# I REALLY need help understanding this concept

1. Dec 15, 2011

### Xyius

This is for my Optics class. I am having a lot of trouble understanding this concept and my professor will not answer any of my emails. I do not know where else to turn.

The concept deals with an air gap between microscope slides or something. I believe it has something to do with fringes of equal thickness but I do not understand what this term means. Below is a problem that I have no idea how to solve because it uses this concept. How does the air gap play a role? I REALLY need help!

1. The problem statement, all variables and given/known data
A beam of white light from 400-700 nm is incident at 45 degrees on two parallel glass plates separated by a 0.001cm air gap.The reflected light is admitted into a prism spectroscope. How many dark lines are seen on the entire spectrum.

2. Relevant equations

I believe I am suppose to use this equation somehow.
$$2n_f tcos\theta_r=(m+1/2)\lambda$$
Where $n_f$ is the refractive index of the film. "t" is the thickness of the film and $\theta_r$ is the angle of refraction. I set it equal to (m+1/2)λ so it would be constructive interference.

3. The attempt at a solution
I do not know where to start as I don't know how to deal with the air gap! Can anyone help? :(???

2. Dec 15, 2011

### technician

This is about INTERFERENCE.... thin film interference...do you have any knowledge of this topic

3. Dec 15, 2011

### dacruick

4. Dec 15, 2011

### Xyius

Yes I do. Forgive me but I feel as though that comment was belittling and rude.

I do not have any problems dealing with anti reflection coating problems in which the bottom layer is a substrate with a high index of refraction. But for some reason these problems with the air gap throw me for a loop. I just do not under stand how to do the problem if the thickness of the glass isn't given. Could anyone give me a "push" in the right direction to the solution to this problem?

EDIT: I am looking at that website now

5. Dec 15, 2011

### technician

Xyius:
Believe me there is no intention on my part to be rude or to belittle you !!!!!
I am a teacher and I want to make learning physics enjoyable and easier !!!!
I have been chastised for giving too much help therefore I feel that I have to ask a very basic question as a first response. !!!!
If you are confident about anti-reflection coatings then this is just an extension of that principle.
In this case with 2 microscope slides the important reflections are from the glass/air interface of the top slide and the air/glass interface of the bottom slide.
There is an air gap and this introduces a path difference between these 2 reflected beams. Depending on this path difference either constructive or destructive interference occurs (bright and dark fringes) Because the slides are at an angle (that is not obvious from your question wording) bright and dark fringes are seen..... hope this helps and be assured most of us here are to help !!!!

6. Dec 15, 2011

### technician

Actually.... it may be because the light is at an angle.... not the slides.... principle is the same

7. Dec 15, 2011

### Xyius

I apologize for misunderstanding. Thank you very much for taking the time to help :)
So this is my thought process now. The beam strikes the top of the glass plate and experiences a phase shift of pi because it is going from a low index of refraction to a high one. The beam travels through the glass and refracts out the other side to the air interface as well as reflecting internally up and out of the top slide. The beam through the air now strikes the bottom plate and experiences another phase shift of pi. This beam now reflects upward and strikes the top plate (another phase shift) and goes up and out the glass.

So there are two beams that are reflected back upward both with a phase shift of pi. Is this correct? Now I just need to set up the equations.... That's I'd where I am having trouble

8. Dec 15, 2011

### dacruick

the phase shift only occurs on reflections not refractions if I remember my optics correctly.

9. Dec 15, 2011

### technician

In practice the only beams you need to take into account are the beams at the air 'wedge' between the 2 slides. So there is no phase shift at the glass/air boundary at the lower surface of the top slide but there is a phase shift of pi at the air/glass boundary of the bottom slide.
You do not need to worry about the thickness of the slides because, in practice, they are very much greater than the wavelength of light.
You should be able to see that the path difference between the 2 reflected beams is
=2t + λ/2 (the λ/2 is the phase shift)
Max occur when 2t + λ/2 = nλ..... hope this helps

10. Dec 15, 2011

### Xyius

I'm confused as to why it is 2t and not 2tcos(theta). Also why is n(lambda) (sorry for not using the latex but I'm on my phone and it's hard enough to type :p ) shouldn't it be (m+1/2)lambda since it is constructive interference we are after?

11. Dec 15, 2011

### technician

I think that you are correct but it means you need to know the refractive index of the glass to sort out the angles.
Path difference = 2tCos∅ looks correct as long as you can determine ∅ (from refractive index)

12. Dec 15, 2011

### Xyius

Aha! I got it! I just got an epiphany or something haha. So I calculated the path difference that takes through the air using the formula involving the cosine. I know that the angle of refraction in the air is 45 since that is what it was in its incidence on the top of the plate. Doing this gave me 14142nm. Then I had 14142 + x/2=nx as you said (I'm using x for wavelength). I then set the wavelength equal to 400 then 700 and found n in each case and took their difference. I got 15 which is the correct answer! I cannot thank you enough for your help. Hopefully that is all the help I will need as of now :)

13. Dec 15, 2011

### technician

BRILLIANT .... well done