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

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To calculate the image size and location with a diverging lens, use the lens formula 1/f = 1/di + 1/do, where the focal length (f) is negative for diverging lenses. After substituting the object distance (do) of 16 cm and focal length of -24 cm, the image distance (di) is found to be -48 cm. The magnification formula hi/ho = di/do gives an image height (hi) of -9 cm, indicating the image is inverted and smaller than the object. The negative sign in both di and hi confirms the characteristics of the image formed by a diverging lens. The calculations align with the expected properties of diverging lenses, confirming the image is virtual and reduced in size.
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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

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