Algebraic Explanation of Image Formation in Convex Lenses

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SUMMARY

The discussion centers on the behavior of images formed by convex lenses when an object is moved within the focal length. The thin lens equation, represented as 1/f = 1/d0 + 1/di, is used to derive that as the object approaches the lens, the image size increases and the image moves farther from the lens, resulting in a virtual image. Numerical examples using focal length (f=5) and object distances (do=4 and do=3) confirm that the image remains on the same side as the object, contradicting the initial ray diagram interpretation. The conclusion emphasizes the importance of correctly applying the thin lens equation to understand image formation.

PREREQUISITES
  • Understanding of the thin lens equation (1/f = 1/d0 + 1/di)
  • Familiarity with the concepts of focal length and object distance in optics
  • Ability to interpret ray diagrams for convex lenses
  • Basic knowledge of virtual and real images in optics
NEXT STEPS
  • Study the derivation and applications of the thin lens equation in various scenarios
  • Learn about ray diagram construction for different lens configurations
  • Explore the characteristics of virtual images versus real images in optical systems
  • Investigate the effects of changing object distance on image properties using simulation tools
USEFUL FOR

Students of optics, physics educators, and anyone interested in understanding the principles of image formation through convex lenses.

Matt M
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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?
 

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