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Thin lens — image position and magnification

A converging lens has f = +15 cm. An object is placed at p = 25 cm from the lens.\nCalculate: (a) the image position q, (b) the transverse magnification m, (c) the type of image.
Given data
f = +15 cm = +0.15 mp = 25 cm = 0.25 m
Review the theory: Ottica
Steps
0 / 2
  1. Calculate the image position q (in cm) using the thin lens equation.
  2. Calculate the transverse magnification m = −q/p.
Full worked solution
  1. Calculate the image position q (in cm) using the thin lens equation.
    1q=1f−1p=115−125\dfrac{1}{q} = \dfrac{1}{f} - \dfrac{1}{p} = \dfrac{1}{15} - \dfrac{1}{25}
    1q=1f−1p=115−125=575−375=275\frac{1}{q} = \frac{1}{f} - \frac{1}{p} = \frac{1}{15} - \frac{1}{25} = \frac{5}{75} - \frac{3}{75} = \frac{2}{75}; q=752=37.5 cmq = \frac{75}{2} = \mathbf{37.5\,cm}. A positive qq indicates a real image formed on the opposite side of the lens.
  2. Calculate the transverse magnification m = −q/p.
    m=−qp=−37.525m = -\dfrac{q}{p} = -\dfrac{37.5}{25}
    m=−qp=−37.525=−1.5m = -\frac{q}{p} = -\frac{37.5}{25} = \mathbf{-1.5}. The negative sign indicates an inverted image; ∣m∣=1.5|m| = 1.5 means the image is 1.5 times larger than the object.
Result:q = 37.5 cm (real image). m = −1.5 (inverted, 1.5× larger).
TheoryOtticaFormula sheetPhysics 2

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