Thin film interference reflection

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SUMMARY

The discussion focuses on calculating the optimal thickness of a coating material with a refractive index of n1 = 1.27 applied to n2 = 1.50 glass for achieving iridescence. The thinnest layer that produces a reflection maximum for normal incidence at both 400 nm and 600 nm is determined to be 472 nm. Additionally, the reflection minimum occurs at a wavelength between 400 nm and 600 nm, requiring the application of the formula 2d = λ/2 for destructive interference, where d is the thickness of the coating.

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  • Understanding of thin film interference principles
  • Familiarity with refractive indices and their implications
  • Knowledge of wavelength and its relationship to light reflection
  • Proficiency in applying interference equations, specifically 2d = λ and 2d = λ/2
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  • Explore the effects of varying refractive indices on interference patterns
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meaowwuff
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Homework Statement


A coating of n1 = 1.27 material is to be added to n2 = 1.50 glass, in order to make it "irridescent."
a) What is the thinnest layer of material which will have a reflection maximum for normal-incidence light at BOTH 400 and 600 nm (vacuum) wavelength?
b) This thickness will have a reflection minimum at some wavelength between 400 and 600 nm. What is the wavelength of the reflection minimum?

2. Homework Equations

2*d=lambda(coating), 2*d=lambda(coating)/2

The Attempt at a Solution



I got the first part correct and the answer was 472 nm but I can't get b right. For b I did 2*d=lambda(coating)/2 since it says that the reflection will be minimum.
 
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2. Homework Equations

2*d=lambda(coating), 2*d=lambda(coating)/2

The Attempt at a Solution



I got the first part correct and the answer was 472 nm but I can't get b right. For b I did 2*d=lambda(coating)/2 since it says that the reflection will be minimum.[/QUOTE]

In order to get minimum reflectance, the path difference between the directly reflected wave and that, which reflects from the glass-layer interface has to be odd number times half of lambda(coating) .
 
Destructive interference occurs at ##2d =\lambda/2##, but also at other path differences...
 
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In your problem light moves from low u to high u.reflection minimum means the part of incident rays reflected will suffer destructive interference .hence 2ud=n(lambda).
 

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