Thin lens equation and image formation

1. The problem statement, all variables and given/known data
An object 4.29 cm high is projected onto a screen using a converging lens with a focal length of 32 cm. The image on the screen is 51.7 cm. Where are the lens and the object located relative to the screen?

ho=4.29 cm
hi=51.7 cm
ƒ=32 cm

2. Relevant equations
1. m=(hi)/(ho)

2. m=ƒ/(ƒ-do)

3. m=-di/do

3. The attempt at a solution
I used equation number 1 to calculate the magnification since I was given the heights of the image and the object. I got 12.0513.

m=(51.7cm)/(4.29cm)=12.0513

Next, I used equation 2 to solve for do. I got 29.345 cm

m=ƒ/(ƒ-do)

(ƒ-do)=ƒ/m

ƒ-ƒ/m=do

32cm-(32cm/12.0513)=do=29.345 cm

Since do is measured as the distance from the surface of the lens/mirror, the object is at a distance of 29.345 cm relative to the lens.

Now that I have values for do and m, I can use another version of the magnification equation to solve for di. I got -353.645 cm

m=-(di/do)

-mdo=di

-(12.0513)(29.345cm)=di=-353.645 cm ←which would be relative to the lens, on the opposite side.

The question asks first for the distance between the lens and the screen that formed the image. Here is where I am not sure if I am thinking about it correctly.

An image would have a height of hi at a distance di from the lens. Since the di I calculated was -353.645 cm, the image was 353.645 cm away from the lens on the right side(taking the left side as the location of the object).

So the lens is 353.645 cm from the screen

The next question asks for the distance of the object from the screen. That would be the length of the distance between the screen and the lens (353.645 cm), plus the length of distance from the lens to the object (29.345 cm).

So the object is 382.99 cm from the screen.

However they are both wrong. The biggest problem with these problems are that mistakes are typically due to incorrect usage of sign conventions, or from misinterpreting the question asked. Right away I see that 382.99 cm is a pretty big number compared to the other values for image height, object height, focal length, etc. Maybe I just haven’t done enough problems to judge whether that is average or not.

Any help would be appreciated.

P.S. I know the answers are wrong because the homework they are on require me to type in the answer in a blank field which I then "submit" by pressing a "Check my work" button. I get 3 attempts for this question, this is my first attempt and it was wrong.

Thanks for the help.

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