Index of Refraction: Find Coating & Thickness for 600nm Wavelength

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SUMMARY

The discussion focuses on calculating the index of refraction for a coating and its required thickness for a glass lens with an index of refraction of 1.40 at a wavelength of 600 nm. The initial calculation provided by a user suggests a coating index of refraction of 1.183 and a thickness of 126.8 nm. Verification of these values can be achieved through the 'λ/4 argument' method, ensuring equal reflection coefficients at both interfaces.

PREREQUISITES
  • Understanding of optical physics principles, specifically refraction and reflection.
  • Familiarity with the concept of wavelength and its measurement in nanometers (nm).
  • Knowledge of the equations governing index of refraction and thickness calculations.
  • Experience with the 'λ/4 argument' in optics for determining coating thickness.
NEXT STEPS
  • Study the principles of index of refraction and how it applies to coatings.
  • Learn about the 'λ/4 argument' and its application in optical coatings.
  • Research the calculation methods for reflection coefficients at different interfaces.
  • Explore practical applications of optical coatings in lenses and other optical devices.
USEFUL FOR

Students in optics or physics courses, optical engineers, and anyone involved in the design and analysis of optical coatings for lenses.

Nymn
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I'm stuck on a homework problem. It states the wavelength is 600 nm and the glass lens has an index of refraction of 1.40. Find the index of refraction for the coating and the required thickness.
I've googled around, but I'm not sure what equations I should be using. I got 1.183 for the coating (square root of 1.4) and 126.8 nm for the thickness. Is that right?
 
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Nymn said:
I'm stuck on a homework problem. It states the wavelength is 600 nm and the glass lens has an index of refraction of 1.40. Find the index of refraction for the coating and the required thickness.
I've googled around, but I'm not sure what equations I should be using. I got 1.183 for the coating (square root of 1.4) and 126.8 nm for the thickness. Is that right?

OK, where do you get the 126nm from and what reasoning did you use? Btw, I am not saying you're wrong at all. You can verify that you are right by just going over the 'λ/4 argument' and getting the reflection coefficient equal at both interfaces.
 

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