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3.2.3b Ray diagrams and the nature of images

Two construction rays fix where a converging lens forms its image

Definition

Real image

An image formed at the place where refracted rays actually meet, so it can be caught on a screen held at that position.

  1. Start the diagram with the principal axis drawn as a horizontal line, and the thin converging lens drawn as a single vertical line crossing it, with the centre of the lens where the two lines meet.
  2. You already know the lens has a principal focus FFF on each side, one focal length fff from the centre, so mark FFF on both sides and then mark 2F2F2F at twice that distance, using the same scale to the left and to the right.
  3. Draw the object as an upright arrow standing on the axis to the left of the lens, and take both construction rays from the tip of the arrow, since that is the point whose image you are locating.
  4. The first construction ray leaves the tip parallel to the principal axis, and the lens refracts it so that it passes through the principal focus FFF on the far side.
  5. The second construction ray leaves the tip and travels straight to the centre of the lens, where it carries straight on without changing direction.
    1. That ray is undeviated because the two glass surfaces are very nearly parallel at the centre, so the bending on the way in is cancelled by the bending on the way out.
  6. Draw the bending as happening once, at the vertical line of the lens: each ray is one straight line up to that line and a second straight line after it.
  7. The point where the two refracted rays cross is the tip of the image, so drop a line from there to the principal axis and draw the image arrow between the two.
  8. Light really does arrive at that crossing point, so a white screen placed there shows a focused picture of the object.
  9. Put arrowheads on both rays pointing away from the object, because a ray diagram has to show the direction the light travels.

Three pairs of words describe an image completely

Definition

Inverted image

An image that points the opposite way to the object, so the top of the object appears at the bottom of the image.

  1. Compare heights first: the image is enlarged if its arrow is taller than the object, same size if the two are equal, and diminished if its arrow is shorter.
  2. Compare directions next: the image is upright if its arrow points the same way as the object's arrow, and inverted if it points the opposite way.
  3. Then say what kind of image it is: real where the refracted rays themselves cross, and virtual where the rays only appear to come from a point when they are traced backwards.
  4. A full description gives one word from each pair, for example real, inverted and diminished.
  5. Magnification puts a number on the size comparison.
    1. m=image heightobject heightm = \frac{\text{image height}}{\text{object height}}m=object heightimage height​
    2. A value of m>1m > 1m>1 means the image is enlarged, m=1m = 1m=1 means same size, and m<1m < 1m<1 means diminished.
    3. Magnification has no unit, because it is one length divided by another, so an object 2.0 cm2.0\ \text{cm}2.0 cm tall giving an image 6.0 cm6.0\ \text{cm}6.0 cm tall has m=3.0m = 3.0m=3.0.

The object's position relative to FFF and 2F2F2F sets the size of the real image

  1. In every case here the object stands beyond the principal focus, the rays cross on the far side of the lens, and the image is real and inverted; only its size and position change.
  2. With the object beyond 2F2F2F, the two refracted rays cross between FFF and 2F2F2F on the other side.
    1. The image is real, inverted and diminished, so m<1m < 1m<1.
    2. This is what happens inside a camera: a person standing several metres away gives a small sharp image a few centimetres behind the lens.
  3. With the object exactly at 2F2F2F, the rays cross at 2F2F2F on the other side.
    1. The image is real, inverted and the same size as the object, so m=1m = 1m=1 and the object distance equals the image distance.
    2. That symmetry is a quick way to find 2f2f2f on the bench: slide the object and the screen until the picture is exactly as tall as the object.
  4. With the object between 2F2F2F and FFF, the rays cross beyond 2F2F2F on the other side.
    1. The image is real, inverted and enlarged, so m>1m > 1m>1.
    2. This is the projector case: a small slide sitting close to the lens throws a large picture onto a screen across the room.
  5. With the object exactly at FFF, the ray through the centre and the refracted parallel ray leave the lens parallel to each other.
    1. Parallel rays never cross, so no image is formed: move a screen anywhere behind the lens and the patch of light spreads out but never sharpens.
  6. Putting the cases together gives one trend: as the object moves towards the lens while staying beyond FFF, the image moves further away and grows larger.
  7. If the object is closer to the lens than FFF, the refracted rays spread apart instead of crossing, so no real image forms and nothing can be caught on a screen.
Practical

Investigation: images formed by a converging lens

  • Aim: to find the focal length of a converging lens and describe the images it forms with the object at different distances.
  • Apparatus: an illuminated object (a torch shining through an arrow-shaped hole), a converging lens, a white screen, a metre ruler and a stand.
  • Method:
    • Clamp the lens upright in the stand near the middle of the bench and lay the metre ruler along the bench so its scale runs through the lens position, measuring every distance from the centre of the lens.
    • Find the focal length first: point the lens at a distant object seen through the window, hold the white screen behind the lens and slide it back and forth until the picture of that distant object is as sharp as you can make it.
    • Read the lens-to-screen distance off the ruler, because that distance is the focal length fff.
    • Mark FFF at a distance fff from the lens on each side, and 2F2F2F at twice fff on each side, so you can see at a glance where the object is sitting.
    • Set up the illuminated object with the torch behind the arrow and the arrow at the same height as the centre of the lens, then stand it beyond 2F2F2F.
    • Move the screen along the bench on the far side of the lens until the arrow looks sharp, then record the object distance, the image distance, the height of the arrow on the screen and whether it points up or down.
    • Repeat the whole measurement with the object at 2F2F2F, then between 2F2F2F and FFF, then at FFF, and finally closer to the lens than FFF.
    • For those last two positions no sharp arrow appears on the screen wherever you put it, so record that no image can be caught.
  • Measure and plot: build a table with columns for object distance, image distance, magnification, whether the image is real or virtual, and whether it is upright or inverted.
  • Result:
    • The distant object gives its image at the focal length, which is what makes the first measurement work.
    • Beyond 2F2F2F the image is real, inverted and smaller than the object.
    • At 2F2F2F it is real, inverted and the same size.
    • Between 2F2F2F and FFF it is real, inverted and larger.
    • At FFF, and anywhere closer than FFF, nothing sharp can be caught on the screen.
  • Watch out:
    • Judging the sharpest focus is a matter of eye judgement and is the main source of inaccuracy, so run the screen past the sharp point and back again to bracket it.
    • Repeat each setting and average the distances rather than trusting one reading.
    • Keep the lens vertical in its stand and measure every distance from the centre of the lens, not from the edge of its mount.
  • Safety: use a distant object seen through the window as the far object and never point the lens at the Sun, because focused sunlight can damage your eyes.

A projector and a camera work at opposite ends of the same rule

  1. A projector puts the slide just beyond FFF, between FFF and 2F2F2F, which throws a real, inverted, enlarged image onto a distant screen.
  2. Because that image is inverted, the slide is loaded upside down, so the picture lands on the screen the right way up.
  3. Focusing a projector means changing the object distance by a few millimetres, turning the lens in or out until the picture on the screen is sharp.
  4. A camera works at the other end of the range: the scene is many metres away, well beyond 2F2F2F, so the real, inverted, diminished image lands on the sensor just outside FFF.
  5. Focusing a camera on a nearer object means moving the lens slightly further from the sensor, because a smaller object distance pushes the image further back.
  6. Both devices work only because the image is real: a screen, a film or a sensor at the crossing point genuinely collects the light.
Example
  • A projector slide carries an arrow 3.0 cm3.0\ \text{cm}3.0 cm tall, and the picture of that arrow on the hall screen measures 1.5 m1.5\ \text{m}1.5 m from tip to tail.
  • Both heights must be in the same unit, so the image height is 150 cm150\ \text{cm}150 cm.
m=image heightobject height=1503.0=50 m = \frac{\text{image height}}{\text{object height}} = \frac{150}{3.0} = 50 m=object heightimage height​=3.0150​=50
  • The image is 505050 times as tall as the slide, so it is enlarged, and since the slide sits between FFF and 2F2F2F that image is also real and inverted.
Common Mistake
  • Do not draw a ray bending twice, once at each glass face: treat a thin lens as a single plane and bend each ray once, at the vertical line you drew for the lens.
  • Do not describe an inverted image as upside down and back to front, because lateral inversion belongs to a plane mirror; inverted means the image points the opposite way to the object.
  • Do not assume a wider lens gives a bigger image, since the diameter only controls how much light gets through and so how bright the image is, while the size is set by the object distance and the focal length.
  • Never forget that a real image can be caught on a screen, so when you are asked to show an image is real, hold a screen where the rays cross and look for the picture on it.
Exam technique
  • Mark FFF and 2F2F2F on both sides of the lens to the same scale before you draw a single ray, because a mis-scaled 2F2F2F throws the whole construction out.
  • Rule both rays past the point where they meet rather than stopping them at the lens, and put the image arrow exactly at that crossing point.
  • Answer a request to describe the image with one word from each of the three pairs, since a size and an orientation with no real or virtual is an incomplete answer.
  • Justify the word real by writing that the refracted rays actually meet, so the image can be formed on a screen.
  • Quote the object position back in any explanation of size, for example that the object lies between FFF and 2F2F2F, so the image comes out enlarged.
Self review
  • Which two rays do you draw from the tip of the object, and what does each one do at the lens?
  • An object stands between 2F2F2F and FFF, so where does the image form and which three words describe it?
  • Why is no image formed when the object sits exactly at FFF?
  • An object 4.0 cm4.0\ \text{cm}4.0 cm tall gives an image 1.0 cm1.0\ \text{cm}1.0 cm tall, so what is the magnification and which size word applies?
  • How would you show on the bench that the image from a converging lens is real?
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3.2.3b Ray diagrams and the nature of images Revision Guide

  1. IGCSE
  2. /Physics
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Revision notes for CIE IGCSE Physics 3.2.3b Ray diagrams and the nature of images: explanations and worked examples.

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