The image formed by a converging lens of focal length is of a distance of 2f from the lens. Calculate the lateral magnification produced
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Correct Answer: Option B
Explanation:
Using the lens formula,
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
Where u, v and f are the object distance, image distance and focal length respectively.
\(\therefore \frac{1}{2f} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{u} = \frac{1}{2f} - \frac{1}{f} \implies \frac{1}{u} = \frac{1}{2f}\)
\(u = 2f\)
\(\therefore Magnification = \frac{\text{image distance}}{\text{object distance}}\)
= \(\frac{2f}{2f}\)
= 1
Using the lens formula,
\(\frac{1}{v} + \frac{1}{u} = \frac{1}{f}\)
Where u, v and f are the object distance, image distance and focal length respectively.
\(\therefore \frac{1}{2f} + \frac{1}{u} = \frac{1}{f}\)
\(\frac{1}{u} = \frac{1}{2f} - \frac{1}{f} \implies \frac{1}{u} = \frac{1}{2f}\)
\(u = 2f\)
\(\therefore Magnification = \frac{\text{image distance}}{\text{object distance}}\)
= \(\frac{2f}{2f}\)
= 1