Location and Magnification of Coin's Final Image

[phunphysics2]

Homework Statement

A converging lens of focal length 8.100 cm is 20.3 cm to the left of a diverging lens of focal length 6.63 cm. A coin is placed 12.4 cm to the left of the convering lens.
Is the location of the coins final image to the left of the convering lens, between the lenses, or to the right of the divering lens?
What is the location and magnification of the coin's final image?

Homework Equations

1/do + 1/di= 1/f
m=hi/ho= -di/do

The Attempt at a Solution

first coin goes through converging lens

1/i + 1/o = 1/f

1/i + 1/12.4 = 1/8.1

i=23.358

this acts as object for next lens

o = 20.3-23.358
o=-3.058

1/i + 1/o = 1/f

1/i + 1/-3.058 = 1/-6.63

i=5.676

so to the right of the diverging lens

m = i1/o1 * i1/o2 = (23.358/12.4)*(5.676/-3.058)=-3.496


***Please help me with which bubble to choose. I think it is between the lenses, but I am not absolutely sure

 

[Simon Bridge]

i=5.676(cm) to which side of the diverging lens?
How far apart are the diverging and converging lenses?

You can check your results by sketching the ray diagrams.

 

[collinsmark]

Homework Statement

A converging lens of focal length 8.100 cm is 20.3 cm to the left of a diverging lens of focal length 6.63 cm. A coin is placed 12.4 cm to the left of the convering lens.
Is the location of the coins final image to the left of the convering lens, between the lenses, or to the right of the divering lens?
What is the location and magnification of the coin's final image?

Homework Equations

1/do + 1/di= 1/f
m=hi/ho= -di/do

The Attempt at a Solution

first coin goes through converging lens

1/i + 1/o = 1/f

1/i + 1/12.4 = 1/8.1

i=23.358

'Looks correct to me.

this acts as object for next lens

o = 20.3-23.358
o=-3.058

1/i + 1/o = 1/f

1/i + 1/-3.058 = 1/-6.63

i=5.676

so to the right of the diverging lens

That also looks correct to me.

***Please help me with which bubble to choose. I think it is between the lenses, but I am not absolutely sure

Wait, what?

As Simon Bridge advised, sketching ray diagrams might be useful here.

Let me point out something that you might find useful in the process.

For the moment, let's remove the diverging lens, but mark the diverging lens' location and focal points for reference (yes, it's a diverging lens, but it still, technically has focal points.) With the converging lens alone, the image that is formed is real, and is at a location in between the position of the diverging lens (if it were still there) and the diverging lens' rightmost focal point.

Compare this situation with two other hypothetical situations, either by asking yourself, or better yet sketching them: What would happen if the image formed by the first lens alone coincided with the position of the diverging lens' rightmost focal point exactly? And what would happen if it was to the right of the diverging lens' rightmost focal point?

When the diverging lens is put back in place, each situation will produce a different result. One will produce a real, final image. One will produce an image at infinity. And one will produce a virtual image.