Calculating Phase Difference and Interference with Laser Light

[duchuy]

Homework Statement

Light diffraction and phase

Relevant Equations

g(t, x) A.sin ( w ( t - x/c ) )

Hi, I need help finding the phase difference between the two rays converging at F'.

I was given this figure, the distance between the lenses is 10cm, the thickness of the lense is 1micrometer and the indexes are on the figure. The light is a laser in which lamda = 500nm.
I tried to write down the two expressions of the two phases of each ray, and I found :
phi1 = (2pi x . 1,1) / 500x10^-9 and phi2 = (2pi x . 1,35) / 500x10^-9.
So i don't understand in this case is x a constant or a variable and how to do I calculate the phase difference of these two rays because when i tried to calculate phi2 - phi1 I found a really weird anwer. And also, is the unit of phase difference rad/m?
And then with the answer, how would I know if it's a constructive or destructive interference?
Thank you all for your help cheers!

 

[kuruman]

I think it is easier to think in terms of the path difference between the rays in terms of wavelengths instead of phases.

How many wavelengths can you fit in the path followed by Rayon 1? What about Rayon 2? What must be true for the path length difference in wavelengths when you have constructive interference? What about destructive interference?

 

[haruspex]

the thickness of the lense is 1micrometer

Judging from the diagram, you mean that each ray passes through a block of length 1micrometer.

 

[duchuy]

I think it is easier to think in terms of the path difference between the rays in terms of wavelengths instead of phases.

How many wavelengths can you fit in the path followed by Rayon 1? What about Rayon 2? What must be true for the path length difference in wavelengths when you have constructive interference? What about destructive interference?

I am sorry if I have misused the vocabulary because I'm a french medical student . But if my translation is correct, the professor asked to determine the phase difference between the two rays at F' and then determine whether the interference is destructive or constructive.
The formula that we have for the phase is Phase : phi = wx/c = 2pfx/c = 2px/lambda. Hope this the formula helps. If it isn't clear, between the two lenses is air and we only consider the rays after they went through the first lens. Thank you for your help.

 

[haruspex]

how would I know if it's a constructive or destructive interference?

If it comes out to a whole number of half wavelengths it should be obvious.
Otherwise, figure out whether the resulting amplitude is more or less than the amplitude of the individual rays. What's another way of writing $\sin(\theta)+\sin(\theta+\phi)$?

 

[duchuy]

Judging from the diagram, you mean that each ray passes through a block of length 1micrometer.

Yes that is correct, but the indexes are different, (i'm not sure if I'm using the correct term but what i mean by index is the number in which the speed of light in a vacuum is divided by to obtain the speed of light in the environment. And the rays are only considered once they have passed the first lens. Thanks for your help.

 

[haruspex]

The formula that we have for the phase is Phase : phi = wx/c

Sure, but that assumes the wave is traveling at speed c. In this case the waves are slowed through the blocks, effectively increasing the path length, x. So you have φ_i =2π wxn_i/c.

 

[duchuy]

If it comes out to a whole number of half wavelengths it should be obvious.
Otherwise, figure out whether the resulting amplitude is more or less than the amplitude of the individual rays. What's another way of writing $\sin(\theta)+\sin(\theta+\phi)$?

I understand the that a constructive interference is equal to 2kpi and a destructive one (2k+1)pi. But i can't get to such expression because when i try to calculate phi I get a really absurd number like 12566370,6, so i really need help finding the value of the phase.
I guess another way of writing that would be 2 ((sin 2theta + phi) / 2 ) cos (phi/2)?

 

[haruspex]

I get a really absurd number like 12566370,6

You didn't seem to understand what x is, so I don't know how you are getting an answer at all.
It's the 1μm length of the blocks.

constructive interference is equal to 2kpi and a destructive one (2k+1)pi

That's for maximally constructive or destructive. It might only be partially so.

another way of writing that would be 2 ((sin 2theta + phi) / 2 ) cos (phi/2)

Right. So regarding that as a wave function of θ, what is its amplitude?

 

[duchuy]

Sure, but that assumes the wave is traveling at speed c. In this case the waves are slowed through the blocks, effectively increasing the path length, x. So you have φ_i =2π wxn_i/c.

Yes, i tried to calculate phi with the n1 index for ray 1. And i found :
phi = omega . x / c = 2.pi.x/(lamba/n1) = 2.pi.x.1,1/500.10^-9.
And I don't understand what the value of x is going to be, but i tried to calculate x = 1 and i found phi= 12566370,61.. when I am supposed to get somthing that looks like 2kpi or (2k+1)pi.
Thanks

 

[duchuy]

You didn't seem to understand what x is, so I don't know how you are getting an answer at all.
It's the 1μm length of the blocks.That's for maximally constructive or destructive. It might only be partially so.

Right. So regarding that as a wave function of θ, what is its amplitude?

I don't how you're talking about amphitude because we have a formula g(x,t) = A sin ( w ( t - x/c)) in which A is the amplitude and therefore the phase is equal to wx/c.
x and t here are space time variables and so i tried to put x = 1 because i gussed that time and space are constant at F' but yeah it seemed kinda off.

 

[duchuy]

That's for maximally constructive or destructive. It might only be partially so.

[/QUOTE]
And also i think since this is beginner biophysics, we are only considering maximally destructive or constructive.
I think you lost me when you introduced theta and phi at the same time. Is theta an angle? I thought that it was already part of phi since phi = 2pix / lambda?
Sorry, bio students are really scared of maths )):

 

[haruspex]

I don't understand what the value of x is

In the equation you wrote in post #1, x is the physical length of the path. Since we are only concerned with the portion of the path that traverses one of the blocks, that's 1 micrometer.
It could not make sense just to set it to "1". It must be a length, so has dimension.

I don't how you're talking about amphitude

That's what constructive and destructive interference are all about: how addition of two waves of the same frequency is equivalent to changing the amplitude.
In the equation that you wrote at the end of post #8, setting φ=0 gives 2 sin(θ), i.e. a doubled amplitude; setting φ=π/2 gives 0, complete destruction.
In between those extremes you can get any amplitude between 0 and 2. You can regard it as (partially) constructive if > 1 and (partially) destructive if < 1.

 

[duchuy]

Ok, but I'm supposed to calculate the phase difference between the two rays, and I'm yet to find the value of phi since i don't know how to find it because i don't know how how to set the value of x in the formula :
phi = 2pix.1,1 / lambda for the first ray.
I see that after finding phi, I am going to be able to find wheter it's contructive or destructive but i don't think I am there yet?

 

[haruspex]

i don't know how how to set the value of x in the formula

I have already told you in posts #9 and #13. It's the length of the block the light passes through in your diagram. It is shown as being 1μm.

 

[duchuy]

I have already told you in posts #9 and #13. It's the length of the block the light passes through in your diagram. It is shown as being 1μm.

But i thought that 1 micrometer is the thickness of the lense? I don't see how you can use that for the phase since we have to calculate the phase difference at F'. Or i think I am just not understanding the concept of what a phase is?

 

[haruspex]

But i thought that 1 micrometer is the thickness of the lense? I don't see how you can use that for the phase since we have to calculate the phase difference at F'. Or i think I am just not understanding the concept of what a phase is?

My post #3 was to point out that it is not the thickness of the lens. Forget the lenses.
In the diagram, just concentrate on the two blue shaded rectangles. These are where the optical path lengths differ, so are the only relevant part of the apparatus.
Each of these blocks is physically 1μm long, as labelled in the diagram (“e=1μm"). This is the x in your formula.
Because they have different indices, they have different optical path lengths. The optical path length is the physical length multiplied by the index. This is what creates the phase difference.

 

[duchuy]

That's what constructive and destructive interference are all about: how addition of two waves of the same frequency is equivalent to changing the amplitude.
In the equation that you wrote at the end of post #8, setting φ=0 gives 2 sin(θ), i.e. a doubled amplitude; setting φ=π/2 gives 0, complete destruction.

Oh wow that's amazing, i used x = 1micrometer and i found a multiple of pi. You're the best.
So now that I used it, phi2 - phi1 = pi.
But now determining the interference at F', since the phase shifted of a value of pi, so 180deg or 0deg. If i replace this value, i would get 2 sin (theta) so a doubled amplitude right?

 

[haruspex]

I'll try to explain a bit more about the phase difference.
The refractive index tells you how much light is slowed in going through the medium. If c is the in vacuo speed then in the medium the speed is c/n.
The two rays maintain the same frequency when going through the blocks, so the slower one must execute more cycles. So when they emerge from the other end, they are at different points (phases) within their cycles. You can visualise this by drawing a sine wave through one block, then copying it, but slightly compressed lengthwise, in the other.
When they emerge, they will have the same wavelength again but the acquired phase difference persists.

You can experiment with interference by graphing two waves of the same frequency and amplitude but different phases in a spreadsheet. Graph their sum. You should see another sine wave. Vary the phase difference to see how the amplitude of the sum changes.

 

[haruspex]

the phase shifted of a value of pi, so 180deg or 0deg

If you found a phase shift of π, that's not zero. 2π would be equivalent to zero.

 

[duchuy]

If you found a phase shift of π, that's not zero. 2π would be equivalent to zero.

Ohh right so in this case pi = 3pi so it's equal (2k+1)pi, k=1 so it's a destructive interference?
You are a life saver, thank you so so much I finally get it now. Thank you for everything and you're patience to bare with me!

 

[haruspex]

Ohh right so in this case pi = 3pi so it's equal (2k+1)pi, k=1 so it's a destructive interference?
You are a life saver, thank you so so much I finally get it now. Thank you for everything and you're patience to bare with me!

Happy to bear with you, but if you don't mind, I won't bare with you.

 

[kuruman]

To @duchuy:
Now that the problem has been solved let me show you the method I originally suggested for your benefit and for the benefit of those who may encounter a similar question. It is based on two simple ideas
1. Every time the traveling plane electromagnetic wave advances by one wavelength, its phase advances by 2π.
2. When light of wavelength λ₀ in vacuum travels through a medium of index of refraction $n$, its wavelength is given by $\lambda=\lambda_0/n$.

In this problem from point F to F' the two rayons advance by the same phase while in vacuum but by different phases when they go through the 1 μm materials because their wavelengths are no longer equal. The equation that gives the phase advancement when the light travels through thickness $d$ is $\phi=2\pi \dfrac{d}{\lambda_0/ n}=2\pi \dfrac{nd}{\lambda_0}$.
Rayon 1 advances by $2\pi \dfrac{1.10\times 10^{-6}}{500\times 10^{-9}}=2\pi(2.2)$.
Rayon 2 advances by $2\pi \dfrac{1.35\times 10^{-6}}{500\times 10^{-9}}=2\pi(2.7)$.
The phase difference between the two is $\Delta \phi=2\pi(2.7-2.2)=2\pi(0.5)=\pi.$
This means destructive interference.

 

[duchuy]

Yes, I didn't really understand what x was in the beginning because I thought that it was the optic path of both rays, so I tried to use trigonometric relations to calculate the distance from the second lens to F', must admit that was pretty dumb.
But since you've commented, may I ask, if the phase difference is between [0,pi], this means that the interference is partially destructive and vice versa if it's between [pi,2pi]?
Thank you so much for the precision.

 

[haruspex]

But since you've commented, may I ask, if the phase difference is between [0,pi], this means that the interference is partially destructive and vice versa if it's between [pi,2pi]?
Thank you so much for the precision.

No, it's a question of whether the resulting sine wave has a larger or smaller amplitude than the originals, i.e. whether the factor $2 \cos (\frac 12\phi)$ in $2 \sin (\theta + \frac 12\phi) \cos (\frac 12\phi)$ has magnitude more or less than 1.
If greater than 1 it is partially constructive. This will happen for $n\pi-\frac{\pi}3<\phi<n\pi+\frac{\pi}3$ for some integer n.

If the two originals have different amplitudes I am not sure how one would classify it.

 

[kuruman]

If the two originals have different amplitudes I am not sure how one would classify it.

You call it partially constructive, I call call it in-between interference. The different amplitude or phase should not faze you. If you write $$A \sin(\theta)+B \sin(\theta+\delta)=C \sin(\theta+\phi),$$expand the sines on both sides, equate sine and cosine coefficients (noting that the equality must hold for all $\theta$) and solve, you get $$C=\sqrt{A^2+B^2+2AB\cos\delta}~;~~\phi=\arctan\left(\frac{B\sin\delta}{A+B\cos\delta}\right)$$which yields the combination of two sines into one: $$A \sin(\theta)+B \sin(\theta+\delta)=\left(\sqrt{A^2+B^2+2AB\cos\delta}\right)~\sin\left[\theta+\arctan\left(\frac{B\sin\delta}{A+B\cos\delta}\right)\right].$$Needless to say that the phase of the sine on the right determines the kind of interference one gets while the amplitude determines the maximum vlaue at constructive interference.

 

[haruspex]

The different amplitude or phase should not faze you.

The algebra is not the issue. If the constituent waves have amplitudes A and B and the combined wave has an amplitude between the two, would I, in my terminology, call it partially constructive or partially destructive?

 

[kuruman]

The algebra is not the issue. If the constituent waves have amplitudes A and B and the combined wave has an amplitude between the two, would I, in my terminology, call it partially constructive or partially destructive?

I suppose I do not understand your terminology. The algebra shows that if one adds two sinusoidals, one gets an intensity that varies from zero to I_max depending on the phase of the sinusoidal. By all accounts, the interference is constructive when the intensity is at some maximum value I_max and destructive when it is zero. For me anything between these two extremes is "in-between" and I don't worry about partial constructive or partial destructive.

I did some plots and found that the amplitude $C$ of the resultant is not necessarily between $A$ and $B$ Shown below is a plot of the coefficients with $A = 1$, $B=1.5$ and $C$ as a function of $\delta$ $(-\pi \leq \delta \leq \pi).$ I can see that partial constructive/destructive might be when $C$ is above/below both $A$ and $B$ but there is also the nopersonsland where $C$ falls between the two. What do you think?

 

[haruspex]

I did some plots and found that the amplitude C of the resultant is not necessarily between A and B

Well, of course not, since they can largely reinforce.
In principle, you could get any amplitude between A+B and |A-B|. So if A>2B you can't get anything less than B.
The trouble with limiting "constructive" to mean totally in phase (and similarly "destructive") is that in practice that will never happen.

 

[kuruman]

The trouble with limiting "constructive" to mean totally in phase (and similarly "destructive") is that in practice that will never happen.

Yes, I see it now.