13.1 Reflection

1. AA line drawn along a wavefront, joining all points of the light wave that are in step
   A student who confuses rays with wavefronts picks this. The line joining points in step is the wavefront itself; the ray crosses the wavefronts at right angles.
2. BA straight line at right angles to the wavefronts, in the direction the wave moves Correct
   A ray is a model of the direction in which light travels: a straight line perpendicular to the wavefronts, pointing the way the wave is traveling. It is a drawing tool, not a physical object.
3. CA thin physical beam, such as a laser's; the light from a lamp contains no such rays
   A student who thinks rays are real thin beams picks this. A ray is a line representing a direction of travel; light from a lamp can be represented by rays just as well as a laser beam can.
4. DA wavy line tracing the up-and-down path that the light follows as it travels
   A student who reads the sine-curve drawing of a wave as the path of the light picks this. In one material light travels in straight lines; the sine curve is a graph of how the wave varies, not a path.

CED 13.1.A.1 · Read this in Fix

1. AArrows pointing out of her eye toward the mirror, then on to the lamp
   A student who thinks seeing involves something traveling out of the eye picks this. The eye receives light; the arrows point from the source, via the mirror, into the eye.
2. BArrows pointing from the lamp to the mirror and then from the mirror to her eye Correct
   A ray diagram shows the path of the light before and after it interacts with matter, with arrows in the direction the light travels. Light leaves the lamp, reflects at the mirror and enters the eye.
3. CRays that end at the mirror's surface, where the image of the lamp is formed and seen
   A student who thinks light travels only as far as an image on the mirror picks this. The light reflects at the mirror and continues to the eye; the diagram must show its path after the reflection too.
4. DWavy lines showing the light moving up and down as it travels to the eye
   A student who thinks light follows a wavy path picks this. Rays are straight lines; light in air travels in straight lines between the lamp, the mirror and the eye.

CED 13.1.A.2 · Read this in Fix

1. AHer eyes send out light that strikes the page and then returns to them.
   A student who thinks seeing involves something traveling out of the eye picks this. Eyes do not send out light; they receive light that comes from the page.
2. BThe page is lit up, so she sees it without light going from it to her eyes.
   A student who thinks an object is seen just because it is lit picks this. Being lit is not enough: she sees the page only because light reflected from it reaches her eyes.
3. CSome of the lamplight reflected by the page travels into her eyes. Correct
   The page is not a source of light. Light from the lamp that is incident on the page is reflected from it in many directions, and the part that enters her eyes lets her see the page.
4. DThe page absorbs the lamp's light and then gives off light of its own.
   A student who thinks only shiny surfaces reflect light picks this. The page does not glow: it reflects the lamp's light, in many directions, which is why it can be seen but shows no image.

CED 13.1.B.1 · Read this in Fix

1. AIt makes 30° with the normal, on the other side of the normal.
   A student who uses the 30° angle to the surface as the angle of incidence picks this. Angles in the law of reflection are measured from the normal: θ_i = 60°, so θ_r = 60°.
2. BIt makes 60° with the normal, on the other side of it. Correct
   The ray makes 30° with the surface, so it makes 90° − 30° = 60° with the normal: θ_i = 60°. By the law of reflection θ_r = 60°, on the other side of the normal, so the reflected ray also makes 30° with the mirror.
3. CIt goes straight back from O along the incident ray's path.
   A student who thinks light bounces straight back from a mirror picks this. That happens only for a ray along the normal; this ray has θ_i = 60°, so it leaves at 60° on the other side of the normal.
4. DIt passes through O and continues below the mirror at 30° to it.
   A student who thinks light continues behind a mirror toward the image picks this. The light is reflected at the surface and stays in front of the mirror.

Working The ray makes 30° with the surface, so θ_i = 90° − 30° = 60°. By the law of reflection θ_r = θ_i = 60°, on the other side of the normal; the reflected ray therefore makes 30° with the mirror.

CED 13.1.B.2 · Read this in Fix

1. AThe rough surface sends rays off at random, with no law of reflection.
   A student who thinks the law of reflection holds only for mirrors picks this. Each ray still obeys θ_i = θ_r; the directions vary because the normals do.
2. BThe paper absorbs the light and then glows, giving off its own light in all directions.
   A student who thinks dull surfaces do not reflect picks this. White paper reflects most of the light that strikes it; it does not glow.
3. CThe light fills the whole room, so the eye need not receive light from the paper.
   A student who thinks a lit object is seen without light traveling to the eye picks this. The patch is seen from a position only if light reflected from it reaches that position; diffuse reflection sends light to all of them.
4. DEach ray obeys the law of reflection, but the normal varies from spot to spot. Correct
   The paper's surface is rough on a small scale, so the normal points in different directions at different points. Each ray reflects according to θ_i = θ_r about its own normal, and the reflected rays leave in many directions: diffuse reflection.

CED 13.1.B.3 · Read this in Fix

1. AP only, since the reflected ray also makes 40° with the normal Correct
   A smooth mirror gives specular reflection: the beam leaves in one direction, with θ_r = θ_i = 40° on the other side of the normal. P lies on that line; Q and R do not, so only P receives the reflected light.
2. BQ only, since the reflected ray makes 40° with the mirror
   A student who measures the angle of reflection from the mirror's surface picks this. Q's line makes 50° with the normal; the reflected ray makes θ_r = 40° with the normal and passes through P.
3. CR only, since the beam bounces right back to the laser
   A student who thinks reflected light bounces straight back picks this. The beam strikes at 40° to the normal, not along it, so it leaves on the other side of the normal.
4. DP, Q and R, as a mirror spreads light in all directions
   A student who expects light reflected from any surface to spread out picks this. That is true of a rough surface; a smooth mirror reflects the parallel light of the beam in a single direction.

Working θ_i = 40° (measured from the normal, as marked). Specular reflection: θ_r = 40° on the other side of the normal, a single direction; that line passes through P. Q lies on the line at 40° to the mirror, which is 50° from the normal; R is beside the laser, on the incident side.

CED 13.1.B.4 · Read this in Fix

1. AArrow A
   A student who thinks a ray runs along a wavefront picks this. Arrow A is tangent to the wavefront; a ray crosses the wavefront at 90°.
2. BArrow C
   A student who thinks a small source sends all its light straight ahead, across the page to the right, as parallel rays picks this horizontal arrow. Light from a small source spreads out in all directions, so at that dot the ray points up and to the right, directly away from S.
3. CArrow B Correct
   A ray is a straight line perpendicular to the wavefront, pointing in the direction of travel. For circular wavefronts around S that direction is straight out from S, along a radius, which is how arrow B is drawn.
4. DArrow D
   A student who thinks light follows a wavy path picks this. Arrow D points in the right direction, directly away from S, but a ray is a straight line; the light does not move up and down as it travels.

CED 13.1.A.1 · Read this in Fix

1. A7.5 cm Correct
   Light travels in straight lines from the lamp. The rays grazing the disk's edges spread out from the lamp, so by similar triangles the shadow's diameter is 3.0 cm × (1.00 m/0.40 m) = 7.5 cm.
2. B4.5 cm
   A student who scales by the disk-to-screen distance picks this: 3.0 cm × (0.60 m/0.40 m) = 4.5 cm. The rays meet at the lamp, so both distances in the ratio must be measured from the lamp.
3. C1.2 cm
   A student who expects the shadow to shrink with distance, as distant objects look smaller, inverts the ratio: 3.0 cm × (0.40 m/1.00 m) = 1.2 cm. The edge rays diverge from the lamp, so the shadow is larger than the disk.
4. D3.0 cm
   A student who thinks light from the lamp travels as parallel rays picks this: the shadow would be the same size as the disk. Rays from a small source spread apart, so the shadow is larger.

Working Rays from the lamp past the disk's edges form similar triangles with their vertex at the lamp. Shadow diameter = 3.0 cm × (1.00 m/0.40 m) = 7.5 cm.

CED 13.1.A.1.i · Read this in Fix

1. A×1
   A student who thinks the lamp's light travels as parallel rays expects a shadow the size of the disk wherever the screen is. The edge rays diverge from the lamp, so the shadow grows as the screen moves away.
2. B×½
   A student who expects a shadow to shrink with distance, as distant objects look smaller, picks this. The edge rays diverge, so the shadow on the farther screen is larger.
3. C×4
   A student who applies a square law to the diameter picks this. By similar triangles the diameter is proportional to the distance from the lamp; it is the shadow's area that increases by a factor of 4.
4. D×2 Correct
   Rays from the lamp past the disk's edges spread out in straight lines, so the shadow's diameter is proportional to the lamp-to-screen distance. Doubling that distance doubles the diameter.

Working Shadow diameter = d × (lamp-to-screen distance)/(lamp-to-disk distance). The lamp-to-screen distance goes from D to 2D with everything else fixed, so the diameter is multiplied by 2.

CED 13.1.A.1.i · Read this in Fix

1. AThe wave nature of light matters here, and a ray model cannot account for the spreading. Correct
   A ray model predicts a band no wider than the slit, since the rays of the narrow beam continue in straight lines. Spreading far wider than the slit is diffraction, which needs the wave model of light.
2. BRays from the slit's two edges fan out along straight lines toward the screen.
   A student who explains all light behavior with rays picks this. The rays of the beam are parallel and pass straight through the slit; nothing makes them fan out, so rays predict a narrow strip, not a wide band.
3. CLight waves reflect off the edges of the slit, which sends them out to the sides.
   A student who attributes the spreading to reflection from the slit's edges picks this. Narrowing the slit makes the spreading wider, which reflection from the edges would not do; the spreading is diffraction, a wave effect that reflection does not explain.
4. DOnly a few rays can fit through the slit, and they move apart as they travel.
   A student who thinks light is made of a countable number of rays picks this. Rays are only lines drawn to show directions of travel, and ray reasoning cannot explain why they would move apart; the spreading is diffraction.

CED 13.1.A.1.ii · Read this in Fix

1. AIts light is bright enough to reach the mirror, while a flashlight's light fades out.
   A student who thinks light travels only a limited distance, set by its brightness, picks this. A flashlight's light easily reaches a mirror across a room; it simply spreads out over a wide area.
2. BA flashlight's light is not made up of rays, so its path cannot be drawn.
   A student who thinks a ray is a physical thin beam picks this. Light from any source can be represented by rays; a flashlight's beam needs many diverging rays, which is less convenient but not impossible.
3. CLaser light is not a wave, so no wave effects can disturb its path.
   A student who takes the ray model of a laser for the laser light itself picks this. Laser light is a wave; its wave nature, studied later, can simply be neglected in this demonstration.
4. DIts narrow, single-color beam can be modeled as one ray that is easy to trace. Correct
   A laser gives a single narrow, monochromatic beam that spreads very little, so it behaves like one ray whose path, and angles, can be seen and measured. A flashlight's wide beam also consists of light that can be represented by rays, but many of them, spreading apart.

CED 13.1.A.1.iii · Read this in Fix

1. A2α+2θ Correct
   Turning the mirror through α turns its normal through α, so the fixed incident ray now makes θ + α with the normal. The reflected ray makes the same angle, θ + α, on the other side of the normal, so the angle between the two rays is 2(α + θ) = 2α + 2θ.
2. Bα+2θ
   A student who thinks the reflected ray turns with the mirror, through α, picks this. The angle of incidence increases by α, and so does the angle of reflection, so the angle between the rays grows by 2α and the reflected ray turns through 2α.
3. C180°−2α−2θ
   A student who takes the new angle of incidence, θ + α, as the angle to the mirror's surface picks this: each ray would then make θ + α with the surface, leaving 180° − 2(α + θ) between them. Angles of incidence and reflection are measured from the normal, so the rays are 2α + 2θ apart.
4. D0
   A student who thinks light bounces straight back along its incoming path picks this. That happens only when the ray strikes along the normal; here the angle of incidence is θ + α, so the reflected ray leaves on the other side of the normal, 2α + 2θ from the incident ray.

Working Turning the mirror through α turns its normal through α. The incident ray is fixed, so the new angle of incidence (measured from the new normal) is θ + α. By the law of reflection, θ_r = θ_i = θ + α, on the other side of the new normal. The angle at O between the incident ray (traced back toward the source) and the reflected ray is θ_i + θ_r = 2α + 2θ; the reflected ray has turned through 2α. (Errors: turning the reflected ray through α with the mirror gives α + 2θ; treating θ + α as the angle to the mirror's surface gives 180° − 2α − 2θ; bouncing straight back gives 0.)

CED 13.1.B.2 · Read this in Fix

1. A1.46 m
   A student who takes the incident ray's angle to the normal (about 40°) as the reflected ray's angle to the mirror's surface picks this: y = (1.75 m)(1.25/1.50) = 1.46 m. Both angles in the law of reflection are measured from the normal, so the reflected ray also makes about 40° with the vertical and rises 2.10 m.
2. B3.60 m
   A student who measures the reflected ray's horizontal run from the point below the laser, 3.00 m, rather than from the point where the beam reflects, picks this: y = (3.00 m)(1.50/1.25) = 3.60 m. The reflected ray starts at the mirror, so its run is 1.75 m and it rises 2.10 m.
3. C2.10 m Correct
   The normal is vertical, so the incident ray makes an angle with tan θ_i = 1.25/1.50 with the normal. The reflected ray makes the same angle on the other side of the normal and covers 3.00 m − 1.25 m = 1.75 m horizontally, so it rises (1.75 m)(1.50/1.25) = 2.10 m before it reaches the wall.
4. D1.07 m
   A student who sets up the similar-triangle proportion with one ratio inverted, y/1.50 m = 1.25 m/1.75 m, picks this. Equal angles with the normal make run/height the same for both rays: 1.75 m/y = 1.25 m/1.50 m, so y = 2.10 m.

Working The normal to the mirror is vertical. Incident ray: it falls 1.50 m while moving 1.25 m horizontally, so its angle of incidence θ_i (from the vertical normal) has tan θ_i = 1.25/1.50 (θ_i = 39.8°). Law of reflection: θ_r = θ_i, on the other side of the normal. The reflected ray starts at the mirror and covers 3.00 m − 1.25 m = 1.75 m horizontally, so its rise y satisfies tan θ_r = 1.75 m/y = 1.25/1.50, giving y = (1.75 m)(1.50/1.25) = 2.10 m. (The two right triangles are similar: y/1.75 m = 1.50 m/1.25 m.) Errors: angle to the normal reused as the reflected ray's angle to the mirror, y = (1.75 m)(1.25/1.50) = 1.46 m; run measured from the point below the laser, y = (3.00 m)(1.50/1.25) = 3.60 m; sides paired the wrong way, y = (1.50 m)(1.25/1.75) = 1.07 m.

CED 13.1.B.2 · Read this in Fix