PHYSICS HELP - optical images, magnification, etc.?

Question

An object is located 3.5cm from the optical centre of a lens. The lens produces an image of magnification that is +0.667 the size of the object.

- Is this lens a diverging or a converging lens? Explain your answer.

- Find the position of the image and the focal length of the lens using the magnification equation and the thin lens equation.

Answer 1

It's a diverging lens and the image is virtual. Only diverging lenses produce positive, <1 (upright, smaller) magnification.
F = 1/(1/R1+1/R2); M = -R2/R1 ==> R2 = -R1*M
Substituting, F = 1/(1/R1-1/(R1*M))
R2 = 1/(1/F-1/R1)
F =-7.01051051051051D-02
R1 = .035
Image position R2 = -.023345 (same side as object)
M = .667
Image is virtual, upright, and smaller than object.