Algebraic Explanation of Image Formation in Convex Lenses

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When an object is within the focal length of a convex lens and is moved closer to the lens, the image formed increases in size and appears farther from the lens. The thin lens equation confirms that as the object approaches the lens, the image remains virtual and on the same side as the object. Numerical examples using the thin lens equation yield consistent results with the expected behavior of the image. Discrepancies between ray diagrams and algebraic solutions may arise from inaccuracies in the ray diagram. The lens equation effectively describes the relationship between object distance, image distance, and focal length.
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Homework Statement



An object is in the focal length of a convex lens. As the object is moved from the focus towards the lens, what happens to the object?

Homework Equations


1/f=1/d0+1/di

The Attempt at a Solution


I know how to solve this problem using ray diagrams, and the solution is:

The image increases in size and moves farther from the lens

I am trying to conclude this algebraically instead of just with a ray diagram. Just choosing numbers though, I let f=5 and do=4. Plugging this into the thin lens equation, I find that d0=-20, showing a virtual image on the same side of the lens as the object. However, then when I move the image closer and let f=5 and d0=3, the thin lens equation gives that d0=-7.5. Hence it seems to me that as the object is moving closer, it is still past the focal point, but moving closer to the lens as opposed to farther. Why is the numerical example leading to different results than the ray diagram? Thanks in advance!
 
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Matt M said:

Homework Statement



An object is in the focal length of a convex lens. As the object is moved from the focus towards the lens, what happens to the object?

Homework Equations


1/f=1/d0+1/di

The Attempt at a Solution


I know how to solve this problem using ray diagrams, and the solution is:

The image increases in size and moves farther from the lens

I am trying to conclude this algebraically instead of just with a ray diagram. Just choosing numbers though, I let f=5 and do=4. Plugging this into the thin lens equation, I find that d0=-20, showing a virtual image on the same side of the lens as the object. However, then when I move the image closer and let f=5 and d0=3, the thin lens equation gives that d0=-7.5. Hence it seems to me that as the object is moving closer, it is still past the focal point, but moving closer to the lens as opposed to farther. Why is the numerical example leading to different results than the ray diagram? Thanks in advance!

Because you drew the wrong ray diagram?

The result given by the lens equation is intuitively right. If you have a magnifying glass, and you're looking at something small (say, an ant) through the lens, then
  1. The image is larger than the object.
  2. The image appears to be on the same side of the lens as the object (so it's a virtual image).
  3. As you move the object closer, the image appears to get closer, as well.
That's exactly what the lens equation tells you. So what ray diagram is telling you something different?
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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