Question

A converging lens of focal length 8.000 cm is 19.8 cm to the left of a diverging lens of focal length -6.00 cm. A coin is placed 12.2 cm to the left of the converging lens.

(a) Find the location of the coin's final image relative to the diverging lens. (Include the sign of each answer. Enter a negative value if the image is to the left of the diverging lens.)

(b) Find the magnification of the coin's final image.

Answer 1

solution:

v = image distance
m = magnification
Real is Positive sign convention.

Converging Lens:
1 / 12.2 + 1 / v = 1 / 8
v = 23.24 cm.
m = - 23.24 / 19.8 = -1.17
real image 23.24 cm. left of the converging lens.  
real image becomes a real object (23.24-19.8 =3.44cm.) left of the diverging lens.

1 / 3.44 + 1 / v = - 1 / 6
v = - 2.19 cm.

m =-v/u =-(-2.19)/(23.24) =.094

coin's final image is -2.19cm left of diverging lens