Question

12) Two lenses are placed 13 cm apart. The first lens is a converging lens with focal length 4cm. The second lens is a diverging lens with focal length 9cm. If an object is placed at a distance of 13 cm from the first lens, A) What is the position of the image? (6 points) B) What is the image type (virtual or real)? (2 points) C) What is the magnification? (2 points)

Answer 1

-ua I a fia tycm f2 = -9cm +4cm focal length for converging lens is the fa is diverging is ove fa= -9 cm When object is left side of lens then according to sign-convention it is negative u,= -13 Lens formula -+-+ (-3) (4) 5.77cm u= u, fi 4,+f, -13 +4 V=5.77cm k image from converging lens This image is act as a object for Liverging eens object distance ug for diverging eens Ug=d-k= 13-5.77=7.23 ug is negative Ug =-7.23cm Lola ㅗ fa (-7.23) (-a) -4cm -7.23-a k=-4 cm k image forom diverging lens final image. y = Ug fa Ugt fa

B) y is negative means it is left side of diverging lens So image is virtual Ffrom diverging cens image is virtual always magnification m= image distance forom eens object distance from lens for Liverging eens ma Vg uz 0 SS4 -4 -7.2 3 Foor converging lens m, -0.443 5.77 -13 Total magnification mam, xma (-0.443)* 10.554) -0.245 245