Optics, interference, very unusual problem

[fluidistic]

Homework Statement

See pictures for the data.
Basically there's a transparent prism looking like an inclined plane and making an angle $$\alpha$$ with the horizontal. Its base is 3 cm large and if we shine a vertical light of wavelength $$\lambda$$, we can observe an interference pattern as seen in the figure. If the interference pattern is 1 cm large, what is $$\alpha$$ worth?
Give the intensity of light in the interference pattern.
Figure out why do we only see an interference pattern in 1 cm despite shining light over the whole prism.

Homework Equations

No idea.

The Attempt at a Solution

I've absolutely no idea how to start this. In fact I don't understand where the interference pattern is (can we see it because of a diffusion somewhere in the prism?!). Is it at the base of the prism? Why is there interference?!
I know there's refraction of light through the inclined surface and then there's a reflection (Changing the phase of $$\pm \pi$$ rad or not? I know reflection causes a change in angle's phase, but in this case it's internal reflection so $$n_i=n_f$$ and I've read that for a change of phase in reflection, the transmitted material must have a higher refractive index than the incident one, which isn't the case in this exercise!).
I'm totally troubled and fail to understand what the heck is going on in this interesting exercise. Any help is greatly appreciated.

More details about the pic: I do not see any difference of intensity in any of the fringes of interference and they are equally distanced. Which lead me to think that the interference does not occur of the surface of the prism but I might be wrong. And sincerely I've never heard of such an interference where there's no difference of intensity in all fringes and suddenly the interference "stops" at a certain place. This problem was in my optics exam that I miserably failed.
P.S.: the direction of the rays in the sketch 1) is vertical. In 2), it wasn't indicated (hence my enormous confusion of where we see the interference!) but I'm guessing the rays should enter the page.

 

[ehild]

Trace the rays which enter at different distance from the edge of the prism. One part of the incident light will be reflected, the other refracted and reflected again from the inner surface(s) before leaving the prism.
You get rays which traveled different lengths so have different phases both in the reflected or transmitted light beam. They interfere. Find the condition for seeing an interference maximum or minimum.

ehild

 

[fluidistic]

Trace the rays which enter at different distance from the edge of the prism. One part of the incident light will be reflected, the other refracted and reflected again from the inner surface(s) before leaving the prism.
You get rays which traveled different lengths so have different phases both in the reflected or transmitted light beam. They interfere. Find the condition for seeing an interference maximum or minimum.

ehild

Thanks for the tip. I had already done this, though I didn't realize that the optical path difference would cause the interference, I thought that waves having an odd number of reflections would interfere with waves with an even number of reflections.
Ok I understand what you mean, but let me ask you a little question: So the waves interfere through almost all the prism, right? How come we can see interference pattern? Is it by diffusion inside the whole prism? Is it only at the base of it?
And by the way, if I trace an entering ray almost on the right edge of the prism, it will be reflected inside the prism, at the base of the prism but more at the left (due to Snell's law) that the point where it incided. Thus, I don't understand why there seems to have interference on the whole right part of the prism instead of the left part.

 

[ehild]

It would be nice of you to present a picture of those rays entering and reflected from the surfaces of the prism. It is difficult to understand what you mean without seeing it.
Forget about "diffusion". The light can travel in the prism along straight lines, and the laws of reflection and refraction apply whenever its strikes an interface.
The incident light is partly reflected from the front surface. Other parts are reflected inside the prism. When they leave the prism at the front, their wave-front is united with that of the first reflected ray, and interference happens.Try this place.

http://www.newagepublishers.com/samplechapter/001585.pdf

ehild

 

[fluidistic]

It would be nice of you to present a picture of those rays entering and reflected from the surfaces of the prism. It is difficult to understand what you mean without seeing it.
Forget about "diffusion". The light can travel in the prism along straight lines, and the laws of reflection and refraction apply whenever its strikes an interface.
The incident light is partly reflected from the front surface. Other parts are reflected inside the prism. When they leave the prism at the front, their wave-front is united with that of the first reflected ray, and interference happens.


Try this place.

http://www.newagepublishers.com/samplechapter/001585.pdf

ehild

Wow. I would have never though that. I get what you say.
I find your link very interesting, though I didn't have time to fully check it.
As of now I won't bother sending my sketches, I have only 2 remaining doubts that I hope I can get rid off without having to send my sketches.
First doubt is, page 9 of the link, it reads

Due to reflection an additional phase difference of $$\lambda /2$$ is introduced

. Shouldn't it be $$\pm \pi$$ rad?! Is he talking about internal reflection (as I suppose he is), in which case the reflection is over a material with lower refractive index (namely the boundary between glass/plastic and air) and no phase difference should be observed between incident and reflected wave, contrarily to the case in which the reflection is over a surface with greater refractive index (such as the boundaries between air and glass/plastic).

I'll try to solve my other doubt by myself tomorrow (it's why do we see interference on this side of the prism and not the other instead, I guess it has to see with simple geometric optics).
By the way I got 0 point in my exam (that I very badly failed) with this exercise and my professor strongly criticized me when I started writing "since the wavelength of the light is in the visible range, I can use geometrical optics to trace the rays." Though now I realize I should have said that the object was much larger than the wavelength of the light! And I did exactly the same rays as in page 9 of your link. Seems like my professor wanted a completely wavelike treatment, or I don't know. In all cases I'll show him how this problem can be solved starting with my supposition.

 

[ehild]

The path difference s between to waves corresponds to a phase difference 2pi s/lambda, and vice versa. It is not usual to replace the phase change at reflectance by a path difference, but it can be done.
The text of the problem with the wedge of prism is very unclear, the incident ligth should not be vertical.

When calculating interference on thin layers, we assume plane waves with infinite wave fronts. The normal of the wave front represents a ray. Geometric optics is valid if the object which reflects and refracts the light is much bigger than the wavelength. You can apply geometric optics for the rays, and so you can calculate the path difference between those rays which belong to the same original wave front.
The contrast of interference depends on the relative amplitude of the interfering light waves. You have really dark fringes if the amplitudes are equal. Imagine you have a very thin glass slab. At a glass-air interface, the intensity ratio of the of the reflected/incident waves is about 0.04. Most of the light enters into the glass, arrives at the back interface, and reflects there. This wave has almost the same intensity when leaving the glass at the front as the first reflected beam. So they will produce really dark and bright fringes. On the contrary, if you see through the slab the directly transmitted wave has to interfere with that which was reflected twice inside the slab, so it is very week. There is interference, but the fringes are weak with respect to the overall transmitted light intensity.

ehild

 

[fluidistic]

Thanks a lot for the information.
You said

The text of the problem with the wedge of prism is very unclear, the incident ligth should not be vertical.

but if you look at page 9, figure 1.10, the incident rays are not vertical.
I've read that there's no change of phase when a wave is reflected over a surface with lower refractive index than the incident medium one. Thus the reflection inside the prism does not change the phase of the wave, but the reflection of the incident ray over the surface of the prism causes the reflected ray to have a change of phase of $$\pm pi$$ rad compared to the incident ray.

I've tried to redo the exercise. I've considered a point P, the observer point.
I've drew 2 rays reaching point P. One is a ray that has been reflected over the surface of the prism. The other is a ray that entered the prism (refracted), reflected over the internal surface of the prism (no change in phase) and refracted again when leaving the prism to reach the air and point P.
I've drew "a" and "d", sides of a triangle.
If I consider $$\alpha \approx 0$$ then $$a\approx \alpha d$$. The optical path taken by the ray inside the prism is thus approximately equal to $$2 n_2 \alpha d$$ where $$n_2$$ is the refractive index of the prism's material.
If I consider $$b \approx c$$ (compared to the wavelength... I'm not sure I can do this approximation).
The total electric field reaching point P of these 2 rays would be $$\vec E = \vec E_0 [\cos (\omega t \pm \pi)+\cos (\omega t + 2 k_2 n_2 \alpha d)]$$. If I fix time to $$t=0$$, I get the condition $$\frac{n_2 \alpha d}{\lambda}=2 \pi m$$ where $$m \in \mathbb{Z}$$. I could isolate $$\alpha$$ and set d=2cm in order to get the answer to the question

If the interference pattern is 1 cm large, what is $$alpha$$ worth?

but this does not convinces me at all. What would be m?!
Further I'm not even dealing with intensity but with electric fields. I know that the intensity is related to the electric field (the square of it and averaged over a full period according to Hecht and it multiplied by its complex conjugate according to my professor. I'm really confused about how to get the intensity, if you could clarify a little bit I'd be glad) but I'm not sure how exactly.

What do you think about what I've thought?

 

[ehild]

I meant that the text of the problem is very unclear. You do not get that interference strips if the light falls vertically to the wedge with horizontal base. Those strips are seen on the surface of the wedge, that means that light rays arrive to your eye from them and the intensity of the light waves arising from different spots will be different. These rays can be imagined as not too extended plane wave-fronts, with normals identical to the direction of propagation. At point P in your drawing, you have two distinct waves, not a single one. The concept of light intensity is assigned to the light waves. The electric field can be calculated at P, but there are two separate waves, propagating in very different directions which can not be united to a single one.When we speak about interference on a thin layer, we have two parallel waves: one reflected from the front surface of the layer, the other enters into the layer and reflects from the back surface. When stepping out from the film, that wave meets a ray reflected from the same spot and their wave fronts merge. The electric fields of both waves add together according to the phase difference between the waves. This phase difference stays the same forever after the waves left the surface of the layer. See the figure 8 in the page.
The intensity of the light wave is the time average of the magnitude of the Poynting vector. It is proportional to the real amplitude of the electromagnetic wave.

In case of the wedge, the directly reflected wave and the one traveling through the wedge and reflected from the back, will not be quite parallel. But the angle alpha is very small and the wedge is very thin - there is no interference otherwise. In this case the two rays are almost parallel, their wave fronts are united and you get a wave with intensity dependent on the phase difference between the waves. See figure 10. The phase difference is proportional to the thickness of the wedge at that spot where these rays meet at the surface. It is not 2 cm, from where did you get it?

ehild
.

 

[fluidistic]

Thank you very much for the explanation.
My problem is that the ray entering the prism and leaving it with refraction points toward the left in my sketch while the rays reflected over the surface of the prism point toward the right of the prism and so they can never be parallel. However if I understood you well, if $$\alpha$$ is tiny, then the rays can be almost parallel and so we can sum the electric field and get the interference pattern, am I right on this?

Another question, you said

You do not get that interference strips if the light falls vertically to the wedge with horizontal base. Those strips are seen on the surface of the wedge, that means that light rays arrive to your eye from them

, do you mean I see the fringes by a diffusion over the surface of the prism?

 

[ehild]

What do you mean on diffusion? The light does not diffuse.
There is no interference if the light falls onto the wedge vertically. You see a spot on an object if a light beam enters into your eye from that spot, and you see it in the direction of the incoming light wave.
You can consider the incoming light beam as a bunch of parallel light rays. The rays are normal to the wavefronts of the incoming wave (green lines). Wavefront means a plane where the electromagnetic field is the same at every point.

Choose a ray (the blue one) which represents a wavelet, part of the incoming beam. Part of this wavelet is reflected from the surface, the other part enters into the prism, is refracted, reaches the back surface, reflects from there than reaches the front surfaces, and part of it leaves the prism at point P. There is an other ray (red) which reaches the wedge at the same point and is directly reflected. These two reflected wavelets do not travel quite parallel, but almost, and they make a common wavelet. The intensity depends on the phase difference and this is determined by the difference of lengths both rays travel from their last common wavefront to point P. You get similar pair of interfering wavelets all over the wedge. At some points, the phase difference of the outgoing wavelets is odd multiple of pi, you see a dark strip at that point. At other points, there is constructive interference between them.

(The statement that the beams are nearly parallel so they can interfere is a bit oversimplification. But the light beams are always a bit conical, even the laser beams. So those rays in the figure are really cones of rays and they contain wavelets which are really parallel.) ehild

 

[fluidistic]

Thank you once again.
So the problem of my final was badly worded? It clearly stated that the light was hitting the wedge vertically. We should see no interference then? Wow. I'll try to explain it to my professor.

Little question regarding your sketch. If alpha is small, can I neglect the optical path of the red ray starting with the second intersection of the green ray and up to the point when it reaches the prism? If yes then the optical path difference would be only n2 times the distance the blue ray travel into the prism. However I'm not sure I can neglect the small path of the red ray.
Almost same question, if alpha is small, can I say that the incoming and reflected blue ray inside the prism travel the same optical path before leaving the prism?

Last question: If the incoming rays are vertical, we get no interference because the refracted+reflected+refracted rays leave the prism in an opposite direction compared to the reflected rays, right? Their $$\vec k$$ is too different to talk of an interference. Am I understanding well?

 

[ehild]

The angle of this wedge is usually very small in order to get interference. You never see interference strips on a common prism, do you? If the angle is very small, the incident light can be nearly vertical, but even then it should come from the left with respect to the wedge normal.

The green lines in my drawing are wavefronts. You have to count the path traveled by the rays from their intersection point with the last common wavefront. The blue ray traveled the blue line inside the prism, you have to multiply this length by the refractive index of the prism, (n₂) to get the optical path. The red line traveled in air between the intersection with the green wavefront to point P where it hit the wedge. You can not ignore this piece in general. If the light hits the wedge along the normal, then the surface is a wavefront and the small piece of the red line disappears. But even this normal incidence is not favourable for interference.

If the angle of the wedge is very small, and the incidence is nearly normal, the optical path difference between the blue and red rays is about 2n₂d. Otherwise the two pieces of the path are different. See the derivation in the paper I sent. It is always better to derive something exactly and then find the limit case.

This wedge problem is usually treated in such a way that the rays are drawn vertical, both in air and in the prism, (neglecting all laws of reflexion and refraction), and the path difference is said to be 2nd, d = x sin(alpha) where x is the distance of P from the edge of the prism. ehild

 

[fluidistic]

Thank you very much ehild, now you've cleared all my doubts about the interpretation of the problem.

This wedge problem is usually treated in such a way that the rays are drawn vertical, both in air and in the prism, (neglecting all laws of reflexion and refraction), and the path difference is said to be 2nd, d = x sin(alpha) where x is the distance of P from the edge of the prism.


ehild

Ok I get it.
Now the phase difference of the reflected and refracted+reflected+refracted waves would be the path difference times $$k \pm \pi$$, right? with k=(2pi)/lambda.

 

[ehild]

Yes, the phase difference is

(2π)/λ * (n₂L₂-n₁L₁) ±π

where L_i is the length traveled in medium i.

ehild

 

[fluidistic]

Thanks.
Neglecting a small (in comparison to the wavelength of light) optical path, I get $$\delta =\frac{4 \pi n_2 d \alpha}{\lambda} -\pi$$ (I took -pi instead of pi because I get a different result that I don't like, but I'm not sure I can get do it).
Since in the diagram we could see totally dark and bright fridges, I can assume the intensities of the reflected and refracted+reflected+refracted rays to be almost equal and therefore I can use the formula for the irradiance $$I=I_0 \cos ^2 \frac{\delta}{2}$$.
Replacing $$\delta$$, I get that the first dark fringe (minimum of intensity) occur when $$d=\frac{\lambda}{2 n_2 \alpha}$$. And in general the minima of intensities are when $$d=\frac{\lambda(2 \pi + 2m)}{4 \pi n_2 \alpha}$$ with $$m \in \mathbb{Z}$$. Is this right?
While if I choose $$\delta =\frac{4 \pi n_2 d \alpha}{\lambda} +\pi$$, I get that the minima of intensities occur when $$d=\frac{m \lambda}{n_2 \alpha}$$ which gives me totally different results... Where did I go wrong?

I still can't find out how to answer the question about why do we see interference only in a part of the prism while we illuminate it entirely.
There's a distance D at which no more maxima or minima of intensity occur when the intensity remains constant... according to my formula, there's no limitation of m. I could directly count the number of fringes and get the greater m, but I do not understand why there's an upper limit. Could you explain this to me (physically obviously, not mathematically)?

 

[ehild]

$$\delta =\frac{4 \pi n_2 d \alpha}{\lambda} -\pi$$ (I took -pi instead of pi because I get a different result that I don't like, but I'm not sure I can get do it).

It can be both +pi or -pi.

Replacing $$\delta$$, I get that the first dark fringe (minimum of intensity) occur when $$d=\frac{\lambda}{2 n_2 \alpha}$$. And in general the minima of intensities are when $$d=\frac{\lambda(2 \pi + 2m)}{4 \pi n_2 \alpha}$$ with $$m \in \mathbb{Z}$$. Is this right?

No.
You get minimum intensity if δ=(2m+1)π, m integer. For d it means that d=mλ/(2n₂α). The first minimum occurs when m=0, at d=0, at the edge of the prism.

I still can't find out how to answer the question about why do we see interference only in a part of the prism while we illuminate it entirely.

I do not understand, either.

 

[fluidistic]

Thank you very much for everything. Problem solved.