A lens produces a 5.00 cm tall image of an 8.00 cm tall object when the object is placed 10.0 cm from the lens. Using m equals h_i over h_o, then d_i equals negative m times d_o, what is the apparent image distance?

1. Magnification is m = h_i / h_o = 5.00 / 8.00 = 0.625. 2. The positive magnification means the image is upright, so it must be virtual, giving d_i a negative sign. 3. Rearranging m = -d_i / d_o gives d_i = -m d_o = -(0.625)(10.0 cm). 4. Carrying out the multiplication gives d_i = -6.25 cm. 5. 6.25 cm without the negative sign would wrongly describe a real image, which does not match an upright magnification. 6. 3.12 cm and -3.12 cm both come from using half of the correct magnification value. _Source: OpenStax Physics (CC BY 4.0), Ch 16 "Mirrors and Lenses", section 16.3 Lenses_