# Thin film intereference

## Homework Statement

[/B]
Scientists are testing a transparent material whose index of refraction for visible light varies with wavelength as n = 30.0 nm^(1/2)/λ^(1/2), where λ is in nm.

A 310-nm-thick coating is placed on glass (n = 1.50) what visible wavelengths will the reflected light have maximum constructive interference?

## Homework Equations

λ_constructive = (2nd)/m where m = 1,2,3,4
Δφ = 2π*(2nd)/λ

d is thickness of film[/B]

## The Attempt at a Solution

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I am really confused by the index of refraction notation n = 30.0 nm^(1/2)/λ^(1/2), but what I got out was that in the visible spectrum, n < 1. So you have air where n = 1. the film where n < 1, and glass where n = 1.5 . So light will reflect off the film with no phase difference, and light reflects off the glass with a phase shift of π. So the light has to travel 2d and have a phase shift of mπ (m = 1,3,5...) to have constructive interference.

π = 2π*(2 * (sqrt(30nm)/(sqrt(λ))) * 310nm) / λ

using a calculator I got λ = 358.6nm

using 3π, I got λ = 191.4nm which isn't in the visible spectrum

This answer is sadly wrong because the online answer box says there are two wavelengths that will have constructive interference.

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## Answers and Replies

Charles Link
Homework Helper
Gold Member
2020 Award
The middle of the visible is ## \lambda=550 \, nm ##. There ## n \approx \frac{30}{23.5} \approx 1.3 ## by this formula. ## \\ ## And an additional hint: The index ## n ## of the film does obey ## 1<n<1.5 ##. This means that a ## \pi ## phase change will occur upon reflection at both the thin film/glass interface and also at the air/thin film interface, so that it can be ignored. ## \\ ## And your equation is correct for constructive interference: ## 2nd=m \lambda_{air} ##.

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