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

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To solve the problem of finding the index of refraction for a coating and its required thickness for a 600 nm wavelength, the index of refraction for the coating is calculated as 1.183 using the square root of the glass lens's index of refraction of 1.40. The thickness of the coating is determined to be 126.8 nm. Verification can be done using the 'λ/4 argument' to ensure the reflection coefficients are equal at both interfaces. The discussion emphasizes the importance of using the correct equations and reasoning in optical calculations. Understanding these principles is crucial for accurately solving similar problems in optics.
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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