# An object is placed in front of a converging lens in such a position that the lens (f = 13.2 cm) produces a real image located 23.5 cm from the lens. Then, with the object remaining in place, the lens is replaced with another converging lens (f = 18.8 cm). A new, real image is formed. What is the image distance of this new image?

I know that I need to use the thin lens equation, but I'm not sure how to start this problem. Please help!

## 5.71

## To solve this problem, you can use the thin lens equation, which relates the object distance (denoted as "do"), the image distance (denoted as "di"), and the focal length of the lens (denoted as "f").

The thin lens equation is given as:

1/f = 1/do + 1/di

Given that the focal length of the first lens (f1) is 13.2 cm and the image distance (di1) is 23.5 cm, you can plug these values into the equation:

1/13.2 = 1/do + 1/23.5

To find the object distance (do), you can rearrange the equation:

1/do = 1/13.2 - 1/23.5

Now, you want to find the image distance (di2) when the lens is replaced with another lens (f2 = 18.8 cm). Since the object distance remains the same, you can use the thin lens equation again:

1/f2 = 1/do + 1/di2

Plugging in the values, you have:

1/18.8 = 1/do + 1/di2

Now, you can solve for di2 by rearranging the equation and plugging in the value for do:

1/di2 = 1/18.8 - 1/do

Substitute the value of 1/do from the previous equation into this equation, and solve for 1/di2. Finally, take the reciprocal of 1/di2 to find the value of di2, which will be the image distance formed by the second lens.