Pfund's method for refraction

[ace]

TL;DR Summary

Pfund's refraction method makes no sense to me

I have looked through all the sources to ever exist for Pfund's method, yet I cannot understand why in the situation of having water on the Petri Dish, is there a 2nd black ring after the second white illumination (first in the center from the laser i assume?). What stops the second white illumination (caused by total reflection between water-glass boundary) to stop and for total reflection between water-air boundary to cause a change in the illumination...

 

[Ibix]

I'm a bit confused. Where's the white light coming from if there's a laser source?

Perhaps you could draw a diagram of the setup with the features you don't understand marked. Pictures can be attached to posts using the Attach files button below the reply box.

 

[ace]

I'm a bit confused. Where's the white light coming from if there's a laser source?

Perhaps you could draw a diagram of the setup with the features you don't understand marked. Pictures can be attached to posts using the Attach files button below the reply box.

Hey, here's the diagrams for the situation. I don't understand why it becomes dark again after acheiving total internal refleciton. As the angle increases, it should become more brighter, as now we have total internal refleciton from the glass-water boundary, i dont understand why it becomes darker first, then white.

 

[Ibix]

Ah, I see. The "white" is actually bright green washing out the sensor a bit.

As well as the critical rays, add a few more to the schematic diagram between them. They should go up to the water surface and come back down, and if you space them fairly evenly where they are emitted you will find that they come down further and further apart. That tells you that the energy is more and more spread out, so the light gets dimmer and dimmer the further out you go.

You are correct that (in principle) the intensity never falls to zero. (In practice the petri dish walls mess this up.) However, all detectors have a minimum detection value. IIRC, for your eye it's about 1% of the brightness of the brightest thing you can see - anything below that is black as far as you are concerned, whatever theory has to say.

So if you take a radial section from the center of your photo, you're looking at the sum of these three graphs:

The red line is, as you said, direct illumination from the laser and is concentrated in the central spot. The green line is illumination reflected from the upper surface and has a sharp-at-the-left peak that decays away to the right without ever falling to zero. The blue line is the same from the lower surface. You therefore see a darker band between the peak of the green and blue lines.

I would strongly recommend at least sketching the ray diagram I described in my second paragraph.

 

[ace]

Thank you so much! I have drawn this, please tell me if this is what you mean by them arriving further apart:

I had 3 more questions which I am confused about:

1. Is it the scale inaccuracy or does the last purple ray actually surpass the critical angle glass-liquid boundary ray?
2. I have been confused about this: When the red or purple rays go to the liquid boundary, why do they only refract and not reflect back down as well? Like the thin-film interference:

3. In your graph, is the x-axis the distance from the centre and the y-axis the intensity?

 

[Ibix]

please tell me if this is what you mean by them arriving further apart

Yes - your purple rays roughly trisect the angle between the two black ones, but the spacing of their return to the bottom increases to the right.

1. Is it the scale inaccuracy or does the last purple ray actually surpass the critical angle glass-liquid boundary ray?

I'm not quite sure what you mean here. The black ray that reflects from the glass/liquid boundary is marked as at the critical angle, and the purple rays clearly come in steeper than that, as I would expect for rays that pass through the boundary.

2. I have been confused about this: When the red or purple rays go to the liquid boundary, why do they only refract and not reflect back down as well?

They do reflect - this is how you can see yourself reflected in a window. But the fraction of light reflected at a boundary is $\left(\frac{n_2-n_1}{n_2+n_1}\right)^2$, where the $n$ are the refractive indices of the media. For media with very similar refractive indices, like water and glass, that will be a fraction of one percent.

Thin film media reflect much better because they are thin enough to use interference to limit the light passing through (or vice versa, depending on what you designed them to do), but that's not significant for a millimeter or two of medium.

3. In your graph, is the x-axis the distance from the centre and the y-axis the intensity?

Yes - sorry, should have labelled it.

 

[ace]

Yes - your purple rays roughly trisect the angle between the two black ones, but the spacing of their return to the bottom increases to the right.

I'm not quite sure what you mean here. The black ray that reflects from the glass/liquid boundary is marked as at the critical angle, and the purple rays clearly come in steeper than that, as I would expect for rays that pass through the boundary.

They do reflect - this is how you can see yourself reflected in a window. But the fraction of light reflected at a boundary is $\left(\frac{n_2-n_1}{n_2+n_1}\right)^2$, where the $n$ are the refractive indices of the media. For media with very similar refractive indices, like water and glass, that will be a fraction of one percent.

Thin film media reflect much better because they are thin enough to use interference to limit the light passing through (or vice versa, depending on what you designed them to do), but that's not significant for a millimeter or two of medium.

Yes - sorry, should have labelled it.

Oh, thank you so much!! I meant that my right-most purple ray went ahead of the black ray for the glass-liquid boundary, is that what really happens?

For your point about reflection, what is that formula called?
Also, if this were the case then how can the rays of light be scattered in all directions when it hits the paint (which has a refractive index similar the that of the glass?)

source: [Reference link redacted by the Mentors]
"If the lower surface is painted, the light scatters from the paint and reenters the glass at all angles, because the indices of refraction of the paint and the glass are comparable." page 3

 

[Ibix]

I meant that my right-most purple ray went ahead of the black ray for the glass-liquid boundary, is that what really happens?

Yes - as you were saying originally, the reflection from the upper surface of the liquid extends out to infinity, at least in principle. It just gets very, very faint. Imagine drawing a purple line only very slightly anticlockwise of the black ray. It'll go through the boundary but will be refracted only very slightly above the horizontal, so it'll go a long, long way to the right before bouncing off the upper surface and a long, long way before returning to the interface.

For your point about reflection, what is that formula called?

I don't actually know. I was reminded of it by another member here when answering a similar question a year or two ago. Wikipedia notes that it can be derived from the Fresnel equations, citing Born and Wolf.

Also, if this were the case then how can the rays of light be scattered in all directions when it hits the paint (which has a refractive index similar the that of the glass?)

Scattering is complicated and can happen a number of different ways. One is a rough surface; another is a collection of fine particles suspended in the paint. Neither of these have anything to do with the interface between the paint and the glass. The point the text is making here is, I think, just that there isn't any significant total internal reflection at the paint/glass boundary - the refractive indices are similar so even if there is total internal reflection the critical angle is only a little less than 90 degrees.

 

[ace]

Aah okay, thank you sooo much for your help!!

 

[ace]

Ah, I see. The "white" is actually bright green washing out the sensor a bit.

As well as the critical rays, add a few more to the schematic diagram between them. They should go up to the water surface and come back down, and if you space them fairly evenly where they are emitted you will find that they come down further and further apart. That tells you that the energy is more and more spread out, so the light gets dimmer and dimmer the further out you go.

You are correct that (in principle) the intensity never falls to zero. (In practice the petri dish walls mess this up.) However, all detectors have a minimum detection value. IIRC, for your eye it's about 1% of the brightness of the brightest thing you can see - anything below that is black as far as you are concerned, whatever theory has to say.

So if you take a radial section from the center of your photo, you're looking at the sum of these three graphs:
View attachment 336508
The red line is, as you said, direct illumination from the laser and is concentrated in the central spot. The green line is illumination reflected from the upper surface and has a sharp-at-the-left peak that decays away to the right without ever falling to zero. The blue line is the same from the lower surface. You therefore see a darker band between the peak of the green and blue lines.

I would strongly recommend at least sketching the ray diagram I described in my second paragraph.

Hello, I have another question now: if an increasing distance caused the rays to become darker again after the brighter circle, why does the total internal reflection for the glass-water boundary trigger another bright circle (This is the top left thing i labelled in my initial picture)? The distance is still increasing...

 

[Ibix]

The distance is still increasing...

Yes, but at that radius you get a new source of illumination, reflection from the glass/liquid interface (the blue line in my sketch graph).

Note that it's not as bright as either the inner ring or the middle spot. Also note that the human eye is roughly logarithmic in its response, so the outer ring is a lot dimmer than it might look.

 

[berkeman]

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