"THIN LENS IMAGERY, MAGNIFICATION, OBJECT, IMAGE DISTANCES. A summary of object/image properties for concave and convex lenses follows. (enter 2 of 3 values and leave unknown as 0, program will calculate.) Lens OBJECT <-------------------- IMAGE -------------------> Location Type Location Orient. rel. size CNCV anywhere virtual |S[I]<|F| erect minfied CNVX ∞>S[O]>2*F real F<S[I]<2*F invert. minified CNVX S[O]=2*F real S[I]=2*F invert. same size CNVX F<S[O]<2*F real ∞>S[I]>2*f invert. magnified CNVX S[O]=F ±∞ CNVX S[O]<F virtual |S[I]|>S[O] erect magnified F = focal length Transverse magnification: M[T] < 0 inverted |M[T]| < 1 minified *** Answer(s) to problem *** (c) PCSCC, Inc., 1993 (a) Set F=0, S[I]=8 and S[O]=38. F=6.61 and M[T]=-.281. Now type (space) F=focal_length (enter), set S[I]=2 and S[O]=0. Object distance is -2.87. M[T]= 0.697. (b) Object is virtual (2.87 in. on other side of lens),erect and minifiedby 0.697. ||An object positioned 38 inches to the right of a positive lens is imaged 8 inches to its left. (a) Where will the object appear if its image is moved to 2 inches from the lens? (b) Describe it. Type comma key to see answer. Type (F2) to return to application file."
