Physics · 3 Waves

3.2.3 Thin lenses

Before you start

Light travels in straight lines, and a ray is drawn with a ruler and an arrow showing its direction.
Refraction (3.2.2): light changes direction when it crosses from air into glass at an angle, and a ray that meets the surface along the normal goes straight on.
The image in a plane mirror (3.2.1) is virtual: you can see it, but no light actually comes from behind the mirror.
Reading and using a scale on a drawing, for example 1 cm on the page represents 2 cm in real life.

Syllabus objectives covered (8)

Describe the action of thin converging and thin diverging lenses on a parallel beam of light
Define and use the terms focal length, principal axis and principal focus (focal point)
Draw and use ray diagrams for the formation of a real image by a converging lens
Describe the characteristics of an image using the terms enlarged/same size/diminished, upright/inverted and real/virtual
Know that a virtual image is formed when diverging rays are extrapolated backwards and does not form a visible projection on a screen
Draw and use ray diagrams for the formation of a virtual image by a converging lens
Describe the use of a single lens as a magnifying glass
Describe the use of converging and diverging lenses to correct long-sightedness and short-sightedness

Part 1 of 8

Two shapes of lens, one axis

A lens is a piece of glass or clear plastic with curved surfaces. Light refracts as it goes into the lens and again as it comes out, so its direction changes twice.
In a thin lens the two bends are so close together that you treat them as one. In a ray diagram you draw each ray bending once, at the centre line of the lens. Do not try to draw the two separate refractions.
A converging lens is thicker in the middle than at the edges. A diverging lens is thinner in the middle than at the edges. The shape tells you which one you have.
Every lens has a principal axis: the line through the centre of the lens, at right angles to the lens. In a ray diagram it is the horizontal line that the lens sits on, and every distance you measure is measured along it.

Part 2 of 8

What each lens does to a parallel beam

Start with a beam of rays that are all parallel to the principal axis, such as light from a very distant lamp. This is the simplest test of what a lens does.
A converging lens bends the rays inwards, towards the principal axis. They all meet at one point on the axis on the far side of the lens. After that point they cross over and spread out again.
Notice why they can all meet at one point. A ray that passes further from the centre of the lens meets a more steeply sloping surface, so it is bent more. The ray along the axis itself meets both surfaces along the normal, so it is not bent at all.
A diverging lens bends the rays outwards, away from the principal axis. They spread apart after the lens and never meet.
Now trace the spreading rays backwards in straight lines, as dashed lines, to the side the light came from. They all meet at one point on the axis. To an eye on the far side, the light seems to come from that point.
ray diagram — not drawn yet
Two panels side by side. Left: a converging lens (thicker in the middle) on a horizontal principal axis; five rays arrive from the left parallel to the axis, bend inwards at the lens and meet at one point on the axis to the right, labelled F; the focal length f is marked as the distance from the centre of the lens to F. Right: a diverging lens (thinner in the middle) on a horizontal principal axis; five rays arrive from the left parallel to the axis and spread outwards after the lens; each spreading ray is extended back to the left as a dashed line, and the dashed lines meet at one point on the axis to the left of the lens, labelled F, with f marked from the centre of the lens to F.
A converging lens brings a parallel beam together at a point on the far side. A diverging lens spreads it out, so the rays only appear to come from a point on the near side.
When you describe this in words, say three things: which way the rays are bent (towards or away from the axis), whether they meet or spread, and where the point is. 'The light is focused' on its own does not say which lens or which side.

Part 3 of 8

Principal focus and focal length

The point where the parallel rays meet has a name. For a converging lens, the principal focus (or focal point) is the point on the principal axis where rays parallel to the principal axis meet after passing through the lens.
For a diverging lens, the principal focus is the point on the principal axis from which rays parallel to the principal axis appear to come after passing through the lens. It is on the same side as the light comes from, because that is where the dashed lines meet.
The focal length is the distance from the centre of the lens to the principal focus. It is measured along the principal axis, in cm or m. The symbol F marks the principal focus on a diagram.
Light can go through a lens in either direction, so a thin lens has a principal focus on each side, both at the same distance from the centre. A converging lens of focal length $10$ cm has an F $10$ cm to its left and another $10$ cm to its right.
In ray diagrams you also mark 2F, the point twice the focal length from the lens, on each side. It is not a new definition, only a useful landmark: the image changes its character as the object passes it.
This gives a quick way to measure the focal length of a converging lens. Light from a very distant object, such as a building seen through a window, arrives almost parallel. Move a screen behind the lens until the image of the distant object is sharp. The distance from the lens to the screen is then the focal length.

Part 4 of 8

Three rays you can always draw

Light leaves every point of an object in all directions, and every ray that reaches the lens helps to form the image. You only need to draw rays whose path you already know. There are three, all drawn from the top of the object.
Ray 1: parallel to the axis, then through F. A ray that arrives parallel to the principal axis is refracted through the principal focus on the far side. This is just the definition of F, used for one ray.
Ray 2: straight through the centre. A ray through the centre of the lens carries on in a straight line without bending. At the centre the two surfaces of a thin lens are parallel to each other, so the ray leaves in the same direction as it arrived.
Ray 3: through F on the near side, then parallel. A ray that passes through the principal focus on the object's side comes out parallel to the axis. It is Ray 1 run backwards, and light can always follow its own path in reverse.
ray diagram — not drawn yet
A converging lens drawn as a vertical line on a horizontal principal axis, with F marked at equal distances on each side. An upright arrow object stands on the axis to the left, further from the lens than F. From the top of the object three rays are drawn with direction arrows: ray 1 travels parallel to the axis to the lens, then passes through F on the right; ray 2 passes straight through the centre of the lens without bending; ray 3 passes through F on the left to the lens, then travels parallel to the axis. All three rays meet at one point below the axis on the right.
Any two of these three rays fix the top of the image. The third one is a check.
Two rays are enough, because two straight lines cross at only one point. Rays 1 and 2 are the easiest to draw accurately. Use Ray 3 as a check if you have time.
Draw every ray with a ruler and put an arrow on it. A freehand ray cannot give a measurement you can trust.

Part 5 of 8

Describing an image: three questions

Every image is described by answering three questions. Answer all three: 'a real image' on its own is only one third of a description.
How big is it? Compare its height with the object: enlarged (taller than the object), same size, or diminished (shorter than the object).
Which way up is it? Upright means the same way up as the object. Inverted means upside down. On a ray diagram an inverted image points to the other side of the axis from the object.
Is it real or virtual? A real image is formed where rays of light actually meet after passing through the lens. Put a screen at that place and a sharp picture appears on it.
A virtual image is formed when the rays leaving the lens are diverging (spreading apart) and never meet. Your eye traces them back in straight lines, and the image is where those backward extensions meet. Drawing a ray back past where it really went is called extrapolating it.
No light actually reaches the place where a virtual image is. So a screen placed there shows nothing: a virtual image does not form a visible projection on a screen. You can still see it, by looking through the lens. The image in a plane mirror is virtual for the same reason.

Part 6 of 8

Drawing a real image

Choose a scale that fits the page, then draw the principal axis and the lens as a vertical line. Mark F and 2F on both sides, measured from the centre of the lens.
Draw the object as an upright arrow standing on the axis, at the right distance and height for your scale. Standing it on the axis means the bottom of the image will also be on the axis, so you only need to find the top.
From the top of the object draw Ray 1 and Ray 2. Where they cross on the far side is the top of the image. Draw the image as an arrow from the axis to that point.
Now use the diagram. Measure the distance from the centre line of the lens to the image, and the height of the image, then convert both back with your scale. That is what 'use' a ray diagram means.
The image depends on where the object is. Object beyond 2F: the image is between F and 2F on the far side, real, inverted and diminished. A camera and the eye work like this.
Object at 2F: the image is at 2F on the far side, real, inverted and the same size as the object.
Object between F and 2F: the image is beyond 2F on the far side, real, inverted and enlarged. A projector works like this.
ray diagram — not drawn yet
Three ray diagrams stacked one above the other, for the same converging lens with F and 2F marked on each side. Top: object beyond 2F on the left; Ray 1 and Ray 2 from its top meet between F and 2F on the right, where a shorter inverted arrow is drawn, labelled 'real, inverted, diminished'. Middle: object at 2F on the left; the rays meet at 2F on the right, where an inverted arrow of the same height is drawn, labelled 'real, inverted, same size'. Bottom: object between F and 2F on the left; the rays meet beyond 2F on the right, where a taller inverted arrow is drawn, labelled 'real, inverted, enlarged'.
As the object moves in towards F, the real image moves further from the lens and gets larger. It is inverted every time.
Object at F: Rays 1 and 2 leave the lens parallel to each other, so they never meet and no image is formed.
Notice the pattern rather than learning five separate facts. As the object moves in from far away towards F, the real image moves away from the lens and grows. A real image made by a converging lens is always inverted.

Part 7 of 8

Object inside F: the virtual image and the magnifying glass

Put the object closer to the lens than F and the method changes at the last step. Draw Ray 1 (parallel, then through F on the far side) and Ray 2 (straight through the centre) exactly as before.
This time the two rays are spreading apart on the far side. They will never cross there, however far you extend them. The lens bends them inwards, but not enough to bring them together.
So extend both rays backwards, as dashed lines, to the object's side of the lens. The point where the dashed lines meet is the top of the image. Draw the image as a dashed arrow from the axis to that point.
The dashed lines are not light. They show where the light appears to come from. Keep the real rays solid and the backward extensions dashed, so it is clear which is which.
ray diagram — not drawn yet
A converging lens drawn as a vertical line on a horizontal principal axis, with F marked on each side. An upright arrow object stands on the axis between the left-hand F and the lens. From the top of the object, a solid ray travels parallel to the axis to the lens, then through F on the right; a second solid ray passes straight through the centre of the lens. On the right the two rays spread apart. Both are extended back to the left as dashed lines, which meet above the axis, further from the lens than the object; a taller, upright, dashed arrow is drawn there as the image. An eye on the right looks back through the lens.
The rays leaving the lens never meet. Traced back, they appear to come from a larger, upright image on the object's own side. That is a virtual image.
Describe the image with the three questions. It is on the same side as the object, and no light reaches it, so it is virtual. It is the same way up, so it is upright. It is taller than the object, so it is enlarged. It is also further from the lens than the object is.
That is exactly how a magnifying glass works. It is a single converging lens. You hold it so that the object is closer to the lens than the principal focus, and you look through the lens from the other side.
Your eye receives the diverging rays and sees an upright, enlarged, virtual image. Because it is virtual, you see it by looking into the lens: you could not catch it on a screen.
The condition matters. If the object is further from the lens than F, the lens forms a real, inverted image instead, and it no longer works as a magnifying glass.

Part 8 of 8

Correcting short sight and long sight

The front of the eye (the cornea and the eye lens) acts as a converging lens. It forms a real, inverted, diminished image on the retina at the back of the eye. You see clearly only when the rays are brought to a focus exactly on the retina.
Short sight. A short-sighted person sees near objects clearly, but distant objects look blurred. Rays from a distant object are brought to a focus in front of the retina, because the eyeball is too long or the eye lens is too strong. By the time the rays reach the retina they have spread out again into a blurred patch.
The eye converges the light too much, so the correction is a diverging lens in front of the eye. It makes the rays spread out slightly before they enter the eye. The eye then brings them to a focus further back, on the retina.
ray diagram — not drawn yet
Two diagrams of an eye, one above the other. Top, labelled 'short sight': parallel rays from a distant object enter the eye, meet at a point in front of the retina, and spread out again to a blurred patch on the retina. Bottom, labelled 'corrected': the same eye with a diverging lens in front of it; the parallel rays are spread slightly by the diverging lens, then the eye brings them to a point exactly on the retina.
Short sight: the focus is in front of the retina. A diverging lens moves it back onto the retina.
Long sight. A long-sighted person sees distant objects clearly, but near objects look blurred. Rays from a near object would be brought to a focus behind the retina, because the eyeball is too short or the eye lens is too weak. They reach the retina before they have met, so again there is a blurred patch.
The eye converges the light too little, so the correction is a converging lens in front of the eye. It makes the rays converge slightly before they enter the eye. The eye then brings them to a focus sooner, on the retina.
ray diagram — not drawn yet
Two diagrams of an eye, one above the other. Top, labelled 'long sight': rays spreading out from a nearby point object enter the eye and would meet at a point behind the retina, shown with dashed lines beyond the retina, leaving a blurred patch on the retina. Bottom, labelled 'corrected': the same eye with a converging lens in front of it; the rays are bent slightly inwards by the converging lens, then the eye brings them to a point exactly on the retina.
Long sight: the focus is behind the retina. A converging lens moves it forward onto the retina.
Remember the pairing by the reason, not by the name. Short-sighted eye: too much converging, so add a lens that diverges. Long-sighted eye: too little converging, so add a lens that converges.
A full answer has three parts: the type of lens, what it does to the rays before they enter the eye, and that the rays are then focused on the retina. Naming the lens alone is only one of the three.

Worked examples

Example 1. An object $4.0$ cm tall stands on the principal axis $24$ cm from a converging lens of focal length $8.0$ cm. Use a ray diagram to determine the distance of the image from the lens and the height of the image. Describe the image.

Show the working

Choose a scale first. The object alone is $24$ cm from the lens, and the image will be somewhere on the far side, so a full-size drawing will not fit. Use 1 cm on the page to represent 2 cm.
On that scale the object is $12$ cm from the lens and $2.0$ cm tall. F is $4.0$ cm from the lens on each side, and 2F is $8.0$ cm on each side. Draw the axis, the lens line, the marks, and the object arrow standing on the axis.
Draw Ray 1 from the top of the object, parallel to the axis, to the lens. From there draw it through F on the far side and carry it on.
Draw Ray 2 from the top of the object straight through the centre of the lens, with no bend.
The two rays cross $6.0$ cm to the right of the lens and $1.0$ cm below the axis. That is the top of the image, so draw the image arrow from the axis down to it. As a check, Ray 3 through the left-hand F leaves the lens parallel to the axis $1.0$ cm below it, and passes through the same point.
Convert back with the scale: each 1 cm on the page is 2 cm in real life.
$6.0 \times 2 = 12 cm 1.0 \times 2 = 2.0 cm$
Now describe it. The rays actually meet there, so it is real. It is below the axis while the object is above, so it is inverted. It is $2.0$ cm tall against an object of $4.0$ cm, so it is diminished.
Sense-check against the pattern. The object is beyond 2F ( $16$ cm), and the image is between F ( $8.0$ cm) and 2F ( $16$ cm) on the far side, real, inverted and diminished. That matches.

Answer: The image is $12$ cm from the lens, on the opposite side from the object, and $2.0$ cm tall. It is real, inverted and diminished.

Example 2. A converging lens has a focal length of $6.0$ cm. An object $1.0$ cm tall stands on the principal axis $4.0$ cm from the lens. Draw a full-size ray diagram and use it to determine the position and height of the image. Describe the image.

Show the working

Spot the key fact first. The object is $4.0$ cm from the lens, which is closer than F at $6.0$ cm. So expect a virtual image on the object's side, and leave room on the left of the page.
Draw the axis and the lens line. Mark F $6.0$ cm from the lens on each side. Draw the object $4.0$ cm to the left of the lens, $1.0$ cm tall.
Draw Ray 1 from the top of the object, parallel to the axis, to the lens. From there draw it through F on the right.
Draw Ray 2 from the top of the object straight through the centre of the lens.
On the right the two rays are moving apart, so they will never meet there. Extend both of them backwards to the left, as dashed lines.
The dashed lines meet $12$ cm to the left of the lens and $3.0$ cm above the axis. Draw a dashed image arrow from the axis up to that point.
Describe it. It is formed by extending diverging rays backwards and no light reaches it, so it is virtual. It is above the axis like the object, so it is upright. It is $3.0$ cm tall against $1.0$ cm, so it is enlarged.
Notice that this is a magnifying glass: object inside F, eye on the far side, upright enlarged virtual image.

Answer: The image is $12$ cm from the lens, on the same side as the object, and $3.0$ cm tall. It is virtual, upright and enlarged.

Example 3. A man can see distant hills clearly, but a newspaper held at a normal reading distance looks blurred. Identify his eye defect, state where light from the newspaper is brought to a focus in his eye, and explain how a lens corrects the defect.

Show the working

Read the symptoms. Distant objects are clear and near objects are blurred. That is long sight.
In a long-sighted eye, rays from a near object would be brought to a focus behind the retina. The eye does not converge them enough before they reach it.
The eye needs more converging, so the correction is a converging lens placed in front of the eye.
Say what the lens does to the rays: it makes them converge slightly before they enter the eye. Then say the result: the eye now brings them to a focus on the retina, so the image of the newspaper is sharp.

Answer: He is long-sighted. Light from the newspaper is brought to a focus behind the retina. A converging lens in front of the eye makes the rays converge slightly before they enter the eye, so they are brought to a focus on the retina and he sees a sharp image.

Where people go wrong

Mistake: Bending the ray that goes through the centre of the lens.

Instead: A ray through the centre of a thin lens carries straight on without changing direction.

At the centre the two surfaces are parallel to each other, so the ray leaves in the direction it arrived. Bending it moves the image to the wrong place, and every measurement taken from the diagram is then wrong.

Mistake: Defining the focal length as the distance from the lens to the image.

Instead: The focal length is the distance from the centre of the lens to the principal focus.

The image is at F only when the light arrives parallel, from a very distant object. For any nearer object the image is somewhere else, but the focal length of the lens has not changed.

Mistake: Putting the principal focus of a diverging lens on the far side.

Instead: It is on the same side as the light comes from.

The rays leaving a diverging lens spread out and never meet. The focus is where they appear to come from, which you find by tracing them back to the near side.

Mistake: Saying a virtual image cannot be seen.

Instead: You can see a virtual image by looking through the lens. What you cannot do is project it onto a screen.

Your eye only needs rays arriving at it, and the diverging rays do reach your eye. No light reaches the image position itself, so a screen placed there shows nothing.

Mistake: Giving a short-sighted person a converging lens.

Instead: Short sight is corrected with a diverging lens. Long sight is corrected with a converging lens.

A short-sighted eye already converges the light too much, so the focus is in front of the retina. A converging lens would move the focus further forward, which makes the blur worse.

Check yourself

A beam of light parallel to the principal axis passes through a thin converging lens. Describe what happens to the rays.
Answer
They are refracted inwards, towards the principal axis, and meet at one point on the axis on the far side of the lens: the principal focus.
The same beam passes through a thin diverging lens instead. Describe what happens to the rays, and state where the principal focus is.
Answer
The rays are refracted outwards, away from the axis, and spread apart. They appear to come from the principal focus, which is on the principal axis on the same side of the lens as the light comes from.
Define focal length.
Answer
The distance from the centre of the lens to the principal focus.
Light from a distant street lamp passes through a converging lens and forms a sharp image on a screen $15$ cm behind the centre of the lens. State the focal length of the lens, and explain how you know.
Answer
$15$ cm. Light from a very distant object arrives almost parallel, and parallel rays are brought to a focus at the principal focus, so the screen is at F.
An object stands $20$ cm from a converging lens of focal length $10$ cm. State where the image is formed and describe it.
Answer
The object is at 2F, so the image is at 2F on the other side, $20$ cm from the lens. It is real, inverted and the same size as the object.
Explain why a virtual image cannot be formed on a screen.
Answer
The rays leaving the lens are diverging and never meet. The image is only where they appear to come from when they are extended backwards. No light actually reaches that position, so nothing appears on a screen placed there.
State where the object must be placed when a converging lens is used as a magnifying glass, and describe the image that is seen.
Answer
Closer to the lens than the principal focus (between F and the lens). The image is virtual, upright and enlarged, and on the same side of the lens as the object.
A long-sighted student needs spectacles. State the type of lens, and explain what it does.
Answer
A converging lens. It makes the rays from a near object converge slightly before they enter the eye, so they are brought to a focus on the retina instead of behind it.

Exam questions

Try each part first. Open the mark scheme after you have written your answer.

Question 1. A converging lens has a focal length of $4.0$ cm. An object $1.5$ cm tall stands upright on the principal axis, $6.0$ cm from the centre of the lens.

(a) Define the principal focus of a converging lens. 2 marks No calculator

Mark scheme

B1 a point on the principal axis
B1 where rays (initially) parallel to the principal axis meet / converge after passing through the lens

(b) Draw a full-size ray diagram to locate the image of the object formed by the lens. 2 marks No calculator

Mark scheme

B1 ray from the top of the object parallel to the axis, refracted at the lens through F on the far side
B1 ray from the top of the object straight through the centre of the lens, undeviated; image drawn from the axis to where the two rays cross. Accept the ray through F on the object side that leaves the lens parallel to the axis as an alternative to either ray (any two correct rays)

(c) Use your diagram to determine the distance of the image from the lens and the height of the image. 2 marks

Mark scheme

B1 image distance 12 cm (accept 11.5–12.5 cm); unit required; ecf from the candidate's own diagram
B1 image height 3.0 cm (accept 2.8–3.2 cm); unit required

(d) Describe the image formed by the lens. 3 marks No calculator

Mark scheme

B1 real; CON if virtual is also given
B1 inverted; CON if upright is also given
B1 enlarged / magnified; CON if diminished or same size is also given

Question 2. A converging lens of focal length $5.0$ cm is used as a magnifying glass to look at a detail on a coin. The detail is $1.2$ cm tall.

(a) State where the coin must be placed for the lens to act as a magnifying glass. 1 mark No calculator

Mark scheme

B1 closer to the lens than the principal focus / less than 5.0 cm from the lens / between F and the lens

(b) The detail is placed on the principal axis, $3.0$ cm from the lens. Draw a full-size ray diagram to show how the lens forms an image of the detail. 2 marks No calculator

Mark scheme

B1 two correct rays from the top of the detail: one parallel to the axis then through F on the far side, one straight through the centre of the lens (accept the ray directed as if from F on the object side that leaves the lens parallel to the axis as an alternative to either)
B1 both emerging rays extended backwards (dashed) to meet on the same side of the lens as the detail, with the image drawn there

(c) Use your diagram to determine the distance of the image from the lens and the height of the image. 2 marks

Mark scheme

B1 7.5 cm (accept 7.0–8.0 cm) from the lens; unit required; the side need not be stated, but CON if stated as on the opposite side from the coin; ecf from the candidate's own diagram
B1 3.0 cm (accept 2.8–3.2 cm); unit required

(d) Explain why this image cannot be formed on a screen. 2 marks No calculator

Mark scheme

B1 the rays of light do not actually meet at / pass through the image position
B1 the rays leaving the lens are diverging and only appear to come from the image when extended backwards, so the image is virtual, owtte

Question 3. A student is short-sighted. She can read a book held close to her, but writing on a distant board looks blurred.

(a) State where, in relation to the retina, light from the distant board is brought to a focus in her eye. 1 mark No calculator

Mark scheme

B1 in front of the retina; do not credit 'behind the retina'

(b) Explain how a lens placed in front of her eye allows her to see the board clearly. 3 marks No calculator

Mark scheme

B1 a diverging lens is used (accept concave lens)
B1 it makes the rays diverge / spread out (slightly) before they enter the eye
B1 so the rays are brought to a focus on the retina / a sharp image forms on the retina

How this is examined

Drawing a ray diagram to find where a real image forms and how big it is: a lens, axis and object given to scale (or distances to draw yourself), followed by a Draw and then a Determine for the image distance or height One mark for each correct ray from the top of the object: parallel to the axis then through F on the far side; straight through the centre with no bend. Then one mark each for the image distance and the image height read from where the rays cross, converted with the scale and given with the unit. Rays drawn with a ruler, with arrows.
Drawing the virtual image made when the object is closer to a converging lens than F: a ray diagram with the object between F and the lens, which may be set as a magnifying glass, followed by a Determine for the position or size of the image The same two rays as for a real image, then both emerging rays extended backwards as dashed lines until they meet on the object's side. The image is drawn there, upright, and its distance is stated as being on the same side of the lens as the object.
How a lens corrects short sight or long sight: a two- or three-mark Explain or Describe about a person who cannot see near or distant objects clearly Three separate points: the right lens (diverging for short sight, converging for long sight); what it does to the rays before they enter the eye ('makes them diverge slightly' or 'makes them converge slightly'); and 'so the rays are brought to a focus on the retina'. Where the uncorrected focus is: in front of the retina for short sight, behind it for long sight.
Describing an image in the three pairs of terms: a Describe after a ray diagram, or a short question about an image on a screen or seen through a lens One word from each group: enlarged, same size or diminished; upright or inverted; real or virtual. Writing both words of one pair (for example 'real or virtual') is a contradiction and scores nothing for that pair.
Defining principal focus and focal length: a one- or two-mark Define, or a label to add to a diagram Principal focus: 'the point on the principal axis' and 'where rays parallel to the principal axis meet after passing through the lens' (for a diverging lens, 'appear to come from'). Focal length: 'the distance from the centre of the lens to the principal focus'.
Using a converging lens as a magnifying glass: a State or Describe about where to hold the lens and what the user sees 'The object is closer to the lens than the principal focus' and 'the image is virtual, upright and enlarged'. A distance stated as more than the focal length contradicts the first point.

Words to know

Converging lens: A lens thicker in the middle than at the edges. It bends a parallel beam inwards so that the rays meet at the principal focus.
Diverging lens: A lens thinner in the middle than at the edges. It bends a parallel beam outwards so that the rays appear to come from the principal focus.
Principal axis: The line through the centre of the lens, at right angles to the lens.
Principal focus (focal point), F: The point on the principal axis where rays parallel to the principal axis meet after passing through a converging lens, or appear to come from after passing through a diverging lens.
Focal length: The distance from the centre of the lens to the principal focus.
Real image: An image formed where rays of light actually meet. It can be formed on a screen.
Virtual image: An image formed where diverging rays appear to meet when they are extended backwards. It cannot be formed on a screen.
Magnifying glass: A single converging lens held so that the object is closer to it than the principal focus, giving a virtual, upright, enlarged image.
Short sight: Near objects are clear but distant objects are blurred, because light from them is focused in front of the retina. Corrected with a diverging lens.
Long sight: Distant objects are clear but near objects are blurred, because light from them would be focused behind the retina. Corrected with a converging lens.

All Physics lessons Next: 3.2.4 Dispersion of light