Calculate the required image size for a stated magnification
Multiply actual length by the desired ratio.
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 specimen is 0.08 mm at ×250. Which is the required image length?
What you will learn
- Use image size = magnification × actual size.
- Check whether a planned drawing fits a given space.
Scale the actual length
Rearranging M = I ÷ A gives I = M × A. A specimen 0.08 mm long drawn at ×250 has image length 250 × 0.08 = 20 mm. Magnification changes the size of the representation, not the physical specimen.
Use the given actual dimension and the requested magnification. The answer is an image length in the same units as the actual length. Distinguish ×250 from 250 mm and distinguish a dimensionless ratio from a measured length.
Plan and verify
If a drawing space is 30 mm wide, a calculated image width 20 mm fits, with room remaining. A requested image width larger than the space needs a smaller magnification; do not change the actual specimen size to make the picture fit.
Check image length by dividing it by actual length. For the example, 20 ÷ 0.08 = 250. A displayed SVG may resize for the screen, so the numerical drawing specification, rather than uncalibrated on-screen millimetres, is the practice data.
Sources for this lesson
Pause and explain
A drawing is too large for the page. What can be reduced without changing the specimen?
Worked example
A 0.08 mm specimen is drawn at ×250. What image length is required?
Show the worked solution
- I = M × A.
- 250 × 0.08 = 20.
- Check 20 ÷ 0.08 = 250.
Answer 20 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.
A 0.12 mm cell is drawn at ×200. Find image length.
Give me a hint
Multiply actual length by ratio.
Compare my reasoning
- 0.12 × 200 = 24.
- Report mm.
24 mm.
Look for these in your work
- Correct product.
- Image-length unit.
A 0.04 mm cell is drawn at ×900. Does its 36 mm image fit in a 40 mm space?
Give me a hint
Calculate before comparing.
Compare my reasoning
- 0.04 × 900 = 36 mm.
- 36 < 40, leaving 4 mm.
Yes; 36 mm image with 4 mm remaining.
Look for these in your work
- Calculate size.
- Compare with available space.
A 0.15 mm specimen must fit within a 45 mm drawing width. What is the largest magnification for that dimension?
Give me a hint
Use the ratio with the limiting image width.
Compare my reasoning
- M = 45 ÷ 0.15.
- M = 300.
×300 maximum for that dimension.
Look for these in your work
- Use the limit.
- Do not alter the specimen size.
Common mistakes
- Changing the representation does not change specimen size.
- Image size is a length, not a ratio.
What can you explain now?
How is image size calculated?
Compare with the explanation
Magnification multiplied by actual size.
I = M × A.
After trying it yourself, choose your next review. This is your self-assessment.
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Make it stick
Calculate a drawing dimension and verify it with the original ratio.
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.