Magnification due to a Diverging Lens

[davidepalmer]

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?

 

[Doc Al]

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?

 

[davidepalmer]

Is it negative since it is to the left of the lens?

 

[Doc Al]

Is it negative since it is to the left of the lens?

Yes.

 

[davidepalmer]

Alright, so I redid the problem with a negative focal length and got a height of .4cm.

 

[Doc Al]

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.)

 

[davidepalmer]

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.

 

[Doc Al]

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.)

 

[davidepalmer]

yay! thank you very much!