How Do You Calculate the Separation Between Two Converging Lenses?

[Jared944]

A coin is located 20.0 cm to the left of a converging lens (f=16.0 cm). A second identical lens is placed to the right of the first lens, such that the image formed by the combination has the same size and orientation as the original coin. Find the separation between the lenses.

So,

f= 16 cm
do = 20 cm

m=-di/do, since the object and image have the same size and orientation,
m=1, so -di=do > -di=20cm > di= -20cm

And that's about where I left it. Does anyone have any input they can contribute?

Much appreciated...

 

[exec]

Real image of the first lens becomes the virtual object to the second lens, vice versa. f = 16.0cm , o=20.0cm. The resulting image of the first lens = V/R?

Treat the real image of the first lens as the virtual object(o = negative) for the second lens. Treat the virtual image of the first lens as the real object(o = positive) for the second lens. Using the equation 1/o + 1/i = 1/f, and there you go.

 

[m2003]

Based on the information provided, the separation between the lenses can be calculated using the thin lens equation:

1/f = 1/do + 1/di

Where f is the focal length of the lens, do is the object distance, and di is the image distance.

Since we know the focal length of the lenses is 16 cm and the object distance is 20 cm, we can rearrange the equation to solve for the image distance.

1/di = 1/f - 1/do
1/di = 1/16 - 1/20
1/di = 0.05
di = 20 cm

Now, to find the separation between the lenses, we can use the following formula:

s = do + di
s = 20 cm + 20 cm
s = 40 cm

Therefore, the separation between the two lenses is 40 cm. This means that the second lens is placed 40 cm to the right of the first lens. By placing the lenses at this distance, the image formed by the combination will have the same size and orientation as the original coin.