Dispersion diagram of light source seen through prism

[D_J]

Homework Statement

Take a photo through a glass prism to view a light source. Notice that in the photo taken, the object seen through the prism has two colored edges, one red and one blue (i.e., “dispersion”). Draw a light ray diagram and find out which color has the higher refractive index through the prism. Note that our eyes were seeing a virtual image of the object through the prism.

Relevant Equations

⇒η = sin[(A+D)2] / sinA2

I know that red light has a lower index of refraction than blue light, but that’s not what I’m seeing. The blue light is where the red light should be.

I can’t afford to join CHEGG. Any chance that someone would help me out for free? I’d really appreciate it.

[Link to chegg removed by the Mentors]

 

[haruspex]

Homework Statement:: Take a photo through a glass prism to view a light source. Notice that in the photo taken, the object seen through the prism has two colored edges, one red and one blue (i.e., “dispersion”). Draw a light ray diagram and find out which color has the higher refractive index through the prism. Note that our eyes were seeing a virtual image of the object through the prism.
Relevant Equations:: ⇒η = sin[(A+D)2] / sinA2

I know that red light has a lower index of refraction than blue light, but that’s not what I’m seeing. The blue light is where the red light should be.

I can’t afford to join CHEGG. Any chance that someone would help me out for free? I’d really appreciate it.

[Link to chegg removed by the Mentors]

Please post a ray diagram, indicating where you observe the red and blue fringes.
Add an explanation of why this seems to show the red light having the greater refractive index.

 

[D_J]

Okay, here’s what I’ve got so far. Diagram A should be correct, but diagram B is what I see when I take the photo. If I let the light shine through the prism, it falls on the paper as in diagram A, but when it hits my retina or the camera, I see what’s shown in diagram B. The blue and red switched places. I don’t get it.

 

[haruspex]

The ray diagrams do not show where the images would be.
To find where an image would be you have to take two rays, same color, same source point, following different paths. E.g. in diagram A, change all the lines to red and draw the image as a single red circle.
Now add blue lines starting out along the same paths as the two red lines but refracting more. Track back the lines they emerge on to find the blue image.

 

[D_J]

Please post a ray diagram, indicating where you observe the red and blue fringes.
Add an explanation of why this seems to show the red light having the greater refractive index.

The ray diagrams do not show where the images would be.
To find where an image would be you have to take two rays, same color, same source point, following different paths. E.g. in diagram A, change all the lines to red and draw the image as a single red circle.
Now add blue lines starting out along the same paths as the two red lines but refracting more. Track back the lines they emerge on to find the blue image.

Okay, they cross over when I trace them back, but that still doesn’t explain the image. In order for the color fringes to be reversed from left to right the rays have to cross over when they emerge from the prism, right?

 

[haruspex]

Okay, they cross over when I trace them back, but that still doesn’t explain the image. In order for the color fringes to be reversed from left to right the rays have to cross over when they emerge from the prism, right?View attachment 299972

No, they only have to cross over in the traceback. The new diagram explains it clearly.

 

[D_J]

Yeah, that’s what I thought, too, but I asked these two color experts, and they both said that my diagram doesn’t explain the reversed spectrum because, for example, an image in a mirror doesn’t require convergence at the image but on your retina. They both said that I need to redraw my ray-tracing diagram to show the rays converging at the observer.

But then I watched this video from Khan Academy, and it still shows the red converging onto our retina from the thinner portion of the apex and the blue from the ticker portion towards the base of the prism.



I’m totally lost. I don’t get it.

 

[haruspex]

these two color experts, and they both said that my diagram doesn’t explain the reversed spectrum because, for example, an image in a mirror doesn’t require convergence at the image but on your retina.

The rays are not converging at the image; the traceback from the rays arriving at your eye converge at the image. That is the principle of all optical images, mirrors included. You perceive a point on an object as being at the point the rays appear to come from (obviously). What happens inside the eye is irrelevant.
Can you draw rays producing an image in a mirror that way and show that to your "experts"?

 

[D_J]

Yeah, I agree with you. It’s just that the experts were David Briggs and Bruce MacEvoy. Their work on color perception is extensive, and they both disagree with me. David is a professor at the University of Technology in Sydney. Here’s their quotes and below is a sketch that Bruce sent me.

“A virtual image, for example your image in a mirror, does not require convergence at the image but on your retina. The key point is that a virtual image cannot do work, no matter where it forms or appears to form. In order to understand how the spectrum image forms you have to trace the separate red and blue content of the dispersion from the source through the prism to the pinhole of your iris.”—Bruce MacEvoy> https://www.handprint.com/LS/CVS/color.html

“Regarding the "reversed spectrum" the essential point is that the apparent position of the blue image is displaced further from the true position than the red image is displaced because the blue ray is bent more than the red ray. I think Bruce MacEvoy's diagram is correct and that strictly speaking you would need to redraw your ray-tracing diagram to show the rays that converge at the same observer.” –David Briggs> http://www.huevaluechroma.com/

That can’t be right though because no matter where you place your eye or the camera, the spectrum is always reversed. If the actual rays crossed over before they entered your eye, the object would be upside down like a pinhole camera, right?

I could ask one of them again, but I don't know how I would frame the question. Hmm...maybe I could just send them a link to this thread.

Thanks for all your help! I really appreciate it!

 

[haruspex]

A virtual image, for example your image in a mirror, does not require convergence at the image but on your retina.

Yes, that's correct. But you do not need to follow the rays all the way to the retina. If the rays that arrive at the eye can be traced back in straight lines to meet at point P then the eye will perceive them as coming from point P. How the eye arranges that they meet at the same point on the retina is another matter.

a virtual image cannot do work

No idea what that means.

trace the separate red and blue content of the dispersion from the source through the prism to the pinhole of your iris

Yes, to the iris. No need to go further.

the apparent position of the blue image is displaced further from the true position than the red image is displaced because the blue ray is bent more than the red ray

Yes.

Bruce MacEvoy's diagram is correct

Yes, but I feel the diagram in post #5 is easier to understand / more convincing because it shows exactly where the images are. But there is no conflict between the two methods, and they lead to the same conclusion.

A better diagram for McEvoy's single ray per colour method would be to show a red ray and a blue ray leaving the same point on the object and both arriving at the centre of the iris. Because of the extra diffraction angle, the blue ray has to take a higher path. This avoids getting tangled up with where rays cross.

 

[D_J]

Yes, that's correct. But you do not need to follow the rays all the way to the retina. If the rays that arrive at the eye can be traced back in straight lines to meet at point P then the eye will perceive them as coming from point P. How the eye arranges that they meet at the same point on the retina is another matter.

That is the main point of contention. Bruce is claiming that to see or photograph a reversed spectrum through a prism, the rays must overlap when exiting from the prism. They're saying that it cannot arise simply from tracing the rays back in a straight line.

The guy from the Khan Academy video has them overlapping in the middle, as well, but he’s saying that they continue to diverge, and the only reason that we see a reversed spectrum is because our eye/camera traces them back in a straight line.

I asked him, too.

Question: Why is the spectrum reversed when viewing a light source through a prism?

Answer: Great question. Since the blue light bends more it appears to have originated from higher in the sky. (If you trace it back it will make sense). And hence when you see it, you see blue at the top and red at the bottom.

 

[haruspex]

Bruce is claiming that to see or photograph a reversed spectrum through a prism, the rays must overlap when exiting from the prism.

It depends which rays you choose to draw.
If you take a red and blue that start on the same path from the object then they will cross over in the traceback, nowhere else.
If you take a blue with a slightly higher path than the red they may cross after entering the prism.
Slightly higher still, they will cross after exiting the prism.
Higher yet and you can make them meet (and cross over) at the iris.
Higher again and they do not cross at all.
The question is, which of these helps to explain clearly what you will see. Meeting at the iris is absolutely clear. It shows (from the traceback!) that the blue image will be higher. The preceding three cases above are unhelpful because you do not know which side of the crossover the images will be.

They're saying that it cannot arise simply from tracing the rays back in a straight line.

Are they? I do not see that in any of the quotes. Is that just your interpretation or are there other statements not quoted?
Tracing back is the way to figure out where you see an image as being. Isn't that blindingly obvious?
Tracing back a single ray of each colour, both meeting at the iris, shows which image will be higher, and that is good enough here, but it does not pinpoint where along those traceback lines the images will be. To do that you need two rays of each colour.

Since the blue light bends more it appears to have originated from higher in the sky

Exactly.

 

[D_J]

They're saying that it cannot arise simply from tracing the rays back in a straight line.

Are they? I do not see that in any of the quotes. Is that just your interpretation or are there other statements not quoted?

You’re right, I could have misinterpreted his answer, but it’s unlikely. This is the diagram that I sent him.

He said, “real images, such as those on the back of your eye, are images that can do work -- affect a photosensor, tickle your retina, expose a film emulsion. the light passed by a pinhole converges at the pinhole, not at the projection screen, but the image on the screen can still do work, for example as a pinhole camera. a virtual image, for example your image in a mirror, does not require convergence at the image but on your retina. the key point is that a virtual image cannot do work, no matter where it forms or appears to form. in order to understand how the spectrum image forms you have to trace the separate red and blue content of the dispersion from the source through the prism to the pinhole of your iris, and once you do that it is clear that you cannot see both the red and blue ends of the spectrum (and everything in between) from the same single "ray," and the error in the diagram you sent me is then easy to see.(see diagram)”

This is his diagram, which clearly shows the rays crossing over when exiting and it even says that the red from one ray and the blue from the other ray doesn’t reach the eye.

He goes on to say, “yes, if you view a light source through a prism, then the blue end of the spectrum will appear toward the pointed corner rather than the flat side of the prism. if you project the light source through the prism onto a wall, then the blue end will appear on the flat side rather than the pointed side of the prism. the reason the spectrum is reversed is because the convergence point of the image is in your eye rather than in the prism.”

 

[rude man]

Yeah, that’s what I thought, too, but I asked these two color experts, and they both said that my diagram doesn’t explain the reversed spectrum because, for example, an image in a mirror doesn’t require convergence at the image but on your retina. They both said that I need to redraw my ray-tracing diagram to show the rays converging at the observer.

But then I watched this video from Khan Academy, and it still shows the red converging onto our retina from the thinner portion of the apex and the blue from the ticker portion towards the base of the prism.



I’m totally lost. I don’t get it.

View attachment 300026

In post 11 bottom right picture that looks wrong, if the intent is to demonstrate dispersion.
Different color rays will not focus on the same spot. E.g. if the screen image distance is adjusted so as to render an exact focus for green light, then a red beam would focus behind the screen (geometrically speaking only of course) while a blue ray would focus in front of the screen.

You should also be aware that some materials might have anomalous dispersion ( n(red) > n(blue). )

 

[haruspex]

You’re right, I could have misinterpreted his answer, but it’s unlikely. This is the diagram that I sent him.
View attachment 300128

He said, “real images, such as those on the back of your eye, are images that can do work -- affect a photosensor, tickle your retina, expose a film emulsion. the light passed by a pinhole converges at the pinhole, not at the projection screen, but the image on the screen can still do work, for example as a pinhole camera. a virtual image, for example your image in a mirror, does not require convergence at the image but on your retina. the key point is that a virtual image cannot do work, no matter where it forms or appears to form. in order to understand how the spectrum image forms you have to trace the separate red and blue content of the dispersion from the source through the prism to the pinhole of your iris, and once you do that it is clear that you cannot see both the red and blue ends of the spectrum (and everything in between) from the same single "ray," and the error in the diagram you sent me is then easy to see.(see diagram)”

This is his diagram, which clearly shows the rays crossing over when exiting and it even says that the red from one ray and the blue from the other ray doesn’t reach the eye.

View attachment 300127

He goes on to say, “yes, if you view a light source through a prism, then the blue end of the spectrum will appear toward the pointed corner rather than the flat side of the prism. if you project the light source through the prism onto a wall, then the blue end will appear on the flat side rather than the pointed side of the prism. the reason the spectrum is reversed is because the convergence point of the image is in your eye rather than in the prism.”

Heavens, what a lot of gobbledegook.
There are two images for each colour: a primary image formed by the bending of the rays through the prism and a secondary image on the retina. We can trust the physiology of eye and brain to form the secondary image suitably and to deduce from it the location of the primary image. The task is to explain what you perceive, the primary image, not how you perceive it. It is quite unnecessary to worry about the secondary one.

The only thing wrong with your diagram in post #13 is that you are assuming the image is at about the same distance as the object. In particular, that it is still on the other side of the prism from the observer. That is why I prefer the diagram with the two rays that land on the centre of the iris. But both work.

Diagrams showing rays that don’t reach the eye achieve nothing.

 

[D_J]

Heavens, what a lot of gobbledegook.
There are two images for each colour: a primary image formed by the bending of the rays through the prism and a secondary image on the retina. We can trust the phyiology of eye and brain to form the secondary image suitably and to deduce from it the location of the primary image. The task is to explain what you perceive, the primary image, not how you perceive it. It is quite unnecessary to worry about the secondary one.

The only thing wrong with your diagram in post #13 is that you are assuming the image is at about the same distance as the object. In particular, that it is still on the other side of the prism from the observer. That is why I prefer the diagram with the two rays that land on the centre of the iris. But both work.

Diagrams showing rays that don’t reach the eye achieve nothing.

But I want to know what physically happens. Why do we see a reversed spectrum?

No matter where you put your eye, e.g., closer, higher, further back, it doesn’t matter, the spectrum is always reversed.

Is it because our eyes trace the rays back in a straight line or is it because the rays overlap each other when exiting the prism before hitting our eyes?

 

[haruspex]

But I want to know what physically happens. Why do we see a reversed spectrum?

No matter where you put your eye, e.g., closer, higher, further back, it doesn’t matter, the spectrum is always reversed.

Is it because our eyes trace the rays back in a straight line or is it because the rays overlap each other when exiting the prism before hitting our eyes?

You see it the way you do because your brain assumes light travels in straight lines from a given source point to each eye. As a result, you perceive a point on an object as being at the place where the trace backs of the received rays from that point meet. That applies to all the rays from the given point that reach the iris of either eye.
How the eye and brain system achieves that is entirely irrelevant. In practice, it is by refocussing the rays to a point on each retina, then within the brain merging the inputs from the two eyes; but one can imagine other systems.

Does this constitute a "reversed spectrum" to be explained? Yes, it is reversed from what you see if you project two rays that start along the same line onto a screen, but it is not a reversal of reality. The reality is that the red and blue rays come from the same point. Projecting onto a screen produces a shift one way, while viewing through the prism produces a shift the other way.

You must get away from the idea that rays crossing is somehow central to this. As I proved in post #12, whether and where rays cross depends on which rays you choose to study. In the upper diagram of "his" picture in post #12, whether the rays cross depends on how close you are to the prism, yet what you perceive does not depend on that.

David Briggs gets it right when he says what matters is the angle of the rays leaving the prism. Since the blue rays are at a steeper angle, they will hit a screen at a lower point but, by the same token, because the primary image is on the other side of the prism from the screen, will appear to come from a higher point.