Calculate actual specimen size by rearranging magnification
Divide image size by magnification and retain length units.
Exam scope and assessment notes
Mapped to the named 0973 section using teaching from the corresponding single-science course. A 0973-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. This qualification reports two grades on the 9–1 scale; confirm the entry route for your zone and exam series.
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.
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?
Worked example
An image measures 42 mm at ×700. Find the actual length in mm.
Show the worked solution
- Actual = image ÷ magnification.
- 42 ÷ 700 = 0.06.
- 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.
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.
An image is 32 mm at ×800. Find actual length.
Give me a hint
Use A = I ÷ M.
Compare my reasoning
- 32 ÷ 800 = 0.04.
- Retain mm.
0.04 mm.
Look for these in your work
- Correct rearrangement.
- Length unit.
A 75 mm image is ×1500. Find actual width and check it.
Give me a hint
Divide then multiply back.
Compare my reasoning
- 75 ÷ 1500 = 0.05 mm.
- 0.05 × 1500 = 75 mm.
0.05 mm.
Look for these in your work
- Correct decimal.
- Successful substitution.
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
- The student used 30 × 400.
- 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.
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.