Q1: Lens makes upright image smaller than object
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_owhere 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