How Does a Distorted Dewar Window Affect Its Focal Length?

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SUMMARY

The focal length of a distorted dewar window, influenced by a pressure difference, can be calculated using the thin lens equation, specifically 1/f = (n-1)/R. In this case, with a radius R1 of 100 meters, a thickness of 3 mm, and an index of refraction n = 4 at a wavelength of 4 µm, the window behaves like a lens when distorted. While the natural focal length of a window is infinite, distortion allows it to focus light, creating an image when viewed from a sufficient distance, such as through binoculars or a telescope.

PREREQUISITES
  • Understanding of the thin lens equation
  • Knowledge of optical properties of materials, specifically index of refraction
  • Familiarity with the concept of focal length in optics
  • Basic principles of light behavior through curved surfaces
NEXT STEPS
  • Research the application of the thin lens equation in optical systems
  • Explore the effects of pressure on optical materials
  • Learn about the design and function of dewar windows in cryogenics
  • Investigate the use of telescopes and binoculars for observing distorted optical elements
USEFUL FOR

Optical engineers, physicists, and students studying optics who are interested in the practical applications of lens behavior in distorted materials.

AaronBurr
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Homework Statement


  1. What is the focal length of a dewar window distorted by a pressure difference? As a result of the pressure difference, the radius of the first surface of the window is R1 = 100 meters. The window is 3 mm thick, with an index n = 4 at l = 4 um.

Homework Equations


I think I can just use the thin lens equation? 1/f=(n-1)/R

The Attempt at a Solution


[/B]
My concern with using the thin lens equation is I'm not sure if a distorted window creates a lens? I know the focal length of a window is naturally infinite. If its bent does it then act as a lens? If not what formula do I use to prove it still acts as a window?Thanks for any help!
 
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This is an excellent problem in that it teaches you practical applications. Basically, you still have a window if you are about 1/2 way, inside of the focal length. And that is, from a practical aspect, a window. However, crank out the numbers and you find that if you are far enough away (and with a good magnifying device, not sure what is applicable, Cause I didn't crunch the numbers either, maybe binoculars, maybe a large telescope), you will see an image, because it will behave like a lens past its focal point (at that distance, you may have trouble finding the focal point, hence the calculation).
 

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