Magnification due to a Diverging Lens

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The discussion focuses on calculating the height of an image formed by a diverging lens with a focal length of -74 cm and an object height of 0.6 cm located at 37 cm. The initial attempt incorrectly used a positive focal length, leading to an incorrect image height of 1.2 cm. After recognizing the need for a negative focal length, the correct image distance was calculated as -24.667 cm, resulting in a magnification of 0.667 and an accurate image height of 0.4 cm. The importance of using the correct sign for the focal length in lens equations is emphasized. The final conclusion confirms the correct image height as 0.4 cm.
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Homework Statement



The focal length of the diverging lens is 74 cm. If an object with height h = 0.6 cm is located at d = 37 cm, what is the height of the image?

Homework Equations



Thin-Lens Equation:

(1/do)+(1/di)=1/f

Magnification Equation:

m=-di/do

The Attempt at a Solution



I know that first I need to find the distance of the image, di. To do this I took: 1/[(1/74cm)-(1/37cm)]. From this I got -74cm.

Next, I plugged this value into the Magnification equation: -(-74cm)/(37cm). From this I received a value of 2. Therefore the image should be 1.2 cm tall. When I entered this answer it was incorrect. I also know this is incorrect because images from a convex lens must be upright, reduced in size, and virtual.

Can anyone help me out on where I am going wrong here?
 
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davidepalmer said:
I know that first I need to find the distance of the image, di. To do this I took: 1/[(1/74cm)-(1/37cm)].
This isn't right. What's the sign of f for a diverging lens?
 
Is it negative since it is to the left of the lens?
 
davidepalmer said:
Is it negative since it is to the left of the lens?
Yes.
 
Alright, so I redid the problem with a negative focal length and got a height of .4cm.
 
davidepalmer said:
Alright, so I redid the problem with a negative focal length and got a height of .4cm.
What did you get for do? (Show how you got it.)
 
So, I redid my equation: 1/[(1/-74cm)-(1/37cm)]. This gave me: -24.667. Then I redid my magnification equation: -(-24.667)/(37). This gave me .667 for my magnification. Then when multiplied by the height i got .4 cm.
 
davidepalmer said:
So, I redid my equation: 1/[(1/-74cm)-(1/37cm)]. This gave me: -24.667. Then I redid my magnification equation: -(-24.667)/(37). This gave me .667 for my magnification. Then when multiplied by the height i got .4 cm.
Actually, that's correct. (For some reason, I thought you meant it was wrong.)
 
yay! thank you very much!
 
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