_{1}= 10 cm and f

_{2}= 15 cm are placed 40 cm apart, as shown on the figure. An object is placed 60 cm in front of the first lens as show in second figure.

a) Find the position of the final image formed by the combination of the two lenses?

b) Find magnification of the final image formed by the combination of the two lenses?

L - Distance between two lenses To determine the image distance, the lens equation can be used. Apply lens equation to first lens d _{i1} = 12 cm First image located 12 cm behind the first lensImage generated from first lens going to be object for the second lens d _{o2} = L – d_{i1
}d_{o2} = 40 cm – 12 cmd _{o2} = 28 cmLets apply lens equation to second lens d _{i2} = 32.31 cmFinal image located at 32.31 cm behind second lens. Lens magnification can be find using m _{1} = - 0.2First lens has magnification of – 0.2 Image magnification in terms of object and image height can be write First lens has magnification of – 0.2, the image is inverted and is 0.2 times of original height. Lets apply image magnification equation to second lens m _{2} = - 1.15Second lens has magnification of – 1.15 Image magnification in terms of object/image height is Image generated from first lens going to be object for the second lens h _{i1} = h_{o2}From this equation we see that total magnification is the product of m _{1} and m_{2}.Total magnification is 0.23, is a positive number, final image is not inverted but it is smaller than original object. |