How Do You Calculate Image Size and Location with a Diverging Lens?

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In summary, a 3 cm tall object is placed 16 cm in front of a diverging lens with a focal length of 24 cm. Using the equation 1/f = 1/di+1/do, the image distance is found to be -48 cm. Plugging this into the equation hi/ho = di/do, the image size is calculated to be -9 cm. However, since the focal length of a diverging lens should be taken as negative, the image appears smaller than the actual size.
  • #1
Robin_oto
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okay the problem reads
-an object 3 cm tall is placed 16 cm in front of a diverging lens with a focal length of 24cm. Find the location and size of the image. What kind of image is it?



1/f = 1/di+1/do

hi/ho = di/do


I'm not to sure how to solve this i used the equation; 1/f = 1/di+1/do and solved for di, getting -48cm. then pluged it into the equation hi/ho = di/do, and got -9cm. It doesn't seem right to me because when i draw out the problem it shows the image as smaller then the actual size, not bigger. please help!
 
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The focal length of a diverging lens should be taken negative: f=-24 cm

ehild
 

1. How do I know if I have a diverging lens problem?

In a diverging lens problem, you will be given the focal length of the lens, the distance of the object from the lens, and the height of the object. You may also be given the distance of the image from the lens. These are all important parameters to solve the problem.

2. How do I calculate the image distance?

The image distance can be calculated using the formula 1/f = 1/do + 1/di, where f is the focal length, do is the distance of the object from the lens, and di is the distance of the image from the lens. Make sure to use the correct signs (+ or -) for the distances.

3. What is the difference between a converging and a diverging lens?

A converging lens is thicker in the middle and causes parallel light rays to converge at a focal point. A diverging lens is thinner in the middle and causes parallel light rays to diverge. In a diverging lens problem, the focal length will always be negative, while in a converging lens problem, the focal length will be positive.

4. Can I use the thin lens equation in a diverging lens problem?

Yes, the thin lens equation, 1/f = 1/do + 1/di, can be used in both converging and diverging lens problems. Just make sure to use the correct signs for the distances as mentioned in the formula.

5. How do I know if the image is real or virtual in a diverging lens problem?

In a diverging lens problem, the image will always be virtual and located on the same side of the lens as the object. This means that the image cannot be projected onto a screen, but can only be seen by looking through the lens.

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