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Mirror and lens image investigations

Key idea: Locate a mirror image with a search pin by removing parallax. Find a real lens image by moving a screen to sharp focus, then measure its position and compare its size and orientation with the object.

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  • SEC G3 Combined Science Physics component 2027
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Take a moment to read at your own pace. Choose the text size and spacing that feel comfortable for you.

Pause after each idea, then explain it in your own words before moving on.

Narrower lines take effect when the screen has room.

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Where exactly is an image? For a real image formed by a converging lens, you can catch it on a screen and measure where the screen is. The image in a plane mirror cannot be caught on a screen, so you find it by lining up a pin with it until the two stay together as you move your head. Both methods let you describe the image from what you actually see. Optical pins are sharp, and ray-box lamps get hot.

Find the image in a plane mirror

A plane-mirror image is virtual: the reflected rays do not actually meet at the apparent image position, so a screen cannot catch it. Find it by removing parallax: a change in apparent alignment when you move your eye. If the search pin and image are at different depths, their sideways alignment changes as you move; matching depths removes that relative change.

  1. Stand the mirror on a line on paper and put an object pin a measured distance in front of it.
  2. Look at the image of the pin in the mirror. Use a short mirror so that a taller “search” pin behind it is visible directly above its edge. Place the search pin where the image seems to be; compare its alignment with the image below it.
  3. Move your head from side to side. If their sideways alignment changes, adjust the search pin. When the alignment stays the same as you move your head, the pin locates the depth of the apparent image.
  4. Measure the object and image distances at right angles to the mirror line.

Illustrative position readings:

object distance / cmimage distance / cm
4.04.1
6.05.9
8.08.0
10.010.2

In these illustrative readings, object and image distances differ by no more than 0.2 cm. The ideal plane-mirror model predicts equal distances. It is virtual, upright, the same size as the object and laterally inverted.

Constructing a plane-mirror imageA vertical plane mirror is between an upright object on the left and its equally sized virtual image on the right. Two rays from the object's top reflect into different parts of the same pupil. Every reflected path obeys the reflection law and carries an arrow pointing away from the mirror. Dashed extensions behind it meet at the top of the virtual image. The object and image have equal perpendicular distances from the mirror.Two rays from one object point locate its virtual image.Plane mirrorObject pointObjectVirtual image pointVirtual imageObserver's eyeNormalirDistance dSame distance dSolid rays carry light; dashed extensions locate where it appears to come from.
Constructing a plane-mirror imageA vertical plane mirror is between an upright object on the left and its equally sized virtual image on the right. Two rays from the object's top reflect into different parts of the same pupil. Every reflected path obeys the reflection law and carries an arrow pointing away from the mirror. Dashed extensions behind it meet at the top of the virtual image. The object and image have equal perpendicular distances from the mirror.Two rays from one object point locate its virtual image. irObject pointVirtual image pointPlane mirrorNormalObserver's eyeObjectVirtual imageDistance dSame distance dSolid rays carry light; dashed extensions locate where it appears to come from.
Two rays from the same object point reflect into the eye. Their backward extensions meet at the virtual image point, at the same perpendicular distance behind the mirror. The normal and angles are shown for the upper ray.

Mirror-line error. In many mirrors the reflecting silver layer is on the back of the glass. If you draw the line along the front of the glass, the reference line does not identify the reflecting plane and systematic discrepancies can result. Prefer a front-surface mirror if available; otherwise follow the apparatus instructions for locating the reflecting plane and mark it consistently.

Find the image formed by a converging lens

A converging lens can form a real image, which a screen can catch.

  1. Estimate the focal length: focus a distant window on the screen. The lens-to-screen distance is roughly the focal length, approximately 15 cm for the illustrative lens used below. Never use the Sun or look at the Sun through a lens.
  2. Put a lit object (such as a torch shining through a shape cut in card) a measured distance u from the lens. Start far enough away to find a sharp screen image.
  3. Move the screen until the image is as sharp as possible. Measure the image distance v from the centre of the lens.
  4. Compare the screen image with the object: record its height, orientation and whether it is larger or smaller. A sharp image on a screen is real.
  5. Repeat for several object distances. If the screen shows no image, check alignment, illumination and screen position. You may need to look through the lens at the object to see a virtual image; the absence of a screen image alone is not enough to classify it.
Measure object and real-image distances from the lens centreAn illuminated upright arrow is left of a converging lens. Two rays from its tip reach the lens and meet at one image point on a screen to the right. The image arrow is inverted and smaller. The measured object distance u runs from the object to the lens centre; image distance v runs from the lens centre to the screen. A rule lies beneath the apparatus. This is an ideal schematic, not the readings in the table.Locate the sharp image with a screenilluminatedarrow objectlensscreeninverted imageoptical centreobject distance uimage distance vmetre rule
Measure object and real-image distances from the lens centreAn illuminated upright arrow is left of a converging lens. Two rays from its tip reach the lens and meet at one image point on a screen to the right. The image arrow is inverted and smaller. The measured object distance u runs from the object to the lens centre; image distance v runs from the lens centre to the screen. A rule lies beneath the apparatus. This is an ideal schematic, not the readings in the table.Locate the sharp image with a screenilluminatedarrow objectlensscreeninverted imageoptical centreobject distance uimage distance vmetre rule
Focus a sharp image on the screen, then measure both distances from the lens’s optical centre. Compare image size and orientation with the object.

Illustrative observations with the same lens, not a table of cases to memorise:

u / cmv / cmimage
10.0none on screenvirtual, upright, magnified (seen by looking through the lens)
20.058.5real, inverted, magnified
25.037.8real, inverted, magnified
30.030.3real, inverted, approximately the same size
40.023.8real, inverted, diminished
50.021.6real, inverted, diminished
60.020.1real, inverted, diminished

You do not need to learn which object distance gives which image. Describe the image you actually see.

Guided practice 1

Reading the lens results

About 4 min

Problem

For the rows with a screen image, describe how image distance changes as the object moves away from the lens. In the first row, the rays leaving the lens diverge; their backward extensions meet where the upright, enlarged image appears. Use that evidence to explain why a screen cannot catch this image.

Describe the trend and explain the first row

Hints

Hint 1

trend

What happens to v as u goes from 20.0 cm to 60.0 cm? What value does it seem to approach?
Hint 2

first row

Which lines meet at the apparent object image: the actual rays or only their backward extensions?
Show solution step by step

  1. Describe the trend

    Method

    As u increases, v decreases, quickly at first and then more slowly, towards about 15 cm.

    Reason

    Rays from each point of a distant object arrive almost parallel and form an image near the focal plane.

    Working

    Answer

    At u = 20.0 cm, the image distance is 58.5 cm. At u = 60.0 cm, it is 20.1 cm.

  2. Explain the first row

    Method

    The first row shows an upright, magnified image through the lens. The given actual rays diverge, and only their backward extensions meet.

    Reason

    The rays leaving the lens spread out and never meet, so there is no real image to catch. They only seem to come from a virtual image on the same side as the object.

    Working

    Answer

    That image is seen by looking through the lens: virtual, upright and magnified.

Judging the sharpest image. The range of screen positions that looks sharp depends on the lens, illumination, object contrast and observer. Approach best focus from both directions, mark the ends of the interval that appears sharp, and use its midpoint. Half that interval gives an estimate of focus-setting ambiguity; scale reading and alignment can add further uncertainty. Work in a dimmed room so that the image is easy to see.

Check your understanding 1 Use actual rays to distinguish real and virtual images

Which image can be caught on a screen?
Show answer

The lens image formed where actual refracted rays meet can be caught on a screen: it is real. The plane-mirror image and a lens image where only backward extensions meet are virtual. If the screen is blank, first check illumination, alignment and available screen positions. Then use the rays or the observed image to decide whether it is virtual.

Check your understanding 2 Read positions rather than a holder’s edge

On an optical bench, an upright object of height 2.0 cm stands at the 45.0 cm mark. The lens’s optical centre is at 70.0 cm; its holder’s right-hand edge is at 72.0 cm. A sharp inverted screen image, measured as 3.0 cm high, appears over screen marks 107.4 cm to 108.2 cm. Use the midpoint to estimate object distance and image distance. Give the midpoint, object distance and image distance to 1 decimal place. Find the image-height ratio and half the sharp-position interval. Describe the image and correct the holder-edge measurement in the next question.

Show the steps Optional working
Unit: cm
Unit: cm
Unit: cm
Show the steps Optional working
Unit: cm
Unit: cm
Show the steps Optional working
Unit: cm
Unit: cm
Show answer

The screen midpoint is (107.4 + 108.2)/2 = 107.8 cm. Measure from the lens’s optical centre: u = 70.0 − 45.0 = 25.0 cm, v = 107.8 − 70.0 = 37.8 cm. The screen image is real, observed inverted, and magnified, because 3.0/2.0 = 1.5. Half the sharp-position interval is 0.4 cm; this is one focus-setting ambiguity, not the total uncertainty in v. Using the holder edge gives 35.8 cm, systematically 2.0 cm too small. More repeats with that reference retain the error: locate the optical centre, align the apparatus and approach focus from both directions.

Check your understanding 3 Describe the image and correct the measurement

Describe the image’s three characteristics from the observations. A learner measures image distance from the holder edge at 72.0 cm and says repeated readings fix that error. Explain the correction, and what the focus-setting interval tells you.

Show answer

The image is real because it is caught on a screen, inverted as observed, and magnified because its height is 1.5 times the object’s height. Measuring from 72.0 cm gives 35.8 cm instead of 37.8 cm, a systematic 2.0 cm underestimate that repeats retain. Measure from the optical centre and check alignment. Half the sharp-position range, 0.4 cm, estimates focus-setting ambiguity; reading and alignment add other uncertainties.

Common mistakes

  • Drawing the mirror line along the front of the glass when the reflecting layer is at the back.
  • Measuring object and image distances along a slant instead of at right angles to the mirror.
  • Measuring the image distance from the edge of the lens holder instead of the centre of the lens.
  • Calling an image virtual only because the screen shows nothing. Check the alignment and screen position first.
  • Taking the first sharp-looking screen position instead of approaching focus from both sides and using the middle.

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Syllabus and review details

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