Answers · Lenses and images

Virtual images — answer key

Check meaning, not exact wording. These are teaching points, not official Cambridge marks.

1. OQ13 · Practice

Model answer. After the lens the rays spread out, and they never meet on the far side, so R and S cannot be the principal focus. Extrapolated backwards, the spreading rays all meet at P, 10 cm in front of the lens on the side the light comes from: that is the principal focus of the diverging lens, and its focal length is 10 cm. Q is the centre of the lens.

A full answer includes

I chose P.
I traced the spreading rays backwards to where they meet, on the side the light comes from.
I did not choose R or S, on the far side of the lens.

Also accept

Watch for

“R, because the principal focus is behind the lens.” — For a converging lens it is. The rays from a diverging lens never meet behind it; they only seem to come from a point in front.
“Q, because the rays bend at the lens.” — Q is the centre of the lens. The principal focus is a focal length away from it.
“S, where the beam has spread out the most.” — The rays never meet on the far side, so no point there is the principal focus. Trace them backwards instead.

2. OQ14 · Practice

Model answer. A converging lens gives an upright, enlarged image only when the object is closer to the lens than its principal focus. The rays from each point of the lamp are then still spreading out after the lens, so they never meet; traced backwards, they seem to come from an image on the lamp’s side. That image is virtual: no light passes through it, so the screen shows only a blurred patch of light at every distance.

A full answer includes

I chose: no sharp image anywhere, because the rays never meet after the lens.
I linked an upright, bigger image to a virtual image.
I did not choose a sharp image at some distance.

Also accept

No sharp image anywhere: the rays never meet after the lens.

Watch for

“A sharp, upright image at one place, where all the rays meet.” — A sharp image on a screen needs rays that really meet, and such an image from a converging lens is inverted. Upright and bigger means the image is virtual.
“A sharp image far away, because spreading rays meet in the end.” — Rays that leave the lens spreading out spread further the further they go; they never meet.
“No sharp image: the lens stops the light reaching the screen.” — Light does reach the screen: it shows a patch of light. The rays just never come together.

3. OQ15 · Practice

Options as numbered on the pupil sheet: 1 — Virtual; 2 — Real.

Model answer. The ceiling light on the paper and the phone screen on the wall are real: a lens brings the rays together there, and the paper or the wall shows them. The splinter seen through the glass held close to it is virtual: the rays reaching your eye are spreading out and only seem to come from the image. The round bowl of water works like a converging lens with the goldfish closer to it than its principal focus, so the bigger goldfish is a virtual image too. The diverging spy-hole lens spreads the rays out, so the people seen through it are a virtual image. The same magnifying glass gives both kinds: real when the object is far away, virtual when the object is closer than its focal length.

A full answer includes

I matched the ceiling light on the paper to real.
I matched the splinter seen through the glass to virtual.
I matched the goldfish seen through the bowl to virtual.
I matched the view through the spy-hole lens to virtual.
I matched the phone screen on the wall to real.
I did not call the splinter’s image real just because it is seen clearly.

Also accept

ceiling light real; splinter virtual; goldfish virtual; spy-hole virtual; phone screen real

Watch for

“The splinter’s image is real, because you can see it clearly.” — You can see virtual images clearly too. A screen at the splinter’s image would show nothing.
“The goldfish’s image is real, because the goldfish is really there.” — The goldfish is real, but its bigger image is where the rays leaving the bowl only seem to come from. No light is there.
“The spy-hole image is real, because it is small.” — Size does not decide. A diverging lens spreads the rays out, so they never meet: its image is always virtual.
“The ceiling light on the paper is virtual, because it is upside down.” — Upside down is a clue for real, not virtual: the rays really meet on the paper.

4. OQ16 · Practice

Model answer. (a) Y is the diverging lens. Its patch only grows as the screen moves away, so the rays after Y are spreading out and never meet. X brings the rays together to a spot (the patch then grows again because the rays cross and spread out beyond it), so X is converging. (b) Extrapolated backwards in straight lines, the rays leaving Y all seem to come from one point on the principal axis, on the lamp’s side of the lens. That point is the principal focus of Y, even though the rays leaving Y never really meet there: it is a virtual focus.

A full answer includes

(a) I identified Y as the diverging lens.
(a) I gave the reason: the patch only grows, so the rays after Y are spreading out.
(b) I wrote that the spreading rays, traced backwards, seem to come from one point.
(b) I wrote that this point, on the lamp’s side of Y, is its principal focus.
I did not identify X as the diverging lens because its patch grows again.

Also accept

(a) Y: its patch only grows, so its rays spread out (b) traced back, the rays seem to come from a point on the lamp’s side: its principal focus

Watch for

“X is the diverging lens, because its patch grows again.” — X’s patch grows only after the rays have met at the spot and crossed. Only a converging lens brings rays to a spot.
“Y has no principal focus, because its rays never meet.” — The principal focus of a diverging lens is the point the spreading rays seem to come from, on the near side.

5. OQ18 · Practice

Model answer. (a) Pupil A’s lamp is further than 2f = 24 cm, so its rays meet at a real image between f and 2f on the other side, further than 12 cm from the lens (in fact 16 cm). The screen is too close: moving it further away, through the blur, finds a sharp, inverted, diminished image. (b) Pupil B’s lamp is closer than f, so the screen shows no sharp image at any distance; looking through the lens from the other side, B sees an upright, enlarged, virtual image. (c) In A’s case the rays really meet, only not where the screen is. In B’s case they never meet, because they are still spreading out after the lens; they only seem to come from a point behind the lamp.

A full answer includes

(a) I wrote that A should move the screen further from the lens.
(a) I wrote that A will find a sharp, real, inverted image there.
(b) I wrote that B finds no sharp image at any distance.
(b) I wrote that through the lens B sees an upright, enlarged image.
(c) I wrote that A’s rays really meet, further from the lens than the screen.
(c) I wrote that B’s rays are still spreading out after the lens.
(c) I wrote that B’s image is virtual.
I did not write that both screens are just in the wrong place.

Also accept

(a) move the screen further away: a sharp, real, inverted image (b) no sharp image anywhere; through the lens an upright, enlarged image (c) A’s rays meet beyond the screen; B’s never meet

Watch for

“Both screens are in the wrong place.” — Only A’s. B’s rays never meet, so there is no right place for a screen.
“A should move the screen closer to the lens.” — A screen at 12 cm is at the principal focus, where only a very distant object is focused. A’s lamp is nearer than that, so its image is further away.

6. OQ19 · Practice

Model answer. C is the tip of the object, and B is the principal focus on the far side, where the parallel ray crosses the axis. After the lens the two rays spread apart and never meet, but there is still an image: it is where they seem to come from. Extrapolated backwards, the dashed lines meet at A. The image is upright, enlarged and virtual, on the same side as the object. D is just a point between the spreading rays.

A full answer includes

I chose A.
I found it where the two dashed lines, the rays extrapolated backwards, meet.
I did not choose B, where only one ray crosses the axis.

Also accept

Watch for

“B, because the rays cross the axis there.” — Only one ray passes through B. The image is where both rays, extrapolated backwards, meet.
“D, because the image is on the far side of a lens.” — A real image is. These rays are spreading out after the lens, so the image is virtual, on the object’s side.
“No image, because the two rays never cross after the lens.” — They never cross after the lens, but the dashed lines, the rays extended backwards, do meet. The image is there: a virtual image.

7. OQ20 · Practice

Model answer. Mark F 5.0 cm from the lens on each side and draw the object 3.0 cm from the lens, 1.0 cm tall. From its tip, the ray parallel to the axis is refracted through F on the far side, and the ray through the centre goes straight on. After the lens they spread apart, so extrapolate both backwards with dashed lines: they meet 7.5 cm from the lens on the object’s side, 2.5 cm above the axis. (a) The image is 7.5 cm from the lens. (b) It is 2.5 cm tall. (c) It is upright, enlarged (2.5 times) and virtual, on the same side of the lens as the object.

Checked automatically

(a) Distance of the image from the lens: 7.5 cm (within ±0.8 cm)
- Common wrong answer 5 : That is the focal length. Find where the dashed lines meet.
- Common wrong answer 3 : That is the object distance. The image is further from the lens than the object.
(b) Height of the image: 2.5 cm (within ±0.3 cm)
- Common wrong answer 1 : That is the object’s height. Measure the image, where the dashed lines meet.

A full answer includes

I drew the parallel ray, then through F on the far side.
I drew the ray through the centre of the lens straight on.
I extrapolated both rays backwards with dashed lines.
(a) Method: I measured from the lens to where the dashed lines meet.
(a) Answer: I wrote about 7.5 cm.
(b) Method: I measured from the axis up to where the dashed lines meet.
(b) Answer: I wrote about 2.5 cm.
(c) I wrote upright, enlarged and virtual.
(c) I wrote that it is on the same side of the lens as the object.
I did not write that there is no image because the rays never cross.

Also accept

(a) 7.5 cm (b) 2.5 cm (c) upright, enlarged, virtual, on the object’s side

Watch for

“There is no image, because the rays never cross.” — They never cross after the lens, but extrapolated backwards they meet: that is the virtual image.
“The image is on the far side of the lens.” — The rays only seem to come from a point on the object’s side. That is where the virtual image is.

8. OQ21 · Practice

Model answer. Row D puts the ant 5.0 cm from lens P, beyond its principal focus, 4.0 cm: the rays meet, and the image is real and inverted, not upright. In the other rows the ant is inside the principal focus, so the image is upright, enlarged and virtual. Rows B and C have the ant at the same distance, 3.5 cm, but lens P has the shorter focal length and magnifies more: a ray diagram gives an image 8 times the height of the ant with lens P, but less than 2 times with lens Q. Rows A and C use the same lens P. At 3.5 cm the ant is much closer to the principal focus than at 1.0 cm, so its image is taller. Row C.

A full answer includes

I placed 5.0 cm beyond the principal focus of lens P, so that image is inverted.
I used the idea that, at the same distance, the lens with the shorter focal length magnifies more.
I used the idea that, inside F, the image grows as the ant gets closer to F.
I chose row C.
I did not choose row D, with the ant beyond the principal focus.

Also accept

Row C

Watch for

“Row A.” — Lens P is the right lens, but at 1.0 cm the ant is far inside its principal focus. The image grows as the ant gets closer to F, while still inside it.
“Row B.” — At the same distance, a lens with a longer focal length magnifies less, and lens Q has the longer focal length.
“Row D.” — At 5.0 cm the ant is beyond the principal focus of lens P, 4.0 cm: the rays meet, and the image is inverted, not upright.

9. OQ22 · Practice

Model answer. (a) At 5 cm the page is closer to the lens than its principal focus, so the rays from each letter are still spreading after the lens: the pupil sees an upright, enlarged, virtual image. (b) At 12 cm the paper is beyond the principal focus, between f and 2f. The rays from each letter now leave the glass coming together, heading for a real, inverted image about 24 cm above the glass. With her eye close to the glass, the rays reach it before they have met, and an eye cannot bring rays that are already coming together to a sharp point on its retina, so the letters blur. (c) With her eye well beyond that real image, the rays have met and are spreading out again when they reach her eye, so she sees the image sharply: the letters are upside down, and the image is real.

A full answer includes

(a) I wrote upright, enlarged and virtual.
(a) I gave the reason: the paper is closer than the principal focus.
(b) I wrote that the paper is now beyond the principal focus.
(b) I wrote that the rays from each letter leave the glass coming together.
(b) I wrote that they reach her eye before they have met.
(b) I wrote that her eye cannot focus rays that are already coming together.
(c) I wrote that the letters look upside down.
(c) I wrote real.
I did not write that the lens stops working when it is too far from the paper.

Also accept

(a) upright, enlarged, virtual (b) beyond F the rays leave coming together and reach the eye before they meet (c) upside down; real

Watch for

“The lens stops working when it is too far from the paper.” — The lens still works, but now it forms a real, inverted image instead of an upright, virtual one.
“The letters blur because the rays leave the glass parallel.” — Parallel rays are what a relaxed eye focuses most easily. The blur starts when the paper is beyond F and the rays leave the glass coming together.
“The sharp letters in (c) are upright, as before.” — Beyond the principal focus the rays cross, so the image they form is inverted.

10. OQ23 · Practice

Options as numbered on the pupil sheet: 1 — Near objects; 2 — A diverging lens; 3 — A converging lens; 4 — Distant objects.

Model answer. A short-sighted person sees near things clearly but distant things blurred, because the eye brings the rays from them together in front of the retina; a diverging lens corrects it. A long-sighted person sees distant things clearly but near things blurred, because the rays from near objects would meet behind the retina; a converging lens corrects it.

A full answer includes

I matched “a short-sighted person sees these blurred” to distant objects.
I matched “a short-sighted eye is corrected with” to a diverging lens.
I matched “a long-sighted person sees these blurred” to near objects.
I matched “a long-sighted eye is corrected with” to a converging lens.
I did not match short sight to near objects.

Also accept

short: distant blurred, diverging; long: near blurred, converging

Watch for

“A short-sighted person cannot see near things.” — Short-sighted people see near things clearly. It is distant things that are blurred.
“A long-sighted eye is corrected with a diverging lens.” — Rays from near things would meet behind the retina, so they need bringing together sooner: a converging lens.

11. OQ24 · Practice

Model answer. Ravi is short-sighted, so his eye brings the rays from the distant chart together too soon, in front of the retina. Lens J helps, so it spreads the rays a little first: J is diverging. Lens K makes things worse, so it does the opposite: K is converging. It starts bringing the rays together before they enter his eye, so they meet even sooner, further in front of the retina, and the chart blurs more. The problem is where the rays meet, not how big the letters look.

A full answer includes

I chose: K is converging, so the rays meet even further in front of the retina.
I used the idea that a short-sighted eye already brings the rays together in front of the retina.
I did not choose an answer about the size of the letters.

Also accept

K is converging, so the rays meet even further in front of the retina.

Watch for

“K is converging, so the rays meet later, further behind the retina.” — A converging lens adds to the bending of the eye, so the rays meet sooner, not later.
“K is diverging, so the rays meet still further in front of the retina.” — Spreading the rays first moves the meeting point back towards the retina. That is what J does.
“K is converging, so the letters look bigger and run into each other.” — Blur is not about size: it depends on where the rays meet, compared with the retina.

12. OQ25 · Practice

Model answer. A short-sighted eye sees distant objects blurred, because it brings their rays together too soon, in front of the retina: diagram C shows that, with no lens. A diverging lens spreads the rays out a little before they enter the eye, so they meet on the retina: diagram D shows the sight corrected. A converging lens, as in B, would make rays that already meet too soon meet even sooner, so B cannot be a short-sighted eye. Diagram A shows a phone held close and focused with no lens, which a short-sighted eye manages without help: nothing is corrected there.

A full answer includes

I looked for a distant object, because those are the ones a short-sighted eye sees blurred.
I looked for a diverging lens in front of the eye.
I checked that the rays meet on the retina.
I chose diagram D.
I did not choose diagram B, with the converging lens.

Also accept

Diagram D

Watch for

“Diagram A.” — A short-sighted eye sees near things clearly without help, so nothing has been corrected. Its problem is with distant objects.
“Diagram B.” — A converging lens brings rays together sooner. A short-sighted eye already brings the rays from a distant hill together too soon, so this cannot be a short-sighted eye.
“Diagram C.” — That is a short-sighted eye: the rays from the distant hill meet in front of the retina. But there is no lens, so its sight is not corrected.

13. OQ17 · Practice

Model answer. Leila sees distant things clearly but near things blurred: long sight. Her eye does not bring the rays from the near ticket together soon enough, so they would meet behind the retina. A converging lens brings the rays together a little before they enter her eye, and then her eye can bring them together on the retina. Row B.

A full answer includes

I named the problem: near things blur, so it is long sight.
I placed the meeting point behind the retina.
I chose a converging lens that brings the rays together a little.
I chose row B.
I did not choose a diverging lens.

Also accept

Row B

Watch for

“Row A.” — That is short sight, where distant things blur. Leila’s blur is with near things.
“Row C.” — The rays do meet behind the retina, but a diverging lens spreads them more and pushes the meeting point further back.
“Row D.” — Rays that meet in front of the retina need spreading, not more bending together. And Leila’s blur is with near things, which is long sight.

14. OT2 · Apply

Model answer. (a) At 4.0 cm the diamond is between f = 2.5 cm and 2f = 5.0 cm, so the rays from each point of it really meet on the far side of the loupe, further than 5.0 cm from it: the image is real, inverted and enlarged, and a card placed there catches it. With the diamond closer than 2.5 cm, the rays were still spreading out after the loupe and never met: the jeweller saw a virtual image, which the rays only seemed to come from, so no card could show it. (b) At twice full size, F is 5 cm from the lens and the diamond 4 cm away. The parallel ray through F on the far side and the ray through the centre spread apart; extrapolated backwards they meet 20 cm from the lens on the paper, which is 20 ÷ 2 = 10 cm from the real loupe, on the diamond’s side. (c) A short-sighted eye brings rays together strongly: the rays from a distant object meet too soon, in front of the retina. The rays from a ring held close are spreading out much more when they enter the eye, so the same eye brings them together on the retina, and she sees the letters clearly. A diverging lens would spread these rays out even more, so they would tend to meet behind her retina, and her eye would have to work harder to keep the letters sharp.

Checked automatically

(b) Supplement: Distance of the image from the loupe (real cm): 10 cm (within ±1 cm)
- Common wrong answer 20 : That is the distance on your paper. Divide by 2 for the real distance.
- Common wrong answer 2.5 : That is the focal length. Find where the extrapolated rays meet.

A full answer includes

(a) I placed 4.0 cm between f = 2.5 cm and 2f = 5.0 cm.
(a) I wrote real, inverted and enlarged.
(a) I wrote that the rays now really meet on the far side of the loupe.
(a) I wrote that with the diamond closer than 2.5 cm the rays never met.
(a) I wrote that the earlier image was virtual.
(b) Method: I measured where the dashed lines meet on the paper and halved it.
(b) Answer: I wrote about 10 cm.
(c) I wrote that her eye brings the rays from near things together on the retina without help.
(c) I wrote that a diverging lens would spread the rays from the ring out even more.
(c) I wrote that the rays would then tend to meet behind her retina.
I did not give 20 cm, the distance on the paper, for (b).

Also accept

(a) real, inverted, enlarged; these rays meet, while before they only seemed to come from the image (b) 10 cm (c) her eye already brings near rays together on the retina; the diverging lens would spread them more, towards behind the retina

Watch for

“At 4.0 cm the image is still upright and virtual.” — At 4.0 cm the diamond is beyond the principal focus, so the rays meet: the image is real and inverted.
“The image is 20 cm away.” — That is the distance on the paper. Divide by 2 for the real distance.
“A short-sighted person cannot see near things clearly.” — Short-sighted people see near things clearly. It is distant things that blur.
“Her glasses would make the letters clearer, because glasses always help.” — Her diverging glasses are for distant things. With near things her eye manages alone, and the lens would spread the rays too much.