The image of a structure measures 30 mm while its actual size is 0.15 mm. What magnification has been used?
Question 1 Report
The image of a structure measures 30 mm while its actual size is 0.15 mm. What magnification has been used?
Answer Details
Use \( \text{magnification} = \frac{\text{image size}}{\text{actual size}} \). Both sizes are in millimetres, so divide directly: \[ \frac{30\ \text{mm}}{0.15\ \text{mm}} = 200 \] The magnification is x200 (no unit). The value x2000 comes from dividing by 0.015 instead of 0.15, a decimal-point slip; check by reversing the sum: \( 200 \times 0.15 = 30\ \text{mm} \), which matches the image size.
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