# Homework Help: Grade 12 physics question regarding (n)

1. Jan 12, 2008

### mrx35g

1. The problem statement, all variables and given/known data
an oil stick 122 nm thick (n=1.40), lying on water, is illuminated by white light incident perpendicular to it's surface. What is the wavelength of the color that appears?

2. Relevant equations
I'm not sure, but i tried n= Lambda(air)/Lambda coating in order to find the lambda in air but it doesn't match.

3. The attempt at a solution

I used the above equation but it didn't work. I'm also having a hard time grasping the concept of index of refraction and how it's related to the thin films. Any strong analogy that you can give me as a tool to understand the concept will be more than helpful to me.

2. Jan 13, 2008

### shawshank

ok you need to know that n of water is 1.33 , so the oil has a higher index than water. btw, it's an oil slick ;)

The index of refraction is simply the ratio of c (speed of light in a vaccum) and the speed of light in the material you're analysing. so it's n = c / v
So basically, you know the maximum speed of light is c so the best ratio you can get is 1 and the higher it gets, the lower the speed will be in the material.

The index is kind of like thickness (most noble and prestigious physicist of this forum are gonna flame me for this), basically light has more trouble going through something with a higher index. Think of like water and tar, it takes more force to push something through tar than water.

For thin films you need to do a diagram, I know where you got this question from and I think their answer is wrong anyways, when i tried it.

Here is the diagram based on the question tho:

http://img47.imageshack.us/img47/9986/diagramvr4.jpg [Broken]

You draw the boundaries and write down the index of each one.

Every light wave refracts (goes through) and reflects so just follow every line and draw one refraction and one reflection. If any wave goes from a fast to slow boundary (therefore from low index to high index), it comes back out of phase. Therefore, any line with a black dot is 180 degrees out of phase, the lines with a box are in phase.

If lines are out of phase the anti-nodes (maximums) cancel out causing destructive interference.

thickness and lambda relate like this:

look at the two waves coming out, and if they are out of phase then at thickness of approaching zero, they would result in destructive interference.

therefore you can say at a thickness of 0, (l / 2) , (l) where l is lambda, destructive interference happens. it's just multiples of a half lambda.

and constructive occurs at a thickness of (l/4) , (3l/4) and onwards with multiples of lambda.

Now this question is really badly worded but I think it wants constructive reflection, if that's the case it's easy, look at the ratios!!

remember, always to do that diagram for every question and only go out of phase when waves go from a fast to slow boundary. Another thing is always use lambda in the material, if u know the lambda in air of the wave then you basically know it's lambda where n = 1 so using n2 / n1 = l1 / l2

u can simplify ln (lambda in material) to ln = la (air) / n

Last edited by a moderator: May 3, 2017