light of wavelength λ is incident normally on the plates. Consider only interference between waves reflected from the bottom of the top plate and the top of the bottom plate.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.96GP
A curved piece of glass wi th a radius of curvature R rests on a flat plate of glass. Light of wavelength A is incident normally on this system. Considering only interference between waves reflected from the curved (lower) surface of glass and the top surface of the plate, show that the radius of the nth dark ring is
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.97GP
BIO The Resolution of the Eye The resolution of the eye is ultimately limited by the pupil diameter. What is the smallest diameter spot the eye can produce on the retina if the pupil diameter ig 4.25 mm? Assume light with awavelength of λ = 550 nm. (Note: The distance from the pupil to the retina is 25.4mm. In addition, the space between the pupil and the retina is filled with a fluid whose index of refraction is n = 1.36.)
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.98PP
What is the minimum angle your eye can resolve, according to the Rayleigh criterion and the above assumptions?
A. 0.862 × 10−4 rad
B. 1.05 − 10−4 rad
C. 1.43 × 10−4 rad
D. 1.95 × 10−4 rad
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.99PP
What is the linear separation between horizontal lines on the screen?
A. 0.0235 mm
B. 0.145 mm
C. 0.369 mm
D. 0.926 mm
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.100PP
What is the angular separation of the horizontal lines as viewed from a distance of 12.0 feet?
A. 1.01 × 10−4 rad
B. 2.53 × 10−4 rad
C. 2.56 × 10−4 rad
D. 12.1 × 10−4 rad
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.101PP
According to the Rayleigh criterion, what is the closest you can be to the TV screen before resolvingthe individual horizontal lines? (In practice you can be considerably closer than this distance before resolving the lines.)
A. 3.51 ft
B. 4.53 ft
C. 11.5 ft
D. 14.0 ft
Solution:
Chapter 28 Physical Optics: Interference and Diffraction
IP Referring to Example 28–2 Suppose we change the slit separation to a value other than 8.5 × 10−5m, with the result that the linear distance to the tenth bright fringe above the central bright fringe increases from 12 cm to 18 cm. The screen is still 2.3 m from the slits, and the wavelength of the light is 440 nm. (a) Did we increase or decrease the slit separation? Explain, (b) Find the new slit separation.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.103IP
IP Referring to Example 28–2 The wavelength of the light is changed to a value other than 440 nm, with the result that the linear distance to the seventh bright fringe above the central, bright fringe is 12 cm. The screen is still 2.3 m from the slits, and the slit separation is 8.5 × 10−5 m. (a) Is the new wavelength longer or shorter than 440 nm? Explain, (b) Find the new wavelength.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.104IP
IP Referring to Example 28–5 The light used in this experiment has a wavelength of 511 nm. (a) If the width of the slit is decreased, will the angle to the first dark fringe above the central bright fringe increase or decrease? Explain, (b) Find the angle to the first dark fringe if the reduced slit width is 1.50 × 10−6 m.
Solution:
Chapter 28 Physical Optics: Interference and Diffraction Q.105IP
IP Referring to Example 28–5 The width of the slit in this experiment is 2.20 × 10−6 m. (a) If the frequency of the light is decreased, will the angle to the first dark fringe above the central bright fringe increase or decrease? Explain, (b) Find the angle to the first dark fringe if the reduced frequency is 5.22 × 1014 Hz.
Solution:
