Answers · Lenses and images

Lenses and real images — answer key

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

1. OQ1 · Practice

Options as numbered on the pupil sheet: 1 — A ray parallel to the principal axis; 2 — The focal length; 3 — The principal axis; 4 — The principal focus.

Model answer. Line 1 passes through the centre of the lens at 90° to it: the principal axis. Point 2 is where the rays parallel to the axis all meet after the lens: the principal focus, F. Distance 3, from the centre of the lens to the principal focus, is the focal length. Ray 4 is one of the rays of the parallel beam, travelling parallel to the principal axis before it reaches the lens.

A full answer includes

I matched 1 to the principal axis.
I matched 2 to the principal focus.
I matched 3 to the focal length.
I matched 4 to a ray parallel to the principal axis.
I did not match 3, a distance, to the principal focus, which is a point.

Also accept

1 principal axis; 2 principal focus; 3 focal length; 4 a ray parallel to the axis

Watch for

“Distance 3 is the principal focus.” — The principal focus is a point, where the rays meet. The distance from the lens to it is the focal length.
“Line 1 is a ray of light, because the rays travel along it.” — Line 1 is the principal axis, a construction line through the centre of the lens; a ray along it would go straight through.
“Point 2 is the focal length.” — The focal length is a distance, so it is drawn as a double arrow. Point 2 is the principal focus.

2. OQ2 · Practice

Model answer. A diverging lens bends rays away from each other, so the parallel beam spreads out after the lens. The rays never meet beyond the lens; extended backwards in straight lines, they all seem to start from one point between the lamp and the lens: the principal focus of the diverging lens.

A full answer includes

I chose: the rays spread apart; traced back, they seem to start at one point on the lamp’s side.
I did not choose a point beyond the lens where the rays meet.

Also accept

The rays spread apart; traced back, they seem to start at one point on the lamp’s side.

Watch for

“The rays meet at one point beyond the lens, then spread apart again after that point.” — That is what a converging lens does: it is thicker in the middle. A diverging lens bends the rays away from each other, so they never meet beyond it.
“The rays stay parallel to each other, but the lens squeezes them into a narrower beam.” — A narrower beam would need rays bent towards each other, and then they would not stay parallel. A diverging lens makes the beam wider.
“The rays stay parallel to each other, because a thin lens hardly bends any of them.” — A thin lens still bends every ray except the one along the principal axis.

3. OQ3 · Practice

Model answer. The lamp is 25 m away, very far compared with the distances on the rule, so the rays from it reach the lens practically parallel, and the lens brings them together at its principal focus. The sharp image on the card is therefore at the principal focus, and the focal length is the distance from the lens to the card. The card is at the lower mark, so subtract it from the lens mark: 62.0 cm − 49.5 cm = 12.5 cm.

Checked automatically

Focal length: 12.5 cm (3 significant figures)
- Common wrong answer 49.5 : That is where the card is on the rule. Find the distance between the card and the lens.
- Common wrong answer 62 : That is where the lens is on the rule. Find the distance between the card and the lens.
- Common wrong answer 111.5 : Find the difference of the two marks; do not add them.
- Common wrong answer 112 : Find the difference of the two marks; do not add them.

A full answer includes

I used the idea that the image of a very distant object is at the principal focus.
Method: I wrote focal length = 62.0 cm − 49.5 cm.
Answer: I wrote 12.5 cm.
I did not give 49.5 cm or 62.0 cm, the marks on the rule.

Also accept

12.5 cm

Watch for

“The focal length is 49.5 cm.” — That is the card’s mark on the rule. The focal length is the distance between the lens and the card: the difference of the two marks.
“A lamp across the room would give the same distance.” — Only a very distant object gives its image at the principal focus; a nearer object’s image is further away.

4. OQ4 · Practice

Options as numbered on the pupil sheet: 1 — Is refracted parallel to the principal axis; 2 — Is refracted through the principal focus on the far side; 3 — Goes straight on.

Model answer. A ray parallel to the principal axis is refracted through the principal focus on the far side of the lens, just as a whole parallel beam is. A ray through the centre of the lens goes straight on, because the middle of a thin lens is like a thin flat sheet of glass. A ray through the principal focus on the near side is refracted parallel to the axis: it is the first ray travelling backwards.

A full answer includes

I matched the parallel ray to “is refracted through the principal focus on the far side”.
I matched the ray through the centre to “goes straight on”.
I matched the ray through the near principal focus to “is refracted parallel to the principal axis”.
I did not match the ray through the centre to a ray that bends.

Also accept

parallel → through F; centre → straight on; through near F → parallel

Watch for

“A ray through the centre of the lens bends towards the principal focus.” — At the centre the two faces of a thin lens are parallel, so the ray goes straight on.
“A ray parallel to the axis stays parallel to it.” — The lens bends it through the principal focus on the far side.

5. OQ5 · Practice

Model answer. Upside down compared with the object is inverted. Shorter than the object is diminished; taller is enlarged, so the 6 cm flame is enlarged. A wall or a sheet of paper can catch the image only where the light really meets, so the image is real.

A full answer includes

(1) I wrote inverted.
(2) I wrote diminished.
(3) I wrote real.
(4) I wrote enlarged.
I did not write “smaller” or “bigger” instead of the syllabus words diminished and enlarged.

Also accept

inverted; diminished; real; enlarged

Watch for

“An image shorter than the object is “smaller”.” — Use the term the syllabus uses: diminished.
“The image on the wall is virtual, because it is only a picture.” — An image that a wall or screen catches is formed where the rays really meet: it is real.
“The image is upright, because the candle is upright.” — The flame on the wall points down, so the image is inverted.

6. OQ12 · Practice

Model answer. Ray A arrives parallel to the principal axis and is refracted through F on the far side: correct. Ray C passes through F on the near side and leaves parallel to the axis: correct. Ray B arrives at the centre of the lens, so it should carry on in the same straight line, downwards; instead it bends upwards. B is wrong, and it misses the point where A and C cross, the tip of the image.

A full answer includes

I chose B.
I gave the reason: a ray through the centre of the lens goes straight on, but B bends at the lens.
I did not choose C, the ray that leaves the lens parallel to the axis.

Also accept

Watch for

“C, because a ray should not travel parallel to the axis after the lens.” — A ray through the principal focus on the near side is refracted parallel to the axis, so C is right.
“A, because the ray should go through the F on its own side.” — A parallel ray goes through the principal focus on the far side of the lens, as A does.

7. OQ6 · Practice

Model answer. 2f = 2 × 15 cm = 30 cm. At 40 cm the object is further than 2f, so the image is real, inverted and diminished, between f and 2f on the other side. At 22 cm the object is between f and 2f, so the image is real, inverted and enlarged, further than 2f on the other side. Both objects are further than f, so both images are real and inverted. Row C.

A full answer includes

I worked out 2f = 30 cm.
I placed 40 cm further than 2f, so its image is diminished.
I placed 22 cm between f and 2f, so its image is enlarged.
I chose row C.
I did not choose an upright, virtual image for 22 cm, which is further than f.

Also accept

Row C

Watch for

“Row A.” — The sizes are swapped. The object further than 2f gives the smaller image; the one between f and 2f gives the bigger one.
“Row B.” — A converging lens gives a real, inverted image of every object further than f, however far away it is.
“Row D.” — 22 cm is further than the principal focus, 15 cm, so the rays meet and the image is real and inverted. Only an object nearer than 15 cm gives an upright, virtual image.

8. OQ7 · Practice

Model answer. Mark F 3.0 cm from the lens on both sides and draw the object as an arrow 3.0 cm tall, 7.5 cm from the lens. From its tip, draw a ray parallel to the axis as far as the lens, then on through F on the far side; draw a second ray straight through the centre of the lens. The two rays cross 5.0 cm from the lens on the far side, 2.0 cm below the axis. (a) The image is 5.0 cm from the lens. (b) It is 2.0 cm tall, shorter than the object, because the object is further away than 2f = 6.0 cm.

Checked automatically

(a) Distance of the image from the lens: 5.0 cm (within ±0.5 cm)
- Common wrong answer 3 : That is the focal length. Find where your two rays cross.
- Common wrong answer 7.5 : That is the object distance. Measure on the other side of the lens, where the rays cross.
(b) Height of the image: 2.0 cm (within ±0.2 cm)
- Common wrong answer 3 : That is the height of the object. Measure the image, below the axis.

A full answer includes

I drew the ray parallel to the axis, then through F on the far side.
I drew the ray through the centre of the lens as one straight line.
(a) Method: I measured from the centre of the lens to where the two rays cross.
(a) Answer: I wrote about 5.0 cm.
(b) Method: I measured from the axis down to where the two rays cross.
(b) Answer: I wrote about 2.0 cm.
I did not give 3.0 cm, the focal length, as the image distance.

Also accept

(a) 5.0 cm (b) 2.0 cm

Watch for

“The image is at F, 3.0 cm from the lens.” — The image is at F only for a very distant object. Here the rays cross further out, at 5.0 cm.
“The image is 3.0 cm tall, the same as the object.” — Same size needs the object at 2f, 6.0 cm. At 7.5 cm the image is shorter than the object.
“The image is on the same side of the lens as the object.” — The object is further than F, so the rays really cross on the far side of the lens.

9. OQ8 · Practice

Model answer. At half size, F is 3.0 cm from the lens on each side, and the object is 4.0 cm from the lens and 1.0 cm tall. The ray parallel to the axis is refracted through F on the far side, and the ray through the centre goes straight on. They cross 12.0 cm from the lens on the paper, 3.0 cm below the axis. With the scale, (a) the image is 24 cm from the lens and (b) it is 6.0 cm tall, three times the height of the object.

Checked automatically

(a) Distance of the image from the lens (real cm): 24 cm (within ±1.5 cm)
- Common wrong answer 12 : That is the distance on your paper. Use the scale: 1 cm on the paper is 2 cm.
(b) Height of the image (real cm): 6.0 cm (within ±0.5 cm)
- Common wrong answer 3 : That is the height on your paper. Use the scale: 1 cm on the paper is 2 cm.

A full answer includes

I drew F 3.0 cm from the lens on the paper.
I drew the object 4.0 cm from the lens on the paper.
I drew the object 1.0 cm tall on the paper.
I drew the parallel ray, then through F on the far side.
I drew the ray through the centre of the lens straight on.
(a) Method: I measured where the rays cross on the paper and doubled it.
(a) Answer: I wrote about 24 cm.
(b) Method: I measured the image on the paper and doubled it.
(b) Answer: I wrote about 6.0 cm.
I did not give 12 cm or 3.0 cm, the lengths on the paper.

Also accept

(a) 24 cm (b) 6.0 cm

Watch for

“The image is 12 cm from the lens.” — That is the distance on the paper. With 1 cm for 2 cm, the real distance is twice as big: 24 cm.
“The image is 2.0 cm tall, the same as the object.” — Same size needs the object at 2f, 12 cm. At 8.0 cm it is between f and 2f, so the image is taller than the object.

10. OQ9 · Practice

Model answer. (a) A ray through the principal focus on the near side is refracted parallel to the principal axis, so after the lens ray 3 travels parallel to the axis, still 1.0 cm below it. (b) Every ray from the tip of the object that passes through the lens arrives at the tip of the image, so the tip of the image is on ray 3 after the lens, where ray 1 crosses it. Every point of that part of ray 3 is 1.0 cm below the axis, so the image is 1.0 cm tall. (c) The tip of the image is below the axis and the tip of the object is above it, so the image is inverted (and diminished: 1.0 cm against 4.0 cm).

Checked automatically

(b) Height of the image: 1.0 cm (within ±0.05 cm)
- Common wrong answer 4 : That is the height of the object. The tip of the image is on ray 3 after the lens.

A full answer includes

(a) I wrote parallel to the principal axis.
(b) Method: I used the idea that the tip of the image is on ray 3, which stays 1.0 cm below the axis.
(b) Answer: I wrote 1.0 cm.
(c) I wrote inverted.
(c) I gave the reason: the tip of the image is below the axis, while the tip of the object is above it.
I did not give 4.0 cm, the height of the object.

Also accept

(a) parallel to the axis (b) 1.0 cm (c) inverted: the tip of the image is below the axis

Watch for

“Ray 3 goes on through the principal focus on the far side.” — That is ray 1, the ray that arrives parallel. A ray through the near principal focus leaves the lens parallel to the axis.
“The image is 4.0 cm tall, the same as the object.” — The height of the image is the distance of its tip from the axis, and the tip is on ray 3, 1.0 cm from the axis.
“The image is upright, because ray 3 is parallel to the axis like ray 1.” — Upright or inverted depends on which side of the axis the tip of the image is: here below, so inverted.

11. OQ10 · Practice

Options as numbered on the pupil sheet: 1 — Slide the screen along the bench until the arrow’s image is sharp.; 2 — Focus the image of a distant flagpole on the screen.; 3 — Measure the distance from the centre of the lens to the screen.; 4 — Check the image: as tall as the arrow, and as far from the lens as the arrow.; 5 — Put the lamp’s arrow at twice that distance from the lens..

Model answer. The image of a distant object is at the principal focus, so focusing a distant flagpole on the screen and then measuring from the centre of the lens to the screen gives the focal length, f. Only then can the lamp go at 2f, twice that distance from the lens. Next, slide the screen until the arrow’s image is sharp. Last, check it: an object at 2f gives a real, inverted image the same size as the object, also 2f from the lens.

A full answer includes

Step 1: I focused a distant flagpole on the screen.
Step 2: I measured from the centre of the lens to the screen.
Step 3: I put the lamp’s arrow at twice that distance from the lens.
Step 4: I slid the screen until the arrow’s image was sharp.
Step 5: I checked the image’s height and distance last.
I did not place the lamp before I had measured the focal length.

Also accept

in the order given

Watch for

“Put the lamp at 2f first, then find the focal length.” — You cannot place the lamp at twice the focal length before you know the focal length: measure it first.
“Check the image as soon as something appears on the screen.” — A blurred patch appears at almost any distance. Check the image only where it is sharp.

12. OQ11 · Practice

Model answer. A converging lens gives a real image the same size as the object only when the object is at twice the focal length, 2f; the image is then at 2f on the other side. (a) So the screen is also 36 cm from the lens. (b) 2f = 36 cm, so the focal length is 36 cm ÷ 2 = 18 cm.

Checked automatically

(a) Distance from the lens to the screen: 36 cm (within ±0.5 cm)
- Common wrong answer 18 : A same-size image is as far from the lens as the object, at 2f.
(b) Focal length: 18 cm (within ±0.5 cm)
- Common wrong answer 36 : With the object at the principal focus there is no image. Same size means the object is at 2f.
- Common wrong answer 72 : The object is at twice the focal length, so halve 36 cm; do not double it.

A full answer includes

(a) I used the rule: a same-size image is at 2f, as far from the lens as the object.
(a) I wrote 36 cm.
(b) Method: I wrote 2f = 36 cm, so f = 36 cm ÷ 2.
(b) Answer: I wrote 18 cm.
I did not write f = 36 cm.

Also accept

(a) 36 cm (b) 18 cm

Watch for

“The focal length is 36 cm.” — At the principal focus there is no image at all. A same-size image means the object is at 2f.
“The screen is 18 cm from the lens.” — A same-size image is as far from the lens as the object: 36 cm.

13. OQ26 · Practice

Model answer. (a) Four results lie between 16.6 cm and 17.0 cm. The result 13.8 cm is about 3 cm below all of them, so it is anomalous: perhaps the card was read where the image was still blurred. (b) Leave the anomalous result out, and take the mean of the other four: (16.6 + 16.9 + 16.7 + 17.0) ÷ 4 = 67.2 ÷ 4 = 16.8 cm.

Checked automatically

(a) The anomalous result: 13.8 cm (exact value)
- Common wrong answer 17 : That result is close to the others. Find the one far from the rest.
- Common wrong answer 16.6 : That result is close to the others. Find the one far from the rest.
(b) The focal length to record: 16.8 cm (3 significant figures)
- Common wrong answer 16.2 : You included 13.8 cm. Leave the anomalous result out of the mean.
- Common wrong answer 13.8 : That is the anomalous result, not the focal length.
- Common wrong answer 67.2 : That is the total of the four results you kept. Divide by 4.
- Common wrong answer 81 : That is the total of all five results. Leave out 13.8 cm, then divide by 4.

A full answer includes

(a) I identified 13.8 cm as the anomalous result.
(b) Method: I wrote (16.6 + 16.9 + 16.7 + 17.0) ÷ 4.
(b) Answer: I wrote 16.8 cm.
I did not include 13.8 cm in the mean.

Also accept

(a) 13.8 cm (b) 16.8 cm

Watch for

“The largest result, 17.0 cm, is the anomalous one.” — Being the largest is not enough: 17.0 cm is close to three other results. The anomalous result is far from all the others.
“Every result goes into the mean.” — An anomalous result is left out, or it pulls the mean away from the true value: with it, the mean would be 16.2 cm.

14. OT1 · Apply

Model answer. (a) 3 m is 750 times the focal length, so the rays from each point of the friend that reach the small lens are practically parallel, and the lens brings them together almost at its principal focus, 4.0 mm behind it. (b) The friend is much further than 2f, so the image is real, inverted and diminished. (c) As the object comes closer to a converging lens, its image moves further from the lens. To keep the image on the sensor, the lens moves away from the sensor.

A full answer includes

(a) I compared 3 m with 4.0 mm: the friend is very far away.
(a) I wrote that the rays from each point arrive almost parallel.
(a) I wrote that parallel rays meet at the principal focus, 4.0 mm behind the lens.
(b) I wrote real, inverted and diminished.
(c) I wrote away from the sensor.
(c) I gave the reason: a nearer object gives an image further from the lens.
I did not write that the image on the sensor is upright.

Also accept

(a) far away, rays nearly parallel, image near F (b) real, inverted, diminished (c) away from the sensor

Watch for

“The image is upright, because the photo on the screen is upright.” — The image on the sensor is inverted; the phone turns it the right way up before it shows the picture.
“The lens moves towards the sensor for a near object.” — A nearer object gives an image further from the lens, so the lens must move away from the sensor.