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IGCSE · Cambridge (CIE) · 0653

Calculate actual specimen size by rearranging magnification

Divide image size by magnification and retain length units.

Size of specimens · Cambridge IGCSE Combined Science 0653 · B2.2

Cross-referenced to the 2028–2029 0653 section using original teaching from the related single-science courses. A 0653-specific objective-by-objective and Core/Supplement review is in progress. Written practice is self-checked and 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

A 42 mm image at ×700 represents which actual length?

What you will learn

  • Rearrange the magnification formula.
  • Check an actual-size result by substitution.
An image length 42 mm at magnification ×700 gives actual length 0.06 mm. Multiplication by 700 checks that 0.06 mm maps to 42 mm in the specified image.Find the smaller specimendimensionImage and magnificationI = 42 mm · M = 700Actual size A = I ÷ M42 ÷ 700 = 0.06 mmCheck by substitution0.06 mm × 700 = 42 mmActual size is smaller
An image length 42 mm at magnification ×700 gives actual length 0.06 mm. Multiplication by 700 checks that 0.06 mm maps to 42 mm in the specified image.

Solve for the specimen dimension

Starting from M = I ÷ A, multiply by A and divide by M to obtain A = I ÷ M. An image of 42 mm at ×700 therefore represents 42 ÷ 700 = 0.06 mm. For an enlarged image, the actual dimension is smaller than the image dimension.

Magnification is dimensionless, so the actual answer retains the image length unit. If the image measurement is in millimetres, the result here is in millimetres. Do not multiply by magnification when finding the smaller actual dimension.

Substitute to check

A = 0.06 mm gives 42 ÷ 0.06 = 700, which recovers the stated magnification. This reverse check helps detect misplaced decimals. Label the result as an actual length or width, rather than a cell count.

A number printed as ×700 is valid only for the image scale it describes. A copied or resized figure may need a new magnification. If a scale bar is provided, measure it on the same displayed copy as the specimen.

Sources for this lesson

Pause and explain

Which substitution checks actual size A?

Put the idea to work

Worked example

An image measures 42 mm at ×700. Find the actual length in mm.

Show the worked solution
  1. Actual = image ÷ magnification.
  2. 42 ÷ 700 = 0.06.
  3. Check 0.06 × 700 = 42.

Answer 0.06 mm.

Try a topic test

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

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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 32 mm at ×800. Find actual length.

Give me a hint

Use A = I ÷ M.

Compare my reasoning
  1. 32 ÷ 800 = 0.04.
  2. Retain mm.

0.04 mm.

Look for these in your work

  • Correct rearrangement.
  • Length unit.
2 · Independent

A 75 mm image is ×1500. Find actual width and check it.

Give me a hint

Divide then multiply back.

Compare my reasoning
  1. 75 ÷ 1500 = 0.05 mm.
  2. 0.05 × 1500 = 75 mm.

0.05 mm.

Look for these in your work

  • Correct decimal.
  • Successful substitution.
3 · Transfer

A student reports actual size 12 000 mm from a 30 mm image at ×400. Diagnose and correct the error.

Give me a hint

Compare multiplication with division.

Compare my reasoning
  1. The student used 30 × 400.
  2. Actual = 30 ÷ 400 = 0.075 mm.

Incorrect multiplication; correct actual length 0.075 mm.

Look for these in your work

  • Identify operation error.
  • Correct result and unit.

Common mistakes

  • Actual size retains a length unit.
  • For enlargement, actual size is smaller than image size.
Recall without your notes

What can you explain now?

How do you find actual size from image size and magnification?

Compare with the explanation

Divide image size by magnification.

A = I ÷ M.

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

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Make it stick

Rearrange from the ratio and check your result by multiplication.

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 Science — Combined

Course scope and assessment

This pathway maps all 121 detailed subject-content sections in the published 0653 syllabus for exams in 2028 and 2029. It adapts original lessons from the related single-science pathways and keeps Combined Science's narrower scope: Biology B1–B16, Chemistry C1–C12 and Physics P1–P5. A 0653-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 separate 2025–2027 syllabus has not been reconciled here; choose the 2028–2029 syllabus only.

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