# Thread: distance with 2 focal points.

1. ## distance with 2 focal points.

An object is placed in front of a converging lens in such a position that the lens (f = 15.0 cm) creates a real image located 21.0 cm from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens (f = 16.0 cm). A new, real image is formed. What is the image distance of this new image?

this is familiar with what i've been working with, but this time there is only one answer and two focal points, so no need to use one focal point, but both! I was unsure how to use both focal points for one answer.

2. Originally Posted by rcmango
An object is placed in front of a converging lens in such a position that the lens (f = 15.0 cm) creates a real image located 21.0 cm from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens (f = 16.0 cm). A new, real image is formed. What is the image distance of this new image?

this is familiar with what i've been working with, but this time there is only one answer and two focal points, so no need to use one focal point, but both! I was unsure how to use both focal points for one answer.
You need the thin lense formula

$\frac{1}{s_1}+\frac{1}{s_2}=\frac{1}{f}$

where $s_1$ is the distance of the object in front of the lense and $s_2$ is the
distance of the image behind the lense and f is the forcal length of the lens
(for a thin converging lens)

In the first case we have:

$\frac{1}{s_1}+\frac{1}{21}=\frac{1}{15}$

and in the second:

$\frac{1}{s_1}+\frac{1}{u}=\frac{1}{16}$

where $u$ is the distance of the new image behind the lens.

Treat $x=1/s_1$ and $y=1/u$ as the variables and we have a pair of simultaneous
linear equations to solve, which will allow us to find $u$.

RonL