13.2 Images Formed by Mirrors

1. AThey converge and all meet at point C.
   A student who takes the center of curvature to be the focal point picks this. C is the center of curvature, twice as far from the mirror as P. Rays through C strike the mirror along a radius and return along themselves; rays parallel to the axis meet at the focal point, P, halfway between the mirror and C.
2. BThey diverge as if they came from point Q.
   A student who expects a concave mirror to spread light, as a concave lens does, picks this. Q is behind the mirror; a concave mirror's surface curves around the incoming light and turns parallel rays in toward the axis, so they converge in front of the mirror, at P.
3. CThey reflect back along their own paths.
   A student who thinks light striking a mirror head-on always reflects straight back picks this. The normal to a curved mirror tilts at points away from the axis, so these rays reflect at an angle and converge at the focal point, P, on the axis halfway between the mirror and C.
4. DThey converge and meet at point P. Correct
   Rays parallel to the principal axis of a concave mirror reflect toward a common point, the focal point. P is on the axis halfway between the mirror and C, so P is the focal point.

CED 13.2.A.1 · Read this in Fix

1. AThey converge to a point in front of the mirror, where they cross.
   A student who thinks every mirror focuses parallel light in front of it picks this. That is true of a concave mirror; a convex mirror's surface bulges toward the light and turns the rays away from the axis, so they never cross.
2. BThey pass through a point behind the mirror, where they cross.
   A student who thinks light actually reaches the focal point behind a convex mirror picks this. No light goes behind the mirror; the reflected rays only appear to come from that point when they are extended backward.
3. CThey diverge as if they came from a point behind the mirror. Correct
   A convex mirror reflects rays arriving parallel to the principal axis so that they diverge; extended backward, they appear to come from a common point behind the mirror, its focal point.
4. DThey go back along their own paths, staying parallel to the axis.
   A student who thinks head-on rays always reflect straight back picks this. Except at the very center, the rays strike the curved surface at an angle to the normal, so they reflect at an angle and spread apart.

CED 13.2.A.2 · Read this in Fix

1. AOn the mirror's surface, as a flat mirror has a focal length of zero
   A student who equates 'no curvature' with a focal length of zero picks this. A flat mirror is the limit of a spherical mirror with an enormous radius of curvature, so f = R/2 becomes infinite, not zero.
2. BInfinitely far from the mirror, as parallel rays stay parallel after reflecting Correct
   A plane mirror reflects parallel rays as parallel rays. They never meet and never appear to come from one point at a finite distance, so the focal point is an infinite distance from the mirror (1/f = 0).
3. CBehind the mirror, as far behind it as an object is in front of it
   A student who thinks the image always forms at the focal point picks this. That location is where a plane mirror's image is; the focal point is a property of the mirror alone and does not depend on where any object is.
4. DIn front of the mirror, where the rays reflected from a distant lamp cross
   A student who thinks every mirror brings parallel rays together in front of it picks this. Light from a distant lamp arrives as parallel rays, and a plane mirror reflects them as parallel rays, which never cross.

CED 13.2.A.3 · Read this in Fix

1. ARs_o/(2s_o − R) Correct
   The focal point is halfway between the mirror and the center of curvature, so f = R/2. Then 1/s_i = 1/f − 1/s_o = 2/R − 1/s_o = (2s_o − R)/(Rs_o), and s_i = Rs_o/(2s_o − R).
2. BRs_o/(s_o − R)
   A student who takes the focal point to be at the center of curvature, f = R, picks this: 1/s_i = 1/R − 1/s_o. The focal point is halfway to C, so f = R/2.
3. C−Rs_o/(2s_o + R)
   A student who gives a concave mirror a negative focal length, f = −R/2, picks this: 1/s_i = −2/R − 1/s_o. A concave mirror's focal point is in front of it, so f = +R/2; the real image here has a positive s_i.
4. D(2s_o − R)/(Rs_o)
   A student who stops at 1/s_i picks this. The expression (2s_o − R)/(Rs_o) is 1/s_i, with units of 1/m; the image distance is its reciprocal.

Working f = R/2. 1/s_i = 1/f − 1/s_o = 2/R − 1/s_o = (2s_o − R)/(Rs_o) ⇒ s_i = Rs_o/(2s_o − R). Check: s_o = R gives s_i = R (object at C, image at C); s_o → ∞ gives s_i → R/2 = f. Distractors: f = R → Rs_o/(s_o − R); f = −R/2 → −Rs_o/(2s_o + R); reciprocal not taken → (2s_o − R)/(Rs_o).

CED 13.2.A.4 · Read this in Fix

1. AThe image no longer exists, as a real image forms only on a screen.
   A student who thinks a real image needs a screen picks this. The screen only scatters light that has already converged there; without it the rays still meet at the image location and continue on.
2. BThe image moves to the mirror's surface, where mirror images are seen.
   A student who thinks mirror images are on the mirror's surface picks this. The image is where the reflected rays meet, in front of the mirror; removing the screen does not change where the rays meet.
3. CAn eye beyond the image location, in the reflected light, still sees the image. Correct
   The reflected rays from each point of the flame still meet at the image location and then spread out beyond it, exactly as if they came from an object there. An eye in that spreading light sees the real image in mid-air.
4. DThe image is still there, but it is now virtual, as no screen catches it.
   A student who defines a real image as one caught on a screen picks this. Whether an image is real depends on whether the reflected rays actually meet there; they still meet at the same place without the screen, so the image is still real.

CED 13.2.A.5 · Read this in Fix

1. AIt is real, as the reflected rays meet at I, behind the mirror.
   A student who reads the dashed extensions as paths of light picks this. No light travels behind the mirror; the dashed lines only show where the reflected rays appear to come from.
2. BIt is virtual, as the reflected rays diverge as if they came from I. Correct
   The reflected rays spread apart and never meet; only their backward extensions (dashed) pass through I. An image at a point from which reflected rays only appear to originate is a virtual image.
3. CIt is real, as an observer in front of the mirror can see it.
   A student who thinks virtual images cannot be seen picks this. Virtual images are seen whenever you look into a mirror; what makes an image real is that the light actually passes through it.
4. DIt is on the mirror's surface, as that is where the rays reflect.
   A student who thinks the image is on the mirror picks this. The reflection happens at the surface, but the reflected rays appear to come from I, behind the mirror, which is where the image is seen.

CED 13.2.A.6 · Read this in Fix

1. AIt is inverted, standing upside down relative to the object.
   A student who reads the sign of s_i as the image's orientation picks this. The sign of s_i gives the image's location (behind the mirror); a virtual image formed by a single mirror is in fact upright.
2. BIt shows an error, as an image distance cannot be negative.
   A student who treats s_i as an unsigned length picks this. In the mirror equation s_i is a signed location, and −6.0 cm is a valid result meaning 6.0 cm behind the mirror.
3. CIt is 6.0 cm behind the mirror, where light meets to form a real image.
   A student who thinks light actually reaches points behind a mirror picks this. No light passes behind the mirror; the rays only appear to come from 6.0 cm behind it, so the image is virtual.
4. DIt is 6.0 cm behind the mirror, so it is a virtual image. Correct
   In this convention a negative image distance locates the image behind the mirror. Light does not pass behind a mirror, so the reflected rays only appear to come from there: the image is virtual.

Working 1/s_i = 1/f − 1/s_o = −1/12.0 − 1/12.0 = −1/6.0 cm⁻¹ ⇒ s_i = −6.0 cm. In the convention in which f is positive for a concave mirror, s_i is positive for an image in front of the mirror and negative for one behind it: the image is 6.0 cm behind the mirror. No light passes behind a mirror, so the image is virtual.

CED 13.2.A.7.i · Read this in Fix

1. A1.0 cm
   A student who inverts the magnification, |M| = s_o/s_i = 15.0/30.0, picks this. The image is farther from the mirror than the object, so it is larger: |M| = s_i/s_o = 2.0.
2. B4.0 cm Correct
   1/s_i = 1/(10.0 cm) − 1/(15.0 cm) = 1/(30.0 cm), so s_i = 30.0 cm. |M| = |s_i/s_o| = 30.0/15.0 = 2.0, so the image is 2.0 × 2.0 cm = 4.0 cm tall.
3. C2.0 cm
   A student who thinks every mirror forms a same-size image picks this. That is true of a plane mirror; for this concave mirror the image is twice as far from the mirror as the object, so |M| = 2.0.
4. D1.3 cm
   A student who places the image at the focal point picks this: |M| = 10.0/15.0. The image is at s_i = 30.0 cm, found from the mirror equation, not at F.

Working 1/s_i = 1/10.0 − 1/15.0 = (3 − 2)/30.0 ⇒ s_i = 30.0 cm. |M| = s_i/s_o = 30.0/15.0 = 2.0. h_i = |M|h_o = 2.0 × 2.0 cm = 4.0 cm.

CED 13.2.A.8 · Read this in Fix

1. AIt is larger, as a convex mirror magnifies, like a magnifying glass.
   A student who expects a convex mirror to magnify, as a convex lens does, picks this. A convex mirror diverges light and forms reduced images; that is why it shows a wide view of the store.
2. BIt is smaller, whatever the customer's distance from the mirror. Correct
   For any real object in front of a convex mirror, 1/s_i = 1/f − 1/s_o with f < 0 gives a virtual image with |s_i| < s_o, so |M| < 1: the image is always reduced, though it grows as the customer approaches.
3. CIt is equal, as a mirror forms images the same size as objects.
   A student who carries over the plane-mirror rule picks this. Only a plane mirror forms same-size images; a convex mirror's images are smaller than the objects.
4. DIt is larger for near customers, smaller for those far away.
   A student who treats a convex mirror like a concave one picks this. The image of a nearer customer is larger than that of a farther one, but it is still smaller than the customer.

CED 13.2.A.9.ii · Read this in Fix

1. A12 cm
   A student who adds the reciprocals, 1/s_i = 1/(20 cm) + 1/(30 cm), picks this. Rearranging 1/s_i + 1/s_o = 1/f gives 1/s_i = 1/f − 1/s_o, which is 1/(60 cm).
2. B30 cm
   A student who applies the plane-mirror rule (image as far from the mirror as the object) picks this. For a curved mirror the image distance depends on the focal length, through 1/s_i + 1/s_o = 1/f.
3. C20 cm
   A student who thinks the image always forms at the focal point picks this. Only the image of a very distant object forms near F; this object is only 1.5 focal lengths away, so its image is much farther out.
4. D60 cm Correct
   From the scale, f = 20 cm and s_o = 30 cm. 1/s_i = 1/f − 1/s_o = 1/(20 cm) − 1/(30 cm) = 1/(60 cm), so s_i = 60 cm, in front of the mirror.

Working Read from the figure: f = 20 cm, s_o = 30 cm. 1/s_i = 1/20 − 1/30 = 3/60 − 2/60 = 1/60 cm⁻¹ ⇒ s_i = 60 cm.

CED 13.2.A.7 · Read this in Fix

1. Ad₂ = 2d₁ Correct
   1/d₁ = 1/f − 1/(3f) = 2/(3f), so d₁ = 1.5f. 1/d₂ = 1/f − 1/(1.5f) = 1/(3f), so d₂ = 3f. Moving the object toward the focal point moves the image away from the mirror: d₂ = 2d₁.
2. Bd₂ = d₁/2
   A student who applies the plane-mirror rule, image distance equal to object distance, picks this: 3f then 1.5f. For a concave mirror, moving the object toward F moves the image away from the mirror.
3. Cd₂ = d₁
   A student who thinks the image always forms at the focal point picks this: d₁ = d₂ = f. The image is at F only for a very distant object; here d₁ = 1.5f and d₂ = 3f.
4. Dd₂ = 4d₁/5
   A student who adds the reciprocals picks this: 1/d = 1/f + 1/s_o gives d₁ = 3f/4 and d₂ = 3f/5. The correct rearrangement is 1/s_i = 1/f − 1/s_o.

Working 1/d₁ = 1/f − 1/(3f) = 2/(3f) ⇒ d₁ = 1.5f. 1/d₂ = 1/f − 1/(1.5f) = 1/(3f) ⇒ d₂ = 3f. d₂/d₁ = 2. Distractors: plane rule 1.5f/3f = 1/2; image at F: ratio 1; added reciprocals: (3f/5)/(3f/4) = 4/5.

CED 13.2.A.7 · Read this in Fix

1. A+60.0 cm
   A student who uses f = +20.0 cm for the convex mirror picks this: 1/(20.0 cm) − 1/(30.0 cm) = 1/(60.0 cm). A convex mirror's focal point is behind it, so f is negative in this convention.
2. B−30.0 cm
   A student who applies the plane-mirror rule (image as far behind as the object is in front) picks this. A convex mirror's image is closer to the mirror than that; the mirror equation gives −12.0 cm.
3. C−12.0 cm Correct
   For a convex mirror f = −20.0 cm. 1/s_i = 1/f − 1/s_o = −1/(20.0 cm) − 1/(30.0 cm) = −5/(60.0 cm), so s_i = −12.0 cm: a virtual image 12.0 cm behind the mirror.
4. D−20.0 cm
   A student who thinks the image forms at the focal point picks this. Only a very distant object has its image at the focal point; this object is 30.0 cm away, and its image is at −12.0 cm.

Working f = −20.0 cm, s_o = +30.0 cm. 1/s_i = 1/f − 1/s_o = −1/20.0 − 1/30.0 = −(3 + 2)/60.0 = −1/12.0 cm⁻¹ ⇒ s_i = −12.0 cm (virtual, behind the mirror). Distractors: f = +20.0 → +60.0 cm; plane rule → −30.0 cm; image at F → −20.0 cm.

CED 13.2.A.7.i · Read this in Fix

1. APoint P Correct
   A plane mirror forms the image of O on the perpendicular from O to the mirror, as far behind the mirror as O is in front: 3 squares, at P. Rays reaching any observer, extended backward, pass through P, so observers 1 and 2 both see the image there.
2. BPoint Q
   A student who thinks the image is straight ahead of each observer, behind the mirror, picks this. Q is opposite observer 2, but the backward extensions of the rays reaching observer 2 pass through P, the same point observer 1 sees.
3. CPoint R
   A student who thinks the image is twice as far behind the mirror as the object is in front picks this. R is 6 squares behind; 6 squares is the object-to-image distance, and the image is 3 squares behind the mirror.
4. DPoint S
   A student who thinks the image is on the mirror's surface picks this. S is where the perpendicular from O meets the mirror; the image is 3 squares behind it.

Working From the grid, O is 3 squares in front of the mirror, so its image is 3 squares behind the mirror on the perpendicular through O: point P. Rays reaching any observer, extended backward, pass through P, so observer 2 sees the image at P, the same point observer 1 sees.

CED 13.2.A.7.ii · Read this in Fix

1. Ah₂ = h₁
   A student who thinks a mirror always forms an image the same size as the object picks this. That holds for a plane mirror; for a concave mirror |M| = |s_i/s_o| changes as the object moves, here from 1/3 to 1.
2. Bh₂ = h₁/3
   A student who uses |M| = s_o/s_i picks this, getting 3 and then 1. The magnification is s_i/s_o, which increases from 1/3 to 1, so the image grows.
3. Ch₂ = 5h₁/3
   A student who adds the reciprocals when finding s_i picks this: s_i = 4f/5 and then 2f/3, giving |M| = 1/5 and then 1/3. With 1/s_i = 1/f − 1/s_o the magnifications are 1/3 and 1.
4. Dh₂ = 3h₁ Correct
   At 4f: 1/s_i = 1/f − 1/(4f) = 3/(4f), s_i = 4f/3, |M| = (4f/3)/(4f) = 1/3. At 2f: 1/s_i = 1/f − 1/(2f), s_i = 2f, |M| = 1. The image height is proportional to |M|, so h₂ = 3h₁.

Working s_o = 4f: 1/s_i = 1/f − 1/(4f) = 3/(4f) ⇒ s_i = 4f/3, |M₁| = 1/3. s_o = 2f: 1/s_i = 1/f − 1/(2f) = 1/(2f) ⇒ s_i = 2f, |M₂| = 1. h₂/h₁ = 3. Distractors: plane rule 1; inverted M: (1)/(3) → 1/3; added reciprocals: s_i = 4f/5 (|M| = 1/5) and 2f/3 (|M| = 1/3) → 5/3.

CED 13.2.A.8 · Read this in Fix

1. AIt passes back through the focal point F.
   A student who thinks every reflected ray passes through F picks this. Only rays arriving parallel to the axis reflect through F. This ray arrives through F, so, by the reversibility of light paths, it leaves parallel to the axis.
2. BIt travels back along its incoming path.
   A student who swaps the rules for the ray through F and the ray through C picks this. Only a ray through C strikes the mirror along the normal and returns along itself; this ray passes through F, halfway to C, so it reflects parallel to the axis.
3. CIt travels parallel to the principal axis. Correct
   The ray passes through the focal point on its way to the mirror, so it is the third principal ray: a ray through the focal point of a concave mirror reflects parallel to the principal axis.
4. DIt bends away from the axis, spreading out.
   A student who expects a concave mirror to spread light, as a concave lens does, picks this. A concave mirror converges light: this ray, arriving through F, leaves parallel to the axis.

CED 13.2.A.9.i · Read this in Fix

1. AIt is real: light itself converges at the image and then spreads out. Correct
   Both reflected rays (not extensions of them) cross at the tip of I, in front of the mirror, so light actually passes through the image: it is real. The diagram also shows it inverted and larger than O.
2. BIt is virtual, since every image formed by a mirror is virtual.
   A student who thinks all mirror images are virtual picks this. That is true for plane and convex mirrors, but here the reflected rays themselves meet in front of the mirror, which makes the image real.
3. CIt is real only when a screen is placed where the rays meet.
   A student who thinks a real image needs a screen picks this. The rays meet at I whether or not a screen is there; a screen would only make the image visible from all directions.
4. DIt is located at F, since a reflected ray passes through F.
   A student who thinks the image forms at the focal point picks this. One reflected ray does pass through F, but the image is where the two reflected rays meet, beyond C.

CED 13.2.A.9 · Read this in Fix

1. AIt is greater at 1.0 m, as the image looks bigger from closer up.
   A student who confuses how big the image looks with how big it is picks this. The image does fill more of the student's view up close, but its height stays equal to the student's height.
2. BIt is smaller at 1.0 m, as less of the student fits in the mirror.
   A student who equates the image with the part of it that fits in the mirror picks this. The image is always full size; in any case, how much of it the student can see in a given mirror does not change with distance.
3. CNeither has a definite height, as a virtual image is not really there.
   A student who thinks a virtual image is imaginary picks this. A virtual image has a definite position and size and can be seen and photographed; here it is as tall as the student at both distances.
4. DThey are equal: the image is always as tall as the student. Correct
   A plane mirror forms an upright, virtual image the same size as the object, at every distance. The image looks larger from 1.0 m only because it is closer to the student's eyes.

CED 13.2.A.9.ii · Read this in Fix

1. A0.25h_o Correct
   For a convex mirror the focal length is −f. Then 1/s_i = 1/(−f) − 1/(3f) = −4/(3f), so s_i = −(3/4)f: a virtual image behind the mirror. |M| = |s_i/s_o| = (3f/4)/(3f) = 1/4, so the image height is 0.25h_o.
2. B0.50h_o
   A student who enters the convex mirror's focal length as +f picks this: 1/s_i = 1/f − 1/(3f) = 2/(3f), so s_i = 1.5f and |M| = 1.5f/3f = 0.50. A convex mirror's focal point is behind it, so its focal length is −f, and the image is virtual, (3/4)f behind the mirror.
3. C4.00h_o
   A student who takes the magnification as s_o/s_i picks this: 3f/(0.75f) = 4.00. The magnification is |s_i/s_o| = 0.75f/3f = 0.25; a convex mirror's image of a real object is always smaller than the object.
4. D1.00h_o
   A student who applies the plane-mirror rule, same-size image, picks this. A curved mirror's image size depends on its focal length and the object's position: here |M| = |s_i/s_o| = 1/4.

Working Convex mirror: focal length −f; s_o = 3f. 1/s_i = 1/(−f) − 1/(3f) = −(3 + 1)/(3f) = −4/(3f) ⇒ s_i = −(3/4)f (virtual, behind the mirror). |M| = |s_i/s_o| = (3f/4)/(3f) = 1/4, so h_i = |M|h_o = 0.25h_o. Errors: +f for the convex mirror → 1/s_i = 1/f − 1/(3f) = 2/(3f), s_i = 1.5f, |M| = 0.50; |M| = s_o/s_i → 3f/(0.75f) = 4.00; plane-mirror rule (same size) → 1.00.

CED 13.2.A.8 · Read this in Fix

1. A0.40R Correct
   The focal length is R/2. An upright image formed by a concave mirror is virtual, behind the mirror, so s_i = −5s_o. Then 1/s_o − 1/(5s_o) = 4/(5s_o) = 2/R, so s_o = (2/5)R = 0.40R, inside the focal point, as a virtual image requires.
2. B0.80R
   A student who takes the focal point to be at the center of curvature, f = R, picks this: 4/(5s_o) = 1/R. The focal point is halfway between the mirror and the center of curvature, so f = R/2 and s_o = 0.40R.
3. C0.60R
   A student who thinks a negative image distance means an inverted image, and so gives the upright image a positive s_i = +5s_o, picks this: 6/(5s_o) = 2/R. At 0.60R the image is five times as tall but real and inverted, 3.0R in front of the mirror. The sign of s_i gives the image's location: an upright image here is virtual, so s_i = −5s_o.
4. D0.10R
   A student who places the image at the focal point, |s_i| = R/2, and then uses |s_i| = 5s_o picks this. The image is at the focal point only for a very distant object; here the mirror equation gives s_i = −2.0R.

Working f = R/2. Upright image from a concave mirror ⇒ virtual, s_i = −5s_o (|M| = 5). 1/s_o − 1/(5s_o) = 4/(5s_o) = 2/R ⇒ s_o = (2/5)R = 0.40R (inside F). Errors: f = R → 4/(5s_o) = 1/R → 0.80R; s_i = +5s_o → 6/(5s_o) = 2/R → 0.60R (real, inverted image); image at F → R/2 = 5s_o → 0.10R.

CED 13.2.A.7.i · Read this in Fix

1. A48.0 cm Correct
   The focal length is half the radius of curvature, 12.0 cm. A real image three times as tall as the object is three times as far from the mirror: s_i = 3s_o. Then 1/(3s_o) + 1/s_o = 1/(12.0 cm) gives s_o = 16.0 cm, so the screen is at s_i = 48.0 cm.
2. B96.0 cm
   A student who takes the focal length to be the whole radius of curvature, 24.0 cm, picks this: 4/(3s_o) = 1/(24.0 cm) gives s_o = 32.0 cm and s_i = 96.0 cm. The focal point is halfway between the mirror and the center of curvature, so f = 12.0 cm and the screen is 48.0 cm away.
3. C16.0 cm
   A student who takes the magnification to be the object distance divided by the image distance picks this: s_o = 3s_i and 1/s_i + 1/(3s_i) = 1/(12.0 cm) give s_i = 16.0 cm. The magnification is s_i/s_o, so an image three times as tall is three times as far from the mirror as the filament: 48.0 cm, with the filament at 16.0 cm.
4. D12.0 cm
   A student who thinks a mirror always forms its image at the focal point picks this. The image is at the focal point only when the object is very far away; for an image three times as tall as the filament, s_i = 3s_o, which puts the screen 48.0 cm from the mirror.

Working f = R/2 = 24.0 cm/2 = 12.0 cm. The image is caught on a screen, so it is real: s_o and s_i are both positive and |M| = s_i/s_o = 3, so s_i = 3s_o. 1/s_i + 1/s_o = 1/f: 1/(3s_o) + 3/(3s_o) = 4/(3s_o) = 1/(12.0 cm), so s_o = 16.0 cm and s_i = 3 × 16.0 cm = 48.0 cm. Check: 1/48.0 + 1/16.0 = 4/48.0 = 1/12.0 cm⁻¹. Errors: f = R = 24.0 cm gives s_o = 32.0 cm, s_i = 96.0 cm; |M| = s_o/s_i gives s_o = 3s_i, 4/(3s_i) = 1/(12.0 cm), s_i = 16.0 cm; image at the focal point gives 12.0 cm.

CED 13.2.A.8 · Read this in Fix