An object is placed in front of a thin lens. A upright image is formed that is one_third the height of the object, if the image is `6.0 cm` from the lens, what is the focal length of the lens? Since the image is upright smaller than object `=>` This is a concave lens, focal length will be negative. Two questions applies: `1/f = 1/d_o +1/d_i` where `d_o` is distance to object, `d_i` is distance to image. `h_i/h_o = (-d_i)/d_o`where `h_o` is object height, `h_i` is image height. `h_i/h_o = (-d_i)/d_o => d_o = -d_ih_o/h_i` Since `h_o/h_i = 3` `d_o = -3d_i` `1/f = 1/d_o +1/d_i = 1/(-3d_i) +1/d_i=2/(3d_i)` `f = (3d_i)/2 = (3(6.0 cm))/2 = 9 cm` For concave lens `f = -9 cm`