Thin film interference reflection

AI Thread Summary
A coating with a refractive index of 1.27 is to be applied to 1.50 glass to achieve iridescence, requiring calculations for reflection maxima at 400 nm and 600 nm wavelengths. The thinnest layer thickness for maximum reflection was determined to be 472 nm. For the reflection minimum, the path difference must be an odd multiple of half the wavelength, leading to the equation 2d = λ/2. The discussion emphasizes the need for destructive interference at this minimum, which occurs when light transitions from a lower to a higher refractive index. Understanding these principles is crucial for accurately determining the required coating thickness.
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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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