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IGCSE · Cambridge (CIE) · 0654

Calculate magnification from image and actual size

Use a ratio with both lengths in millimetres.

Section B2.2 · Size of specimens

Size of specimens · Cambridge IGCSE 0654 · B2.2

Mapped to the named 0654 section using teaching from the corresponding single-science course. A 0654-specific review of every Core and Supplement objective is still in progress. Written practice is self-checked; it does not demonstrate supervised laboratory competence or award an official Cambridge grade.

Before you begin

  • Recall the distinction between plant, animal and bacterial cells. Read the supplied evidence before drawing a conclusion.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which quantity is measured on the drawing?

What you will learn

  • State the magnification formula.
  • Calculate a dimensionless ratio from matching lengths.
The supplied numerical example compares image length 36 mm with actual length 0.09 mm, giving magnification ×400. The screen drawing itself is not a calibrated millimetre ruler.Image size divided by actualsizeMatching dimensions in mmImage length I = 36 mmActual length A = 0.09 mmMagnification M = I ÷ A36 ÷ 0.09 = 400Result: ×400Length units cancelNot 400 mm
The supplied numerical example compares image length 36 mm with actual length 0.09 mm, giving magnification ×400. The screen drawing itself is not a calibrated millimetre ruler.

Compare the same dimension

Magnification = image size ÷ actual size. Image size is the measured length in the drawing or displayed image; actual size is the corresponding length in the specimen. Measure the same dimension in both: do not divide image length by actual width.

When both lengths are in millimetres, the units cancel and magnification has no length unit. An image length of 36 mm and actual length 0.09 mm give 36 ÷ 0.09 = 400, written ×400. It is not 400 mm.

Check the direction of the ratio

An enlarged image has a greater length than the actual specimen, so its magnification exceeds one. Reversing the division would give 0.0025 in the example, which is inconsistent with enlargement. An image can also be reduced, with magnification below one; the formula still applies.

The lengths supplied in this lesson are fixed numerical data. A responsive drawing on a phone is not physically 36 mm long unless calibrated. Use the stated measurements, not an uncalibrated screen ruler or the SVG pixel width.

Sources for this lesson

Pause and explain

A 36 mm image represents 0.09 mm actual length. Which result is valid?

Put the idea to work

Worked example

An image is 36 mm long and the corresponding specimen is 0.09 mm long. Calculate magnification.

Show the worked solution
  1. Both lengths are in mm.
  2. Use 36 ÷ 0.09.
  3. Units cancel.

Answer ×400.

Try a topic test

Choose Easy, Normal or Difficult. Save your answers and open worked review after finishing.

Test this topic →

From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

An image is 24 mm long; actual length is 0.06 mm. Find magnification.

Give me a hint

Divide the larger image length by actual length.

Compare my reasoning
  1. 24 ÷ 0.06 = 400.
  2. Cancel mm units.

×400.

Look for these in your work

  • Correct division.
  • No length unit.
2 · Independent

An image is 54 mm wide; actual width is 0.18 mm. Find magnification.

Give me a hint

Use corresponding widths.

Compare my reasoning
  1. 54 ÷ 0.18 = 300.
  2. State the ratio.

×300.

Look for these in your work

  • Matching widths.
  • No mm on the answer.
3 · Transfer

A student divides an image length of 20 mm by actual width 0.04 mm and gets 500. Why is the calculation not justified even though arithmetic is correct?

Give me a hint

Matching units are necessary but not sufficient.

Compare my reasoning
  1. Length and width are different dimensions.
  2. Need actual length corresponding to the image length.

It compares non-corresponding dimensions; obtain actual length first.

Look for these in your work

  • Identify mismatch.
  • Specify the missing measurement.

Common mistakes

  • Divide image size by actual size, not the reverse.
  • Magnification has no mm unit.
Recall without your notes

What can you explain now?

What is the formula for magnification?

Compare with the explanation

Image size divided by actual size.

Use matching dimensions and matching units.

After trying it yourself, choose your next review. This is your self-assessment.

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Make it stick

Write the ratio, explain the unit cancellation and check its direction.

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Sciences — Co-ordinated (Double)

Course scope and assessment

This 2028–2029 pathway maps all 152 detailed subject-content sections in the published 0654 syllabus to original teaching and staged practice. It adapts teaching from the related single-science pathways and excludes clear 0610-only topics such as classification systems, human excretion, tropic responses, food supply, population dynamics, detailed pollution and biotechnology. Some explanations may include wider single-science context. A 0654-specific objective-by-objective review, Core/Supplement audit, human subject-teacher review, supervised laboratory experience and cumulative assessment remain in progress. Written practice is self-checked and does not predict an official grade. The 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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