Question

An object is placed in front of a converging lens in such a position that the lens (f = 14.0 cm) creates a real image located 22.0 cm from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens (f = 15.0 cm). A new, real image is formed. What is the image distance of this new image?

Answer 1

we will use the standard lens equation,

1/p +1/q = 1/f

where p is the object distance, q the image distance and f the focal length

the first experiment has i = 22 cm and f = 14 cm, so we can calculate p:

1/p + 1/22 = 1/14
1/p = 1/14 - 1/22 = 0.026
p = 38.5 cm

now we have p = 38.5 cm and f = 15 cm, so we solve for q

1/38.5 + 1/q = 1/15
1/q = 1/15 - 1/38.5 = 0.041
q = 24.6 cm

the image is formed 24.6 cm behind the lens