13.4 Images Formed by Lenses

1. AIt spreads away from the axis, as if coming from the near focal point.
   A student who carries over the mirror rule (a convex mirror spreads light out) picks this. A convex lens is thicker in the middle and converges parallel light; spreading as if from the near focal point is what a concave (diverging) lens does.
2. BIt crosses the principal axis at the focal point beyond the lens. Correct
   A convex lens is a converging lens: rays arriving parallel to the principal axis are refracted so that they converge toward a common location on the transmitted (far) side of the lens, the focal point.
3. CIt meets the axis at 2F, two focal lengths past the lens.
   A student who confuses the focal point with the point 2F picks this. Parallel rays converge at the focal point, one focal length beyond the lens; 2F is twice as far away.
4. DIt goes straight through, as it is already parallel to the axis.
   A student who confuses a ray parallel to the axis with the ray along the axis picks this. Only a ray along the axis (or through the center of the lens) goes straight through; a ray above the axis is bent toward the focal point.

CED 13.4.A.1 · Read this in Fix

1. AIt converges and passes through the focal point on the right.
   A student who carries over the mirror rule (a concave mirror converges light) picks this. A lens that is thinner in the middle is diverging: the refracted rays spread apart and never meet on the right.
2. BIt goes straight on, as it is already parallel to the axis.
   A student who confuses a ray parallel to the axis with the ray along the axis picks this. Only a ray along the axis (or through the center of the lens) passes straight through; these rays are above and below the axis, so the lens bends them away from it.
3. CIt spreads out as if it came from the focal point on the left. Correct
   The lens is thinner at its center than at its edges, so it is a concave (diverging) lens. Rays arriving parallel to the principal axis are refracted so that they diverge as if they came from the focal point on the incident side, which is the left.
4. DIt spreads out as if it came from a point 2f to the left of the lens.
   A student who confuses the focal point with the point at twice the focal length picks this. The refracted rays appear to come from the focal point itself, one focal length to the left of the lens.

CED 13.4.A.2 · Read this in Fix

1. ANo image forms any more, because an image needs a screen to form on.
   A student who thinks a real image exists only on a screen picks this. The screen only scatters the light toward observers; the rays meet at the image location whether or not a screen is there.
2. BIt forms instead on whatever surface the light reaches next, such as a wall.
   A student who pictures the image as a picture that travels until it lands on a surface picks this. The rays from each point of the flame meet at one distance only; on a wall farther away the light from each point is spread into a blur.
3. CThe image is still there, but it has now become a virtual image.
   A student who thinks 'real' means 'shown on a screen' picks this. The image is real because the refracted rays actually meet there, and they still do.
4. DRays from the flame still meet there, so the image still forms. Correct
   A real image is formed where rays from each point of the object meet after refraction. Removing the screen does not change the rays, so they still meet at the same place; an eye placed beyond that point and looking toward the lens sees the image.

CED 13.4.A.3 · Read this in Fix

1. AA screen placed at I would show a sharp image of O.
   A student who thinks every image can be caught on a screen picks this. Only the dashed extensions meet at I; no light actually arrives there, so a screen at I would show no image.
2. BThe rays reaching the observer only appear to come from I. Correct
   After the lens the two refracted rays diverge; they meet only when they are extended backward (the dashed lines), at I. So the light reaching the observer never passes through I: it only appears to come from there, and I is a virtual image.
3. CI is a real image, as are all images formed by a converging lens.
   A student who thinks converging lenses always form real images picks this. O is closer to the lens than the focal point, so the refracted rays still diverge and the image is virtual, as in a magnifying glass.
4. DI is a reduced image, which is smaller than the object O.
   A student who expects virtual images to be smaller than the object picks this. The diagram shows I taller than O: a converging lens with the object inside the focal point gives an enlarged virtual image.

CED 13.4.A.4 · Read this in Fix

1. AA screen 15 cm from the lens, on the object's side, would display it.
   A student who thinks a virtual image can be caught on a screen picks this. The negative s_i does place the image 15 cm from the lens on the object's side, but the refracted light never passes through that point; it only appears to come from there, so a screen placed there shows no image.
2. BThe image is inverted, because a negative value means upside down.
   A student who reads the sign of s_i as the orientation picks this. The sign of s_i gives location; a virtual image formed by a single lens is upright.
3. CThe image is virtual, so no screen placed anywhere would show it. Correct
   In this convention a negative s_i places the image 15 cm from the lens on the same side as the object. For a single lens the refracted light leaves on the other side and only appears to come from that point, so the image is virtual: the light never meets there, and no screen position catches the image.
4. DThe lens is diverging, as a converging lens forms real images.
   A student who thinks converging lenses always form real images picks this. A converging lens with the object inside the focal point also gives a negative s_i (a magnifying glass), so the sign alone does not show that the lens is diverging.

Working Convention: s_i < 0 → image 15 cm from the lens on the object's side; the refracted light only appears to diverge from there, so the image is virtual and cannot be caught on a screen. Both lens types can give s_i < 0.

CED 13.4.A.5.i · Read this in Fix

1. A1.5 cm
   A student who uses |M| = s_o/s_i picks this: 3.0 cm × 30.0/60.0. The magnification is s_i/s_o; an image farther from the lens than the object is larger than the object.
2. B6.0 cm Correct
   1/s_i = 1/20.0 cm − 1/30.0 cm = 1/60.0 cm, so s_i = 60.0 cm. |M| = |s_i/s_o| = 60.0/30.0 = 2.0, so h_i = 2.0 × 3.0 cm = 6.0 cm (the image is real and inverted).
3. C1.2 cm
   A student who adds the reciprocals picks this: 1/s_i = 1/20.0 + 1/30.0 gives s_i = 12.0 cm and h_i = 3.0 × 12.0/30.0. The thin-lens equation gives 1/s_i = 1/f − 1/s_o.
4. D2.0 cm
   A student who puts the image at the focal point picks this: s_i = 20.0 cm, so h_i = 3.0 × 20.0/30.0. The image is at the focal point only for a very distant object; here s_i = 60.0 cm.

Working 1/s_i = 1/f − 1/s_o = 1/20.0 − 1/30.0 = 0.0500 − 0.0333 = 0.0167 cm⁻¹ → s_i = 60.0 cm. |M| = s_i/s_o = 60.0/30.0 = 2.0; h_i = |M| h_o = 2.0 × 3.0 cm = 6.0 cm. (s_o/s_i: 1.5 cm; adding: s_i = 12.0 cm → 1.2 cm; s_i = f: 2.0 cm.)

CED 13.4.A.6 · Read this in Fix

1. AThrough the focal point on the right
   A student who thinks every ray through a converging lens goes through the far focal point picks this. That rule is for rays arriving parallel to the axis; a ray through the near focal point leaves parallel to the axis.
2. BStraight on, without changing direction
   A student who swaps the principal-ray rules picks this. The ray that goes straight on is the one through the center of the lens; this ray meets the lens below the center, after passing through the near focal point.
3. CBent outward, away from the principal axis
   A student who carries over the mirror rule (a convex mirror spreads light out) picks this. A convex lens is converging: it bends this ray back toward the axis until it leaves parallel to it.
4. DOut parallel to the principal axis Correct
   The ray passes through the focal point on the incident side before it reaches the lens. This is one of the three principal rays: a ray through the near focal point is refracted so that it leaves parallel to the principal axis.

CED 13.4.A.7.i · Read this in Fix

1. AThe whole image is still there, but it is dimmer. Correct
   Light from every point of the bulb passes through every part of the lens, and all of it is brought to that point's image. The uncovered half still brings light from each point of the bulb to its image point, so the whole image forms, with about half as much light.
2. BHalf of the image is missing from the screen.
   A student who thinks each part of the lens forms one part of the image picks this. Every image point receives light through the uncovered half of the lens, so no part of the image disappears.
3. CThe image is unchanged in both size and brightness.
   A student who thinks each image point is formed only by the ray through the center of the lens picks this. All the light from an object point that passes through the lens forms its image point, so blocking half of the lens halves the light and dims the image.
4. DNo image appears, since some principal rays are blocked.
   A student who thinks an image needs the principal rays picks this. The principal rays are only a convenient way to draw the image; the rays through the uncovered half still meet at every image point.

CED 13.4.A.3 · Read this in Fix

1. A+0.6f
   A student who adds the reciprocals picks this: 1/s_i = 1/f + 1/(1.5f) = 1/(0.6f). Rearranging 1/s_i + 1/s_o = 1/f gives 1/s_i = 1/f − 1/s_o.
2. B−0.5f
   A student who inverts each term separately picks this: s_i = f − 1.5f = −0.5f. The reciprocals must be combined first; the reciprocal of 1/f − 1/s_o is not f − s_o.
3. C+1.0f
   A student who thinks a lens always forms its image at the focal point picks this. Only a very distant object has its image at the focal point; an object at 1.5f has its image at 3.0f.
4. D+3.0f Correct
   1/s_i = 1/f − 1/s_o = 1/f − 1/(1.5f) = (1.5 − 1)/(1.5f) = 1/(3.0f), so s_i = +3.0f. The image is real, on the side of the lens opposite the object.

Working 1/s_i = 1/f − 1/s_o = 1/f − 1/(1.5f) = (1.5 − 1)/(1.5f) = 0.5/(1.5f) = 1/(3.0f) → s_i = +3.0f (positive: opposite side, real). Errors: 1/f + 1/(1.5f) = 2.5/(1.5f) → 0.6f; f − 1.5f = −0.5f; image at F → f.

CED 13.4.A.5 · Read this in Fix

1. A−8.00 cm Correct
   For a diverging lens f = −12.0 cm. 1/s_i = 1/f − 1/s_o = −1/12.0 cm − 1/24.0 cm = −3/24.0 cm, so s_i = −8.00 cm: a virtual image 8.00 cm from the lens, on the same side as the object.
2. B+24.0 cm
   A student who enters the focal length as +12.0 cm picks this: 1/s_i = 1/12.0 − 1/24.0 = 1/24.0. A diverging lens has a negative focal length in this convention, and it forms a virtual image of a real object.
3. C−24.0 cm
   A student who adds the reciprocals picks this: 1/s_i = −1/12.0 + 1/24.0 = −1/24.0. Rearranging the thin-lens equation gives 1/s_i = 1/f − 1/s_o.
4. D−36.0 cm
   A student who inverts each term separately picks this: s_i = f − s_o = −12.0 cm − 24.0 cm. The reciprocals must be combined before the reciprocal of the result is taken.

Working f = −12.0 cm, s_o = +24.0 cm. 1/s_i = 1/f − 1/s_o = −1/12.0 − 1/24.0 = −0.0833 − 0.0417 = −0.125 cm⁻¹ → s_i = −8.00 cm. (f = +12.0 cm: 1/s_i = 0.0417 → +24.0 cm; adding: −0.0833 + 0.0417 = −0.0417 → −24.0 cm; term by term: −12.0 − 24.0 = −36.0 cm.)

CED 13.4.A.5.i · Read this in Fix

1. AIt converges at F_R, the focal point on the right of the lens.
   A student who thinks a lens has one fixed focal point picks this. Parallel light converges on the side where it leaves the lens; for light arriving from the right that is the left side, at F_L.
2. BIt converges at a point twice as far to the left as F_L.
   A student who confuses the focal point with the point 2F picks this. Parallel light converges at the focal point itself, one focal length from the lens.
3. CIt converges at F_L, the focal point on the left of the lens. Correct
   A lens has a focal point on each side. Light arriving from the right leaves the lens on the left, so the convex (converging) lens brings the parallel beam together at the focal point on that transmitted side, F_L.
4. DIt spreads out as if it came from F_R, on the right.
   A student who carries over the mirror rule (a convex mirror spreads light out) picks this. A convex lens is converging: it brings parallel light to a focus on the side where the light leaves the lens, here at F_L.

CED 13.4.A.5.ii · Read this in Fix

1. A0.67 Correct
   At s_o = 2f: 1/s_i = 1/f − 1/(2f) = 1/(2f), so s_i = 2f. At s_o = 4f: 1/s_i = 1/f − 1/(4f) = 3/(4f), so s_i = 4f/3. The factor is (4f/3)/(2f) = 2/3 ≈ 0.67: the image moves closer to the lens, toward the focal point.
2. B2.00
   A student who assumes the image distance is proportional to the object distance, as for a plane mirror, picks this. For a lens 1/s_i = 1/f − 1/s_o, and the image moves from 2f to 4f/3, closer to the lens.
3. C0.50
   A student who treats s_i as inversely proportional to s_o picks this. The 1/f term in 1/s_i = 1/f − 1/s_o means that doubling s_o does not halve s_i; it changes s_i from 2f to 4f/3.
4. D1.00
   A student who thinks the image always forms at the focal point picks this, expecting no change. The image is at 2f at first and at 4f/3 afterward.

Working s_o = 2f: 1/s_i = 1/f − 1/(2f) = 1/(2f) → s_i = 2f. s_o = 4f: 1/s_i = 1/f − 1/(4f) = 3/(4f) → s_i = (4/3)f. Factor = (4/3)/2 = 0.67. (Proportional: 2.00; inverse proportion: 0.50; image at F: 1.00.)

CED 13.4.A.5 · Read this in Fix

1. AIt is upright, since an image keeps the object's orientation.
   A student who thinks the image is a copy of the object carried through the lens picks this. The image is built point by point: the rays from the tip meet below the axis, so the image is inverted.
2. BIt is inverted, because the tip's image point P lies below the axis. Correct
   The image of the tip is where the rays from the tip meet, at P. P is below the principal axis while the tip of the object is above it, so the image is inverted (and, since P is farther from the axis than the tip, enlarged).
3. CIt forms at the right-hand F, where one of the rays crosses the axis.
   A student who thinks images form at the focal point picks this. The parallel ray crosses the axis at F, but the image of the tip is where all three rays meet, at P, beyond 2F.
4. DIt exists at P just when a screen is placed there to catch it.
   A student who thinks a real image exists only on a screen picks this. The rays actually meet at P, so the real image forms there whether or not a screen is present.

CED 13.4.A.7 · Read this in Fix

1. Af/(d + f)
   A student who adds the reciprocals picks this: 1/s_i = 1/f + 1/d gives s_i = df/(d + f). Rearranging the thin-lens equation gives 1/s_i = 1/f − 1/d.
2. B(d − f)/f
   A student who uses |M| = s_o/s_i picks this. The magnification is |s_i/s_o|, with the image distance on top.
3. Cf/(d − f) Correct
   1/s_i = 1/f − 1/d = (d − f)/(df), so s_i = df/(d − f). Then |M| = |s_i/s_o| = [df/(d − f)]/d = f/(d − f).
4. D(d − f)/d
   A student who inverts each term separately picks this: s_i = f − d, of size d − f, so |M| = (d − f)/d. The reciprocals must be combined before inverting: s_i = df/(d − f).

Working 1/s_i = 1/f − 1/d = (d − f)/(df) → s_i = df/(d − f). |M| = s_i/s_o = f/(d − f). (Adding: s_i = df/(d + f) → f/(d + f); s_o/s_i → (d − f)/f; s_i = f − d → |M| = (d − f)/d.)

CED 13.4.A.6 · Read this in Fix

1. AOnly the image formed by lens Y is larger than the object. Correct
   Lens X: 1/s_i = 1/10 − 1/30 = 1/15, so s_i = 15 cm and |M| = 15/30 = 0.5 (reduced). Lens Y: 1/s_i = 1/20 − 1/30 = 1/60, so s_i = 60 cm and |M| = 60/30 = 2 (enlarged). Only Y's image is larger than the object.
2. BOnly lens X forms an image larger than the object.
   A student who uses |M| = s_o/s_i picks this, getting 30/15 = 2 for X and 30/60 = 0.5 for Y. The magnification is s_i/s_o, so X's image is reduced and Y's is enlarged.
3. CBoth images are exactly the same size as the object.
   A student who thinks an image is a same-size copy of the object picks this. The image size depends on where the rays from each point meet; here |M| is 0.5 for X and 2 for Y.
4. DNeither image is larger; both images are reduced.
   A student who puts each image at its lens's focal point picks this, getting |M| = 10/30 and 20/30. The images are at 15 cm and 60 cm, so Y's image is enlarged.

Working X: 1/s_i = 1/10 − 1/30 = 2/30 → s_i = 15 cm, |M| = 0.5. Y: 1/s_i = 1/20 − 1/30 = 1/60 → s_i = 60 cm, |M| = 2. (s_o/s_i: X 2, Y 0.5; images at F: 1/3 and 2/3.)

CED 13.4.A.7.ii · Read this in Fix

1. AD − x
   A student who thinks a lens always forms its image at the focal point picks this, taking f = s_i = D − x. The image is at the focal point only for a very distant object; for an object a distance x away, 1/f = 1/x + 1/(D − x).
2. BD
   A student who inverts each term of the thin-lens equation separately picks this: s_i = f − s_o gives f = s_o + s_i = x + (D − x) = D. The reciprocals must be added first: 1/f = 1/x + 1/(D − x).
3. Cx(D − x)/D Correct
   The image is on the screen, so s_o = x and s_i = D − x. Then 1/f = 1/x + 1/(D − x) = D/[x(D − x)], and f = x(D − x)/D.
4. DD/(x(D − x))
   A student who stops at 1/f picks this. D/[x(D − x)] is the value of 1/f, with units of m⁻¹; the focal length is its reciprocal, x(D − x)/D.

Working The image is on the screen: real, on the far side of the lens, so s_o = x and s_i = D − x (both positive). 1/f = 1/s_o + 1/s_i = 1/x + 1/(D − x) = (D − x + x)/[x(D − x)] = D/[x(D − x)] ⇒ f = x(D − x)/D. (D = 100 cm, x = 25 cm: f = 18.75 cm; check 1/25 + 1/75 = 4/75 ✓.) Errors: image at F → f = D − x; s_i = f − s_o → f = s_o + s_i = D; stopping at 1/f → D/(x(D − x)).

CED 13.4.A.5 · Read this in Fix

1. A1.2 mm
   A student who rearranges the thin-lens equation as 1/s_i = 1/f + 1/s_o picks this: s_i = 4.8 cm and |M| = 0.60, giving 1.2 mm. Subtracting, 1/s_i = 1/f − 1/s_o = −1/(24.0 cm), gives |M| = 3.0 and a 6.0 mm image.
2. B1.0 mm
   A student who solves the thin-lens equation term by term, s_i = f − s_o = 4.0 cm, picks this: |M| = 4.0/8.0 = 0.50, giving 1.0 mm. The reciprocals must be combined: s_i = −24 cm, |M| = 3.0 and h_i = 6.0 mm.
3. C6.0 mm Correct
   1/s_i = 1/(12.0 cm) − 1/(8.0 cm) = −1/(24.0 cm), so s_i = −24 cm: a virtual image on the letter's side of the lens. |M| = |s_i/s_o| = 24/8.0 = 3.0, so the image is 3.0 × 2.0 mm = 6.0 mm tall.
4. D3.0 mm
   A student who places the image at the focal point picks this: |M| = 12.0/8.0 = 1.5, giving 3.0 mm. With the letter inside the focal length, the image is virtual and 24 cm from the lens, so |M| = 3.0 and h_i = 6.0 mm.

Working f = +12.0 cm, s_o = +8.0 cm. 1/s_i = 1/f − 1/s_o = 1/12.0 − 1/8.0 = (2 − 3)/24.0 = −1/24.0 cm⁻¹, so s_i = −24 cm (a virtual image on the letter's side of the lens). |M| = |s_i/s_o| = 24/8.0 = 3.0, so h_i = |M|h_o = 3.0 × 2.0 mm = 6.0 mm. Errors: 1/s_i = 1/f + 1/s_o gives s_i = 4.8 cm, |M| = 0.60, 1.2 mm; s_i = f − s_o = 4.0 cm gives |M| = 0.50, 1.0 mm; image at the focal point gives |M| = 12.0/8.0 = 1.5, 3.0 mm.

CED 13.4.A.6 · Read this in Fix