Thin lens equation

Also called Thin-lens equation, Mirror equation, Lens equation

The thin lens equation relates image distance, object distance and focal length: one over the image distance plus one over the object distance equals one over the focal length. The same printed equation serves mirrors.

1si+1so=1f\frac{1}{s_i} + \frac{1}{s_o} = \frac{1}{f}

The AP Physics 2 sheet prints it once, in the Waves, Sound, and Optics group, and the CED names it in EK 13.4.A.5: for a thin lens, the location of an image depends on the focal length of the lens and the distance between the object and the midline of the lens. EK 13.2.A.7 applies the identical relation to mirrors, so one equation covers both halves of Unit 13.

Signs are where this goes wrong, and the CED does not print the table. EK 13.4.A.5.i and EK 13.2.A.7.i both say only that the locations of the focal point, the object and the image follow sign conventions used to determine those locations relative to the lens or mirror itself. There is no sign convention box on the equation sheet either. The standard scheme, which the rest of Unit 13 is consistent with:

QuantityPositive whenNegative when
ffconverging opticdiverging optic
sos_oreal object in front(not met in AP Physics 2)
sis_ireal imagevirtual image

Solve it, do not eyeball it. The reciprocals do not cancel. Rearranged, si=sof/(so−f)s_i = s_o f/(s_o - f). A converging lens of f=10f = 10 cm with the object at so=15s_o = 15 cm gives si=(15)(10)/5=30s_i = (15)(10)/5 = 30 cm, positive and therefore real, and ∣M∣=∣30/15∣=2\lvert M \rvert = \lvert 30/15 \rvert = 2, so twice the size.

EK 13.4.A.5.ii adds that lenses have a focal point on both sides. Focal length carries the sign that picks the optic.

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