# Thin film interference reflection

#### meaowwuff

1. 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

3. 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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#### ehild

Homework Helper

2. Homework Equations

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

3. 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) .

#### BvU

Homework Helper
Destructive interference occurs at $2d =\lambda/2$, but also at other path differences....

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).

"Thin film interference reflection"

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