physics test 2 extra questions
. A double-slit experiment yields an interference pattern due to the path length difference from light traveling through one slit versus the other. Why does a single slit show a diffraction pattern? (a) There is a path length difference from waves originating at different parts of the slit. (b) The wavelength of the light is shorter than the slit. (c) The light passing through the slit interferes with light that does not pass through. (d) The single slit must have something in the middle of it, causing it to act like a double slit.
(a)
As an object moves from just outside the focal point of a converging lens to just inside it, the image goes from _____ and _____ to _____ and _____. (a) large; inverted; large; upright. (b) large; upright; large; inverted. (c) small; inverted; small; upright. (d) small; upright; small; inverted.
(a)
How do eyeglasses help a nearsighted person see more clearly? (a) Diverging lenses bend light entering the eye, so the image focuses farther from the front of the eye. (b) Diverging lenses bend light entering the eye, so the image focuses closer to the front of the eye. (c) Converging lenses bend light entering the eye, so the image focuses farther from the front of the eye. (d) Converging lenses bend light entering the eye, so the image focuses closer to the front of the eye. (e) Lenses adjust the distance from the cornea to the back of the eye.
(a)
If the distance from your eye's lens to the retina is shorter than for a normal eye, you will struggle to see objects that are (a) nearby. (c) colorful. (b) far away. (d) moving fast
(a)
Light from a green laser of wavelength 530 nm passes through two slits that are 400 nm apart. The resulting pattern formed on a screen in front of the slits is shown in Fig. 24-55. If point A is the same distance from both slits, how much closer is point B to one slit than to the other? (a) 530 nm. (b) 265 nm. (c) 400 nm. (d) 0 nm. (e) It depends on the distance to the screen
(a)
Light passing through a double-slit arrangement is viewed on a distant screen. The interference pattern observed on the screen would have the widest spaced fringes for the case of (a) red light and a small slit spacing. (b) blue light and a small slit spacing. (c) red light and a large slit spacing. (d) blue light and a large slit spacing
(a)
Suppose you are standing about 3 m in front of a mirror. You can see yourself just from the top of your head to your waist, where the bottom of the mirror cuts off the rest of your image. If you walk one step closer to the mirror (a) you will not be able to see any more of your image. (b) you will be able to see more of your image, below your waist. (c) you will see less of your image, with the cutoff rising to be above your waist.
(a)
The image produced on the retina of the eye is _____ compared to the object being viewed. (a) inverted. (c) sideways. (b) upright. (d) enlarged.
(a)
To shoot a swimming fish with an intense light beam from a laser gun, you should aim (a) directly at the image. (b) slightly above the image. (c) slightly below the image.
(a)
You want to create a spotlight that will shine a bright beam of light with all of the light rays parallel to each other. You have a large concave spherical mirror and a small lightbulb. Where should you place the lightbulb? (a) At the focal point of the mirror. (b) At the radius of curvature of the mirror. (c) At any point, because all rays bouncing off the mirror will be parallel. (d) None of the above; you can't make parallel rays with a concave mirror.
(a)
Describe the single-slit diffraction pattern produced when white light falls on a slit having a width of (a) 60 nm, (b) 60,000 nm.
(a) A slit width of 60 nm would produce a central maximum so spread out that it would cover the entire width of the screen. No minimum (and therefore no diffraction pattern) will be seen, because the first minimum would have to satisfy sinθ =λ/D ≈ 10 which is not possible. The different wavelengths will all overlap, so the light on the screen will be white. It will also be dim, compared to the source, because it is spread out. ( b) For the 60,000-nm slit, the central maximum will be very narrow, about a degree in width for the blue end of the spectrum and about a degree and a half for the red. The diffraction pattern will not be distinct, because most of the intensity will be in the small central maximum and the fringes for the different wavelengths of white light will not coincide.
A thin converging lens is moved closer to a nearby object. Does the real image formed change (a) in position, (b) in size? If yes, describe how.
(a) As a thin converging lens is moved closer to a nearby object, the position of the real image changes. As the object distance decreases, the thin lens equation says that the image distance increases in order for the calculation for the focal length to not change. Thus, the position of the image moves farther away from the lens until the object is at the focal point. At that condition, the image is at infinity. ( b) As a thin converging lens is moved closer to a nearby object, the size of the real image changes. As the object distance decreases, the thin lens equation says that the image distance increases, which means that the magnification i o ( / ) m d d = − gets larger. Thus, the size of the image increases until the object reaches the focal point, in which case a real image will no longer be formed.
What happens to the diffraction pattern of a single slit if the whole apparatus is immersed in (a) water, (b) a vacuum, instead of in air.
(a) If the apparatus is immersed in water, then the wavelength of the light will decrease and the diffraction pattern will become more compact. (λ=λ/n) ( b) If the apparatus is placed in a vacuum, then the wavelength of the light will increase slightly and the diffraction pattern will spread out very slightly.
Example 23-16 shows how to use a converging lens to measure the focal length of a diverging lens. (a) Why can't you measure the focal length of a diverging lens directly? (b) It is said that for this to work, the converging lens must be stronger than the diverging lens. What is meant by "stronger," and why is this statement true?
(a) The focal length of a diverging lens cannot be measured directly because the diverging lens cannot form a real image. Therefore, it is very difficult to measure the image distance—some indirect measure (as described in this problem) must be used. ( b) For the converging lens to be stronger, it must have a shorter focal length then the absolute value of the focal length of the diverging lens. If the diverging lens is stronger, then instead of the focal point simply being moved back from its original location in Fig. 23-45, the rays would never focus. The diverging lens would diverge the entering rays so much that, again, a virtual image would be formed, and no direct measure would be possible.
White light strikes (a) a diffraction grating and (b) a prism. A rainbow appears on a wall just below the direction of the horizontal incident beam in each case. What is the color of the top of the rainbow in each case? Explain.
(a) Violet light will be at the top of the rainbow created by the diffraction grating. Principal maxima for a diffraction grating are at positions given by sinθ =mλ/D Violet light has a shorter wavelength than red light so will appear at a smaller angle away from the direction of the horizontal incident beam. ( b) Red light will appear at the top of the rainbow created by the prism. The index of refraction for violet light in a given medium is slightly greater than for red light in the same medium, so the violet light will bend more and will appear farther from the direction of the horizontal incident beam.
For diffraction by a single slit, what is the effect of increasing (a) the slit width, (b) the wavelength?
(a) When you increase the slit width in a single-slit diffraction experiment, the spacing of the fringes decreases. The equation for the location of the minima, sinθ = mλ/D indicates that θ is decreased for a particular m and λ when the width D increases. This means that the bright spots on the screen are more closely packed together for a wider slit. ( b) When you increase the wavelength of light used in a single-slit diffraction experiment, the spacing of the fringes increases. The equation for the location of the minima, sinθ = mλ/D indicates that θ is increased for a particular m and D when the wavelength increases. This means that the bright spots on the screen are spread farther apart for a longer wavelength.
(a) Does the focal length of a lens depend on the fluid in which it is immersed? (b) What about the focal length of a spherical mirror? Explain
(a) Yes, the focal length of a lens depends on the surrounding fluid. The relative values of the index of refraction of the fluid and the index of refraction of the lens will determine the refraction of light as it passes from the fluid through the lens and back into the fluid. The amount of refraction of light determines the focal length of the lens, so the focal length will change if the lens is immersed in a fluid. ( b) No, the focal length of a spherical mirror does not depend on the surrounding fluid. The image formation of the spherical mirror is determined by reflection, not refraction, and is independent of the medium in which the mirror is immersed.
If unpolarized light is incident from the left on three polarizers as shown in Fig. 24-57, in which case will some light get through? (a) Case 1 only. (b) Case 2 only. (c) Case 3 only. (d) Cases 1 and 3. (e) All three cases
(b)
Imagine holding a circular disk in a beam of monochromatic light (Fig. 24-56). If diffraction occurs at the edge of the disk, the center of the shadow is (a) darker than the rest of the shadow. (b) a bright spot. (c) bright or dark, depending on the wavelength. (d) bright or dark, depending on the distance to the screen
(b)
When a nearsighted person looks at a distant object through her glasses, the image produced by the glasses should be (a) about 25 cm from her eye. (b) at her eye's far point. (c) at her eye's near point. (d) at the far point for a normal eye
(b)
When moonlight strikes the surface of a calm lake, what happens to this light? (a) All of it reflects from the water surface back to the air. (b) Some of it reflects back to the air; some enters the water. (c) All of it enters the water. (d) All of it disappears via absorption by water molecules
(b)
A converging lens, like the type used in a magnifying glass, (a) always produces a magnified image (image taller than the object). (b) can also produce an image smaller than the object. (c) always produces an upright image. (d) can also produce an inverted image (upside down). (e) None of these statements are true.
(b, d)
. The colors in a rainbow are caused by (a) the interaction of the light reflected from different raindrops. (b) different amounts of absorption for light of different colors by the water in the raindrops. (c) different amounts of refraction for light of different colors by the water in the raindrops. (d) the downward motion of the raindrops.
(c)
If someone is around a corner from you, what is the main reason you can hear him speaking but can't see him? (a) Sound travels farther in air than light does. (b) Sound can travel through walls, but light cannot. (c) Sound waves have long enough wavelengths to bend around a corner; light wavelengths are too short to bend much. (d) Sound waves reflect off walls, but light cannot
(c)
If you shine a light through an optical fiber, why does it come out the end but not out the sides? (a) It does come out the sides, but this effect is not obvious because the sides are so much longer than the ends. (b) The sides are mirrored, so the light reflects. (c) Total internal reflection makes the light reflect from the sides. (d) The light flows along the length of the fiber, never touching the sides.
(c)
When a CD is held at an angle, the reflected light contains many colors. What causes these colors? (a) An anti-theft encoding intended to prevent copying of the CD. (b) The different colors correspond to different data bits. (c) Light reflected from the closely spaced grooves adds constructively for different wavelengths at different angles. (d) It is part of the decorative label on the CD
(c)
When the reflection of an object is seen in a flat mirror, the image is (a) real and upright. (b) real and inverted. (c) virtual and upright. (d) virtual and inverted
(c)
When you look at a fish in a still stream from the bank, the fish appears shallower than it really is due to refraction. From directly above, it appears (a) deeper than it really is. (b) at its actual depth. (c) shallower than its real depth. (d) It depends on your height above the water.
(c)
You cover half of a lens that is forming an image on a screen. Compare what happens when you cover the top half of the lens versus the bottom half. (a) When you cover the top half of the lens, the top half of the image disappears; when you cover the bottom half of the lens, the bottom half of the image disappears. (b) When you cover the top half of the lens, the bottom half of the image disappears; when you cover the bottom half of the lens, the top half of the image disappears. (c) The image becomes half as bright in both cases. (d) Nothing happens in either case. (e) The image disappears in both cases
(c)
A lens can be characterized by its power, which (a) is the same as the magnification. (b) tells how much light the lens can focus. (c) depends on where the object is located. (d) is the reciprocal of the focal length.
(d)
Blue light of wavelength passes through a single slit of width d and forms a diffraction pattern on a screen. If we replace the blue light by red light of wavelength we can retain the original diffraction pattern if we change the slit width (a) to d 4. (b) to d 2. (c) not at all. (d) to 2d. (e) to 4d
(d)
The image of a nearby object formed by a camera lens is (a) at the lens' focal point. (b) always blurred. (c) at the same location as the image of an object at infinity. (d) farther from the lens than the lens' focal point
(d)
While you are photographing a dog, it begins to move away. What must you do to keep it in focus? (a) Increase the f-stop value. (b) Decrease the f-stop value. (c) Move the lens away from the sensor or film. (d) Move the lens closer to the sensor or film. (e) None of the above.
(d)
A converging lens, such as a typical magnifying glass, (a) always produces a magnified image (taller than object). (b) always produces an image smaller than the object. (c) always produces an upright image. (d) always produces an inverted image (upside down). (e) None of these statements are true.
(e)
Virtual images can be formed by (a) only mirrors. (b) only lenses. (c) only plane mirrors. (d) only curved mirrors or lenses. (e) plane and curved mirrors, and lenses.
(e)
Which of the following can form an image? (a) A plane mirror. (b) A curved mirror. (c) A lens curved on both sides. (d) A lens curved on only one side. (e) All of the above
(e)
How might you determine the speed of light in a solid, rectangular, transparent object?
A light ray entering the solid rectangular object will exit the other side following a path that is parallel to its original path but displaced slightly from it. The angle of refraction in the glass can be determined geometrically from this displacement and the thickness of the object. The index of refraction can then be determined using Snell's law with this angle of refraction and the original angle of incidence. The speed of light in the material follows from the definition of the index of refraction: / . n c υ= Alternatively, if the angles of incidence and refraction at one surface can be measured, then Snell's law can be used to calculate the index of refraction.
Inexpensive microscopes for children's use usually produce images that are colored at the edges. Why?
A poor-quality, inexpensive lens will not be correctly shaped to fix chromatic aberrations. Thus, the colors you see around the edges of these lenses are from all the different colors of light focusing at different points, instead of all being focused at the same point.
The human eye is much like a camera—yet, when a camera shutter is left open and the camera is moved, the image will be blurred. But when you move your head with your eyes open, you still see clearly. Explain.
All light entering the camera lens while the shutter is open contributes to a single picture. If the camera is moved while the shutter is open, then the position of the image on the film moves. The new image position overlaps the previous image position, causing a blurry final image. With the eye, new images are continuously being formed by the nervous system, so images do not "build up" on the retina and overlap with each other. Your brain "refreshes" the image from the retina about 30 times a second. In other words, your brain and your retina work together in a manner similar to how a motion picture camera and film work together, which is not at all like how a still camera and film work together.
A child looks into a pool to see how deep it is. She then drops a small toy into the pool to help decide how deep the pool is. After this careful investigation, she decides it is safe to jump in—only to discover the water is over her head. What went wrong with her interpretation of her experiment?
As discussed in Question 9, the bottom of the pool and the toy will appear closer to the top of the water than they really are. If the child is not aware of this phenomenon, then she will underestimate the depth of the water.
Why are bifocals needed mainly by older persons and not generally by younger people?
As people get older, their eyes can no longer accommodate as well. It becomes harder for the muscles to change the shape of the lens, since the lens becomes less flexible with age. In general, people first lose the ability to see far objects, so they need corrective diverging lenses to move their "far point" back toward infinity. Then, as people get older, their near point increases and becomes greater than the ideal value of 25 cm. They may still need the diverging portion of their corrective lenses (kept as the upper part of the corrective lens) so they can have a far point at infinity, but now they also need a converging lens to move the near point back toward 25 cm for seeing close objects. Thus, as people get older, their far point is too close and their near point is too far. Bifocals (with two different focal lengths) can correct both of these problems.
Monochromatic red light is incident on a double slit, and the interference pattern is viewed on a screen some distance away. Explain how the fringe pattern would change if the red light source is replaced by a blue light source.
As red light is switched to blue light, the wavelength of the light is decreased. Thus, sind m θ λ = says that θ is decreased for a constant m and d. This means that the bright spots on the screen are more closely packed together with blue light than with red light.
Reading glasses use converging lenses. A simple magnifier is also a converging lens. Are reading glasses therefore magnifiers? Discuss the similarities and differences between converging lenses as used for these two different purposes.
Both reading glasses and magnifiers are converging lenses. A magnifier, generally a short focal length lens, is typically used by adjusting the distance between the lens and the object so that the object is exactly at or just inside the focal point. An object exactly at the focal point results in an image that is at infinity and can be viewed with a relaxed eye. If the lens is adjusted so that it focuses the image at the eye's near point, then the magnification is slightly greater. The lenses in reading glasses typically are a fixed distance from the eye. They are used when the near point is too far away to be convenient—such as when reading a book. These lenses cause the rays from a nearby object to converge somewhat before they reach the eye, allowing the eye to focus on an object that is inside the near point. The focal length of the lens needed for reading glasses will depend on the individual eye. For both reading glasses and magnifiers, the lenses allow the eye to focus on an object that is located closer than the near point.
Explain why chromatic aberration occurs for thin lenses but not for mirrors.
Chromatic aberrations in lenses occur due to dispersion, which is when different colors (or wavelengths) of light are bent different amounts due to the fact that the index of refraction of most materials varies with wavelength. Thus when light passes through a lens, not all of the different colors come out of the lens at the exact same angle, and they are all focused at different positions. A mirror, in contrast, reflects light off of a smooth metallic surface. This surface reflects all different colors of light at the exact same angle, thus there is no refraction, no dispersion, and no chromatic aberrations. Of course, in most mirrors, there is a piece of glass that covers the metallic reflector. Since the two faces of this piece of glass are parallel to each other, even though refraction and dispersion take place inside the piece of glass, when the light emerges from the parallel face, all of the different colors are once again going in the same direction, and there is no dispersion and no chromatic aberration.
What would be the color of the sky if the Earth had no atmosphere?
If Earth had no atmosphere, then the color of the sky would be black (and dotted with stars and planets) at all times. This is the condition of the sky that the astronauts found on the Moon, which has no atmosphere. If there were no air molecules to scatter the light from the Sun, then the only light we would see would be from the stars, the planets, the Moon, and direct sunlight. The rest of the sky would be black.
If the Earth's atmosphere were 50 times denser than it is, would sunlight still be white, or would it be some other color?
If the atmosphere were 50% more dense, then sunlight (after passing through the atmosphere) would be much redder than it is now. As the atmosphere increased in density, more and more of the blue light would be scattered away in all directions, making the light that reaches the ground very red. Think of the color of a deep red sunset, but this might be the color even when the sun was at high elevations.
Can a diverging lens form a real image under any circumstances? Explain.
If the diverging lens is the only piece of optics involved, then it is not possible for it to form a real image. Diverging lenses, by definition, cause light rays to diverge and will not bring rays from a real object to a focal point as required to form a real image. However, if another optical element (for example, a converging lens) forms a virtual object for the diverging lens, then it is possible for the diverging lens to form a real image.
Will a nearsighted person who wears corrective lenses in her glasses be able to see clearly underwater when wearing those glasses? Use a diagram to show why or why not.
No, a nearsighted person will not be able to see clearly if she wears her corrective lenses underwater. A nearsighted person has a far point that is closer than infinity and wears corrective lenses to bring the image of a faraway object to their far point so she can see it clearly. See the pair of diagrams. The object is at infinity. In air, the image is at the (relatively close) far point. If the person's eyes and glasses are underwater, and since the index of refraction of glass is closer to that of water than to that of air, the glasses will not bend the light as much as they did in the air. Therefore, the image of the faraway object will now be at a position that is beyond the person's far point. The image will now be out of focus.
A lens is made of a material with an index of refraction In air, n=1.25 it is a converging lens. Will it still be a converging lens if placed in water? Explain, using a ray diagram.
No. If a lens with 1.25n = is a converging lens in air, then it will become a diverging lens when placed in water, with 1.33.n = The figure on the left shows that as parallel light rays enter the lens when it is in air, at the first surface the ray is bent toward the normal, and at the second surface the ray is bent away from the normal, which is a net converging situation. The figure on the right shows that as parallel light rays enter the lens when it is in water, at the first surface the ray is bent away from the normal, and at the second surface the ray is bent toward the normal, which is a net diverging situation.
Can a light ray traveling in air be totally reflected when it strikes a smooth water surface if the incident angle is chosen correctly? Explain
No. Total internal reflection can only occur when light travels from a medium of higher index of refraction to a medium of lower index of refraction.
How can you "see" a round drop of water on a table even though the water is transparent and colorless?
One reason you can see a drop of transparent and colorless water on a surface is refraction. The light leaving the tabletop and going through the drop on the way to your eye will make that portion of the tabletop seem to be raised up, due to depth distortion and/or magnification due to the round drop. A second reason that you can see the drop is reflection. At certain angles, the light sources in the room will reflect off of the drop and give its location away.
What does polarization tell us about the nature of light?
Polarization demonstrates the transverse wave nature of light and cannot be explained if light is considered as a longitudinal wave or as classical particles.
Can real images be projected on a screen? Can virtual images? Can either be photographed? Discuss carefully.
Real images can be projected onto a screen. In this situation, real light rays actually focus at the position of the image, and if the screen is placed at the image location, then the image can be seen on the screen. Virtual images cannot be projected onto a screen. In this situation, no real light rays are converging at the position of the virtual image, so they cannot be seen on a screen (recall that a plane mirror's virtual image is actually behind the mirror, where no real light rays travel). Both real images and virtual images can be photographed. Real images can be photographed by putting the film at the image location, so the film takes the place of the screen. Both virtual and real images can be photographed in the same way that you can see both virtual and real images with your eye. As long as the camera (or the eye) is positioned so that diverging rays from the (real or virtual) image enter the camera (or the eye), the converging lens is able to make an image from those rays on the film. There is no difference in the way that the camera records rays that came from a virtual image or rays that came from some object directly. See Fig. 23-37c or Fig. 23-39, and simply replace the eye with a camera. The same image will be recorded in both situations.
How can you tell if a pair of sunglasses is polarizing or not?
Take the sunglasses outside and look up at the sky through them. Rotate the sunglasses (about an axis perpendicular to the lens) through at least 180°. If the sky seems to lighten and darken as you rotate the sunglasses, then they are polarizing. You could also look at a liquid crystal display or reflections from a tile floor (with a lot of "glare") while rotating the glasses and again look for the light to be lighter or darker depending on the rotation angle. Finally, you could put one pair of glasses on top of the other as in Fig. 24-44 and rotate them relative to each other. If the intensity of light that you see through the glasses changes as you rotate them, then the glasses are polarizing.
We can hear sounds around corners but we cannot see around corners; yet both sound and light are waves. Explain the difference
The bending of waves around corners or obstacles is called diffraction. Diffraction is most prominent when the size of the obstacle is on the order of the size of the wavelength. Sound waves have much longer wavelengths than do light waves. As a result, the diffraction of sound waves around a corner of a building or through a doorway is noticeable, and we can hear the sound in the "shadow region," but the diffraction of light waves around a corner is not noticeable because of the very short wavelength of the light.
What is the focal length of a plane mirror? What is the magnification of a plane mirror?
The focal length of a plane mirror is infinity. The magnification of a plane mirror is 1.+ As the radius (and focal length) of a spherical mirror increases, the front surface gets more and more flat. The ultimate limit is that as the radius (and focal length) of the spherical mirror goes to infinity, the front surface becomes perfectly flat. For this mirror, the image height and object height are identical (as are the image distance and object distance) and the image is virtual, with a magnification of 1.m = +
Nearsighted people often look over (or under) their glasses when they want to see something small up close, like a cell phone screen. Why?
The glasses that near-sighted people wear are designed to help the viewing of distant objects by moving the focus point farther back in the eye. Near-sighted eyes in general have good vision for close objects whose focus point already lies on the retina. They can actually see close objects better without the corrective lenses. The near point of near-sighted people can be very close to their eye, so they bring small objects close to their eyes in order to see them most clearly.
An object is placed along the principal axis of a spherical mirror. The magnification of the object is -1. Is the image real or virtual, inverted or upright? Is the mirror concave or convex? On which side of the mirror is the image located?
The image is real and inverted, because the magnification is negative. The mirror is concave, because convex mirrors can only form virtual images. The image is on the same side of the mirror as the object; real images are formed by converging light rays, and light rays cannot actually pass through a mirror.
Is the image formed on the retina of the human eye upright or inverted? Discuss the implications of this for our perception of objects.
The images formed on our retinas are real, so they are inverted. The implication of this is that our brains must flip this inverted image for us so that we can see things upright.
Why must a camera lens be moved farther from the sensor or film to focus on a closer object?
The lens equation 1/f= 1/do + 1/di says that as an object gets closer to the lens o(d decreases) and the focal length of the lens remains constant, the image distance id must get larger to create a focused image. Since we cannot move the film or sensor inside the camera, the lens must be moved farther away from the film or sensor. This is unlike the human eye, where the focal length of the lens is changed as the object distance changes.
A photographer moves closer to his subject and then refocuses. Does the camera lens move farther away from or closer to the camera film or sensor? Explain.
The lens moves farther away from the film. When the photographer moves closer to his subject, the object distance decreases. The focal length of the lens does not change, so the image distance must increase, by Eq. 23-8, 1/f = 1/do + 1/di To get the image and the film at the same place, the lens needs to move away from the film.
You look into an aquarium and view a fish inside. One ray of light from the fish is shown emerging from the tank in Fig. 23-48. The apparent position of the fish is also shown (dashed ray). In the drawing, indicate the approximate position of the actual fish. Briefly justify your answer.
The light rays from the fish are bent away from the normal as they leave the tank. The fish will appear closer to the side of the tank than it really is.
Although a plane mirror appears to reverse left and right, it doesn't reverse up and down. Discuss why this happens, noting that front to back is also reversed. Also discuss what happens if, while standing, you look up vertically at a horizontal mirror on the ceiling
The mirror doesn't actually reverse right and left, either, just as it doesn't reverse up and down. Instead it reverses "front" and "back." When you are looking in a flat mirror and move your right hand, it is the image of your right hand that moves in the mirror. When we see our image, though, we imagine it as if it is another person looking back at us. If we saw this other person raise their left hand, then we would see the hand that is on the right side of their body (from our point of view) move.
What type of mirror is shown in Fig. 23-50?
The mirror is concave, and the person is standing inside the focal point so that a virtual, upright image is formed. A convex mirror would also form a virtual, upright image, but the image would be smaller than the object. See Fig. 23-16 and Example 23-4 for similar situations.
Why doesn't the light from the two headlights of a distant car produce an interference pattern?
The reason you do not get an interference pattern from the two headlights of a distant car is that they are not coherent light sources. The phase relationship between the two headlights is not constant—they have randomly changing phases relative to each other. Thus, you cannot produce zones of destructive and constructive interference where the crests and troughs match up or the crests and crests match up. Also, the headlights are far enough apart that even if they were coherent, the interference pattern would be so tightly packed that it would not be observable with the unaided eye.
If Young's double-slit experiment were submerged in water, how would the fringe pattern be changed?
The wavelength of light in a medium such as water is decreased when compared to the wavelength in air. Thus, dsinθ=mλ says that θ is decreased for a particular m and d. This means that the bright spots on the screen are more closely packed together in water than in air.
You can tell whether people are nearsighted or farsighted by looking at the width of their face through their glasses. If a person's face appears narrower through the glasses (Fig. 25-47), is the person farsighted or nearsighted? Try to explain, but also check experimentally with friends who wear glasses.
This person is nearsighted. Diverging lenses are used to correct nearsightedness, and converging lenses are used to correct farsightedness. If the person's face appears narrower through the glasses, then the image of the face produced by the lenses is smaller than the face, virtual, and upright. Thus, the lenses must be diverging; therefore, the person is nearsighted.
Where must the film be placed if a camera lens is to make a sharp image of an object far away? Explain.
To make a sharp image of an object that is very far away, the film of a camera must be placed at the focal point of the lens. Objects that are very far away have light rays coming into the camera that are basically parallel; these rays will be bent and focused to create an image at the focal point of the lens, which is where the film should be in order to record a focused image.
In attempting to discern distant details, people will sometimes squint. Why does this help?
To see far objects clearly, you want your eye muscles relaxed, which makes your lens relatively flat (large focal length). When you f-stop your eye down by closing your eyelids partially (squinting), you are only using the middle of your lens, where it is the most flat. This creates a smaller circle of confusion for the lens, which helps you see distant objects more clearly.
What is the angle of refraction when a light ray is incident perpendicular to the boundary between two transparent materials?
When a light ray meets the boundary between two materials perpendicularly, both the angle of incidence and the angle of refraction are 0°. Snell's law shows this to be true for any combination of the indices of refraction of the two materials.
A ray of light is refracted through three different materials (Fig. 23-49). Which material has (a) the largest index of refraction, (b) the smallest?
When the light ray passes from the left material to the middle material, the ray bends toward the normal. This indicates that the index of refraction of the left material is less than that of the middle material. When the light ray passes from the middle material to the right material, the ray bends away from the normal, but not far enough to make the ray parallel to the initial ray, indicating that the index of refraction of the right material is less than that of the middle material but larger than the index of refraction of the left material. The ranking of the indices of refraction is, least to greatest, left, right, and middle. So the middle has the largest index and the left has the smallest.
If a concave mirror produces a real image, is the image necessarily inverted? Explain.
Yes, if the mirror is the only piece of optics involved. When a concave mirror produces a real image of a real object, both do and di are positive. The magnification equation, m=-di/do results in a negative magnification, which indicates that the image is inverted. If instead a virtual object were involved, then an upright real image would be formed.
When you look down into a swimming pool or a lake, are you likely to overestimate or underestimate its depth? Explain. How does the apparent depth vary with the viewing angle? (Use ray diagrams.)
You are likely to underestimate the depth. The light rays leaving the bottom of the pool bend away from the normal as they enter the air, so their source appears to be more shallow than it actually is. The greater the viewing angle relative to the normal, the greater the bending of the light and therefore the shallower the apparent depth.
Parallel light rays cross interfaces from medium 1 into medium 2 and then into medium 3 as shown in Fig. 23-51. What can we say about the relative sizes of the indices of refraction of these media?
n2 > n1 > n3
