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Magnification due to a Diverging Lens

  • #1

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 recieved 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?
 

Answers and Replies

  • #2
Doc Al
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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?
 
  • #3
Is it negative since it is to the left of the lens?
 
  • #4
Doc Al
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  • #5
Alright, so I redid the problem with a negative focal length and got a height of .4cm.
 
  • #6
Doc Al
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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.)
 
  • #7
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.
 
  • #8
Doc Al
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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.)
 
  • #9
yay! thank you very much!
 

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