Question

Two lenses, one converging with focal length 20.0 cm and one diverging with focal length -10.0 cm are placed 25.0 cm apart. An object is placed 60.0 cm in front of the converging lens. Determine (a) the position and (b) the magnification of the final image formed. (c) Sketch a ray diagram for this system.

Answer 1

Focal length of converging lens is

$f_1 = +20 \;cm$

Focal length of diverging lens is

$f_2 = -10 \; cm$

Distance between the two lenses is

$d = 25.0 \;cm$

(a)

Use the equation

$\frac{1}{v} +\frac{1}{u} = \frac{1}{f}$

Position of object from converging lens is

$u_1 = 60.0 \;cm$

Position of image is calculated as

$\frac{1}{v_1} = \frac{1}{f_1} -\frac{1}{u_1} \\ = \frac{1}{20} -\frac{1}{60} \\ = \frac{1}{30} \\ v_1 = 30 \;cm$

The positive value of image position represents that image is formed on the right side of lens.

The image will work as an object for diverging lens.

The object distance from diverging lens is

$u_2 = 25-30 = -5 \;cm$

The final position of image is

$\frac{1}{v_2} = \frac{1}{f_2} -\frac{1}{u_2} \\ = \frac{1}{-10} -\frac{1}{-5} \\ = \frac{-1}{10} + \frac{1}{5} \\ v_2 = 10 \; cm$

Since the image position is positive, the image is formed beyond the diverging lens.

(b)

The magnification of the final image is

$m=m_1m_2 \\ = (-\frac{v_1}{u_1})(\frac{-v_2}{u_2}) \\ = (\frac{-30}{60})(\frac{-10}{-5}) \\ = -1.0$

The magnification is negative which implies that image is inverted.

(c)

The following figure shows the ray diagram for this system.