# Magnification due to a Diverging Lens

## 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?

Doc Al
Mentor
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?

Doc Al
Mentor
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.

Doc Al
Mentor
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.

Doc Al
Mentor
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!