What is the thinnest film for strong reflection of visible light?

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Homework Help Overview

The problem involves determining the thickness of a thin oil film that produces strong reflection for visible light, specifically with a wavelength of 700 nm. The oil film has a refractive index of 1.25 and floats on water with a refractive index of 1.33.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to calculate the thickness of the film using a formula related to constructive interference but questions the correctness of their approach. Other participants discuss the concept of wavelength in different media and suggest the need for applying the law of refraction to find the wavelength in the oil film.

Discussion Status

Participants are exploring different interpretations of the problem, particularly focusing on the relationship between wavelength in air and in the oil film. Some guidance has been provided regarding the law of refraction and its relevance to the problem, but no consensus has been reached on the correct approach or solution.

Contextual Notes

One participant mentions not being introduced to the law of refraction, indicating a potential gap in foundational knowledge that may affect their understanding of the problem.

nubey1
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Homework Statement


A very thin oil film(n=1.25) floats on water(n=1.33). What is the thinnest film that produces a strong reflection for visible light with a wavelength of 700nm?


Homework Equations


[tex]\lambda[/tex]=(2nd)/(m-0.5)
d=(m-0.5)*([tex]\lambda[/tex]/2n)
thinnest film has m=1

The Attempt at a Solution


d=(0.5*700)/(2*1.25)=140nm
This answer is incorrect. Is there a problem in my technique?
 
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It seems to me the film will just be a half wavelength thick so that light traveling through it to the water surface and back will be a full wavelength delayed and so will constructively interfere with waves reflected off the film surface.

Of course that will be in terms of the wavelength in the film, so you'll need a Law of Refraction formula that let's you computer the wavelength in the film.
 
I was not introduced to the law of refraction. I did look ahead in my physics book catch a glimpse of what it is about. This is my attempt at just hucking the formulas.
[tex]\vartheta[/tex]=arcsin(n2/n1)=arcsin(1.25/1.33)=1.222
[tex]\vartheta[/tex]*[tex]\lambda[/tex]=1.222*700=855.5nm

Is this correct?
 
If I understand this correctly, it says the wavelength is 700 nm in AIR.
In the oil film the wavelength will be reduced by a factor of 1.25 to 560 nm.
The part of the law of refraction you need is that
[tex]\lambda1/[/tex] [tex]\lambda2 = n2/n1[/tex]
Use n1 = 1 for air, n2 = 1.25 for the oil.
 

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