Hard Optics Problem: Calculate Magnification & Image Distance

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SUMMARY

The discussion focuses on calculating the magnification (m) and image distance (Si) for an object placed in front of a converging lens, with a thick glass plate positioned behind it. Participants emphasize the use of the lens formula and lens maker's equation to derive the necessary calculations. It is established that the magnification remains constant regardless of the presence of the glass slab, and the thickness of the slab introduces constraints that must be considered in the calculations. The interaction between the lens and the glass slab is crucial for determining the final image position.

PREREQUISITES
  • Understanding of the lens formula and its application
  • Familiarity with the lens maker's equation
  • Knowledge of optical principles regarding virtual objects
  • Concept of refraction through different media
NEXT STEPS
  • Study the derivation and application of the lens formula in optical systems
  • Explore the lens maker's equation in detail for various lens types
  • Investigate the concept of virtual objects in optics
  • Learn about the effects of thickness and refractive index on image formation in optical systems
USEFUL FOR

Optics students, physics educators, and professionals in optical engineering who are involved in lens design and image analysis.

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1. An object (height o) is placed at an object distance S in front of a converging lens. The lens is placed at a distance d<Si (image distance) in front of a very thick glass plate (index of refraction n, thickness larger than image distance). The surface of the glass plate is perpendicular to the optical axis. Calculate the magnification m and the image distance Si



2. Lens makers equation? I'm not really sure



3. I know the magnification will be the same whether the glass is there or not, but I can't actually get any equations out for it :(
 
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You can find out where the first image will be formed using the lens formula.

That image acts as a virtual object for the glass slab. Here you can use the lens makers formula to find out where the image will be after the first refraction off the glass surface.

Then you subtract/add the thickness of the glass slab to the second image and that's your second virtual object for refraction at the second surface of the glass slab.

There's something else here too, the fact that the glass slab is thicker than the image distance... keep that in mind when you solve it. That'll put a mathematical constraint on your answer somewhere after the first refraction at the slab surface.
 
Thanks for your reply. I had a look at this, but don't understand how you can use the first image as a virtual object for the slab of glass, because it will be inside it. Also, we aren't told the thickness of the glass slab, I think because it is infinitely thick so it doesn't matter.

Any ideas?
 

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