Solving the Mystery of Wavelength Refraction: A Thin 200nm Film of Oil

AI Thread Summary
A discussion centered on calculating the wavelength of light reflected from a thin 200nm oil film on water revealed a misunderstanding of the interference equation. The initial calculation yielded an incorrect wavelength of 800 nm, which is outside the visible spectrum. The error stemmed from misapplying the equation and incorrectly assuming the integer value of m. By adjusting m to 2, the correct wavelength of 533 nm was identified, aligning with the visible spectrum. The conversation highlighted the importance of understanding phase changes and the correct application of the interference formula in optics.
JJK1503
Messages
12
Reaction score
0

Homework Statement


http://lon-capa.bd.psu.edu/res/psuerie/gwb6/physics/optics/e1p4.png
A thin 200nm film of oil (n = 2.0) floats on water (n = 1.33). To a normal human, what wavelength will the film appear when viewed from above (nearly perpendicular to the film)?

Homework Equations



for constructive interference
2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi (in this case)

The Attempt at a Solution



This was a question I missed on the midterm. I want to know where I went wrong for the final. Here is what I did.

There is a poem our prof taught us; high to low, phase no. Low to high, phase change pi.
Based on the picture, which hopefully shows up, above I can see there is 1 phase change of pi where air meets oil. my equation is then

2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi
where m = 1, n = oil = 2, d = 200 nm

I solve for lambda and get the answer of 800 nm.

This is not the correct answer, and I know this right away because light of this wavelength is not in the visible spectrum.

I know the answer is 533 nm but I don't know why. I feel like I am missing something here conceptually.

Please help.So, it looks like the picture didn't make it through to the post it looks something like this the lines separate the different mediums.

air
_________________________

oil
_________________________

water
_________________________
 
Last edited:
Physics news on Phys.org
Ok, I figured it out.

First off my math was bad when I came up with the 800 nm. I only multiplied by d rather than 2 d had I done this I would have found a wavelength of 1600 nm. Still not the right answer but it is the right equation.

It then occurred to me that I have a variable that I can play with which is m. I had assumed that m = 1. Bad assumption because there really isn't any reason m can't be any integer I want. If I make m = 2 the problem is solved.
 
JJK1503 said:
There is a poem our prof taught us; high to low, phase no. Low to high, phase change pi.
I like that!

JJK1503 said:
2 * pi * m = ( 2 * pi * n * 2 * d) / lambda + pi
where m = 1, n = oil = 2, d = 200 nm

I solve for lambda and get the answer of 800 nm.
I don't see how you got that answer. Using the same basic approach, I get the given answer.

Ah... I see you've got it now. (Too late! But good for you.)
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Back
Top