1. When we looked at the red light through a spectroscope we made the following observations: The red part of the spectrum is elongated adn sharper, whereas the other colors on both sides were visible but shortened adn fuzziers.
Joe's comment: I don't think I understand what you mean by "elongated and sharper" vs "shortened and fuzzier." When I look through the spectroscope, I see a multicolored band, which I can make horizontal by turning the spectroscope the right way, and goes through the colors of the rainbow several times both to right and left. I ignore everything but the first set of rainbow colors (either to right or left) -- the others are basically echoes. This first set I will call the spectrum.
I always see the spectrum the same place, no matter what light I am looking at. The pattern is always the same width in the horizontal direction -- blue is always the same place, the red is always the same place. The height of the spectrum is also the same, if the light source is large enough to illuminate the whole slot; or it could be less tall, if I am looking at a small, distant light.
The big difference between different light sources is how bright are the different parts of the spectrum. Some parts may be so dim that they are missing altogether.
We believe that the red is more visible (longer and sharper) because the filter blocked out some of the color wavelengths and if we could isolate only the red frequency without any outside light, then we believe we would only see red through our spectroscope. We still saw the other colors of the spectrum using the filter because we couldn't block all of the other wavelengths using the red screen. For the red seen on the computer, ther was peripheral light from the computer screen and room light.
Joe's comment: Another reason is that the filters are not perfect -- the red filter lets some green light leak through. Using a double red filter shows how important this part is.
The eye is not very good at measuring brightness. So a spectrum with most of the blue and green removed doesn't look so different from a spectrum with equal mixtures. We tend to notice that the colors are there, rather than noticing that they are very dim.
We noticed that whatever color light we were looking at, that was the color of the spectrum that stood out the most.
Joe's comment: That works for red or green, but there isn't a white part of the spectrum, and the yellow patch on the computer screen isn't brighter yellow according to the spectroscope.
Least Sure Answer: 3. The diffraction grating has closely spaced rows of little holes, whereas the compact disc grooves are spaced out in a circular pattern. Light passes through the diffraction grating and light is reflected of the compact disc. For one thing we did not need a spectoscope to see the spectrum off the spectroscope. The spectrum we observed through the diffraction grating gives you a slice of a rainbow effect. This is different than what you see using the compact disc. The compact disc reflects the entire spectrum in many directions in a conical pattern. This design allowed you to observe the overlapping of the different wavelengths so that white was observed when placed at a certain angle along with color spectrums on the disc.
I am not exactly sure why the whole disc didn't turn white or had color on it. I am not exactly sure how the grooves on a compact disc separated the white light into different wavelengths.
Joe's comment: Let's not get into how it works! The question to ask is, do the diffraction grating and the compact disk work more or less the same way, or are they completely different? After all, there are many ways to get color effects. The spectra are similar, in that the colors come in the same order -- if you slowly tilt the compact disk away from you, starting with a reflection of the light bulb in the middle, you will for a while see nothing but then a bright blue band begins to show up, followed by the other colors of the rainbow. Blue comes first. This is like the diffraction grating, where blue is the color closest to the center, where the lightbulb is. And when you have finished with red, you may see the beginning of another blue band, just before you have tilted the CD as far as you can.
Again I'm not sure if this is what you were looking for, but please email us and let us know if we need to anwer the questions over.
Please give us some quided questions if we totally missed the mark, but I don't think we did.
Joe's comment: What we are looking for is that you think about the questions and are observant, and give a good discussion of what you observed and decided. By these rules there can be many different answers that are good; there isn't a particular right answer. If the questions are confusing, this may mean that we didn't ask good questions, and we'll try to improve them. But also notice that we don't have to ask a very good question to evoke a good discussion -- I have always claimed that a perfectly valid physics exam question would be "What is your favorite color, and why?" -- with the idea that the person answering it would explain why their favorite color had the best possible spectrum.
(SS)
> >We were sure of #3. The disc and the grating are alike in that they both > >show the spectrum and the spectrum moves as you move both of them. They are > >different in that the disc' spectrum can only be seen as far a the disc' > >surface. The spectrum of the grating can be seen far away from the light > >source.
Joe's comment: It's true that with the disk in one position, you only get to see part of the spectrum, but if you slowly turn it you can see different parts. The two ways to make a spectrum are similar in several ways: if you look straight through the diffraction grating, you see the white light. Then to either side, you see a spectrum, with blue on the inside, closer to the white light. The spectrum goes through the set of rainbow colors Violet/Indigo/Blue/Green/Yellow/Orange/Red, and then after a region in which there is nothing showing, you get the spectrum again... and again, perhaps. if you hold the CD the right way, you see the reflection of the white light. If you now turn it in any way, you will come to a spectrum, starting with blue, and working your way throught the rainbow colors VIBGYOR, followed by a region in which there is nothing showing, and then you get the spectrum again... and again, perhaps.
There are other ways to make a spectrum -- with a prism, for example -- but they don't give the multiple spectrum effect described here.
> We were not sure about #1 or #2. > >
Joe's comment: OK, but what did you think about these questions? The goal here is not to arrive at the Official Right Answer -- it is to practice observing and interpreting, and communicating what you decided or did not understand.
(Pirates and Cardinals)
> Make A Spectroscope > > 1.a. The spectrum was red, but yellow was closest to the source. > b. The green light spectrum had the green light closest. > c. Yes, you could differentiate between the two, because the width of > the spectrum's colors were different.
Joe's comment: The question doesn't specify how we are observing the spectrum. I hope you used the spectroscope, at least some of the time, because it gives a more complete separation of the light.
Something to watch out for is that the eye doesn't pay much attention to brightness of light, and is very interested in color. So all lights seem to have the same spectrum ranging from red to deep blue, until you compare them side-by-side, and then perhaps you notice that in the spectrum of a red light the blue and green are really faint, while in the spectrum of a blue light the blue is somewhat brighter.
> > 2. a. The white light onto the spectrum began with people and tapered down > to red. There was a blank spot > and then it began again. The white light onto a white background > shows a brighter, slightly wider spectrum.
Joe's comment: This question does specify using the spectroscope, and comparing things side by side. What I see when I compare the spectrum made by a red object and by a white object is that the spectrum due to the red object is dimmer in the blue and green and about equally bright in the red.
If the red object makes red light, we don't need to shine white light on it to see red.
If the red object turns white light into red light (in the same sense that a coffee maker makes coffee out of water -- you put a pot of water in and a pot of coffee comes out), then the spectrum of the red object could be brighter some places.
The third choice, that the red object removes the blue and green parts of white light to make red light, best describes what I see. Don't you agree?
> > 3. The CD and Diffraction Grating has similarities and differences. The > similarities are both clear, easy to see, and the > same colors were present. The differences are the spectrum is reversed > between the two. The CD has red, > yellow, green, and blue. The diffraction grating red, yellow, blue and > green. The CD's spectrum's width was > wider than the grating. >
Joe's comment: Please check on the diffraction grating observation -- I think you have blue out of place.
When you say the width of the spectrum, I think you mean that the red part is farther away from the blue part? Yes, that's true.
You mentioned above that in the spectroscope as you look farther and farther to one side of the slit you see a spectrum blue --> red, a dark space, and then it starts over blue --> red. And if you look to the other side, you see the same pattern -- so that the whole picture is
red ......blue dark red .....blue dark slit dark blue....red dark blue....red
The CD does something similar, though it is harder to observe -- but if you are illuminating it with just one light source and then slowly tilt it, you will see it go through the colors. If the colors go from blue to red, then after red comes nothing and then we start over with blue. But if as you tilt you go from red to blue, you might come to the position where the CD is like a mirror, reflecting white light, followed by blue --> red again.
(Perfectionists)
> We decided that we were the most unsure of Question Number 3. It seemed too > easy. The spectrum of the CD and diffracation grating both contained all > the primary and secondary colors of Red, orange, yellow, green, blue, > indigo, and violet. The diffraction grating only contained these colors, > while the CD contained many other colors. These colors seemed to be a > mixture of the primary and secondary colors. We saw pink, light purple, > cyan, magenta, and many other color mixtures. >What! We're not allowed to ask easy questions?
The purpose here was to make you look carefully at the diffraction grating spectrum. The second part, observing the CD, is hard to do (because it wasn't designed for this purpose).
So let's begin with the diffraction grating. Here we want a single small light in an otherwise dim room. "Straight ahead" through the diffraction grating we see the light itself, a bright white light. Then to either side (if that's how the diffration grating is oriented), we see a rainbow, with blue coming first and then on through the other colors to red. Beyond that we encounter another rainbow, again with blue on the inside -- and there is still one more rainbow beyond that. The colors always come in the same order, and are the same order as the rainbow (which comes about in a completely different way, for a completely different reason, so that is interesting).
Now we do something similar with the CD. The most similar thing do to would be to put it in a fixed position in a dimly lit room with one light source and then run around it, seeing what we see from different directions; or we could slowly tilt the CD in various directions. There is a particular direction in which we see the reflection of the light source, bright and white -- that corresponds to "straight ahead" for the diffraction grating. Then to either side we see ... and beyond that ... and the colors always come... ... . Or do they?
I suspect, from your mention of pink and light purple, that your light source was not a single little light bulb -- it was th overhead lights, or an open window. You saw what you saw, and so in this important sense your observation was correct. But I see something different, because I do the experiment a different way. Good scientific communication involves describing the experiment well enough that it is reproducible; good scientific understanding entails knowing what parts of the experiment must be specified.
> > We were the most sure of question number one. The spectrum of very red > light looks mostly red because the red light is reflected and the other > colors are absorbed. Blue and green light are different because they > reflect themselves and absorb the other colors. For example, blue light > reflects and the others are absorbed when looking at a blue light. Green > light is reflected, while the others are absorbed when looking at a green > light. We think that you could determine which color you were observing [in > a black&white photo] by > the position of a bright white spot in the color spectrum. Red, green, and > blue would all appear white in a black and white photograph of their > spectrums. White light is all the colors combined. By knowing the order of > the colors of the spectrum, that red is first, green is after red, and blue > is last, we could determine the color by its position in the spectrum. The > diffraction grating would show a much larger portion of white where each > specific color would be. >
Joe's comment: Even though our eyes cannot see ultraviolet light, the black-and-white film probably can. So we should be able to tell a light source that is making UV from a light source that is not, by seeing if the photograph of the spectrum continues on "beyond blue."
(Newtons)
Spectrum
Question 1 The spectrum of red light looked very bright with wide
stripes, that had more colors.
The blue had a litle red but purple, blue and green were dominant
colors. The green had very little red in it.
I believe we probably could idetify the difference because of the width
of the stripes.
Joe's comment:
I think you are saying that different filters would give different patterns, which is the basic idea.
Note that no matter what kind of light you are looking at, the spectroscope always puts the red part the same place -- farther away from the center than the blue and green. The difference between red light and green light is that when you look where the red light is supposed to be, there isn't much in the green light.
So suppose we put labels on the photograph to indicate where the different
wavelengths were supposed to appear. Then a certain kind of light might
give this picture:
(picture from //www.pa.uky.edu/~straley/labaids/spectrum2/specta.gif )
This light contains red and orange and yellow and green wavelengths, but
no blue or violet. So it might be yellow light, since red + green looks yellow
(it might also be orange, if the red was a little brighter than the green).
Question 2 It doesn't matter where and which direction the compact disc
is moved. It will show a reflection. In the diffraction, you get the
same spectrum, but the yellow is a lot brighter and easier to detect.
Joe's comment:
When I look at a compact disk illuminated by one bright light (for example, outside in the sun) I see a band of color or just one color if it is somewhat far away. There is a particular orientation in which I get the direct reflection (white, very bright). Now as I slowly tip or tilt the CD, I come to blue, and then green, and then yellow....orange.... red .........................and again blue, .... red .................... and again blue ..... red. I'll admit that I see these things because I expect to see them, but I really do think they are there. When I look through the diffraction grating, I see the light straight ahead (white, bright) and then off to either side I see blue, and then green, and then...... .... ... . . The CD is a bit like a mirror with a diffraction grating glued to it.
(Light my fire)
3. (Uncertain) A diffraction grating has very closely spaced grooves on it. The color effects you get from a compact disk are
also diffraction, because there are closely spaced rows of little holes, recording the data. Compare the spectrum made by a
diffraction grating to the spectrum made by reflecting light from a compact disk. Compact Disc (CD) vs. Diffraction Grating
(DG) a. CD-Because the grooves go around in a circular manner, you can get the spectrum to fan out. DG-We
only see a "bar shaped" spectrum. b. CD-The spectrum is refracted and then reflected in order for us to see it.
DG-The light is refracted but not reflected...we the spectrum directly.
Joe's comment: Good point!
c. CD-You can't hold it up to light in order to see the spectrum...the light has to be reflected. DG-You have to hold
it up to the light in order to see the spectrum.
Joe's comment: You have emphasized the differences, but there are also important similarities. The diffraction grating shows not just one spectrum (i.e. a sequence ranging from red to violet) but six -- three to each side of the central view. This is a feature of how the diffraction grating works; the extra copies are basically echoes of the first two -- you can think of them as being what happens if you send the spectrum through the diffraction grating a second and third time (it separates the wavelenghts a little more). The CD does this too -- if you slowly turn it you go from blue to red but then come to blue again, and if you tune through the central direct reflection, you will find blue again. This is different from other ways of separating light into its components -- a prism gives just one copy of the spectrum, and a rainbow gives one or sometimes two (the first rainbow goes from blue on the inside of the arc to red on the outside; the second rainbow is farther out and goes from red on the inside to blue on the outside). This tells us that the diffraction grating and the CD are close relatives, while the prism and the rainbow are something else entirely.
(Dark Brigade)
Question number one is the one we are most comfortable with. The spectrum of red light has a single wavelength which reflects only red light. It differs in that it has the least frequency of the three. Therefore, it has the least energy of the three. Blue having the higher frequency, therefore having the greater energy. We feel that blue would be the most reflective due to its position on the spectrum.Red would be the least reflective due to the lesser amount of energy.Therefore, we think you could differentiate between the shades of black and white photos, due to there degrees of reflective energy.
Joe's comment: I'm not sure what the reflections are doing here.
There are many kinds of red light. Some might have a single wavelength (a laser pointer is a good example), while others might contain many different wavelengths (as you will see if you look at a red stop light or car taillights through your diffraction grating). But they have in common that they are bright in the red part of the spectrum and not very bright elsewhere. Blue light is brighter in the blue part.
The point the question was trying to make (perhaps not very well, yet) is that we don't have to perceive color to know what the spectrum is. The blue part of the spectrum is closer to the center and the red part is farther away, and the red part of the spectrum always shows up the same place, no matter whether it is white light, red light, or yellow light that is being studied. Then from the brightness of the different regions of the black and white photograph we can deduce what the spectrum was, and from that, the color our eyes would perceive.
For example, here is what a photograph might look like. I have put labels at the places that the various different wavelengths appear ( image from http://www.pa.uky.edu/~straley/labaids/spectrum2/specta.gif )
This is the spectrum of some kind of light. The picture is supposed to indicate that light having green, yellow, orange, and red wavelengths is present, but no blue or violet. This would probably look yellow to us, because we see red+green as yellow, and we see yellow as yellow. Or it might look orange, if the green were a little dimmer and the red a little brighter.
The part about energy is interesting, and we will get back to this later. What the statement "blue carries more energy than red" is trying to say is that light actually delivers its energy in lumps, and the blue lumps are bigger than the red ones. But when we talk about brightness of light, we actually are talking about the total amount of energy present. This means that to make a red light and a blue light that look equally bright, we have to have more of those red lumps. But since we don't detect the lumpiness with our eyes, we'll never know.
Where the lumpiness becomes important is when we ask whether the light can cause a physical or chemical change, such as causing sunburn or fading the drapes. Here the situation is that the events take place one molecule at a time, and either the energy lump is big enough to make something happen or it is not. So ultraviolet light will give you a sunburn but blue light won't; blue light might fade the drapes but red light won't.
An analogy: you walk up to a drink machine, put in a coin, and push the button. If you put in a red penny, nothing happens (in fact, you lose the penny). If you put in a blue dollar coin, you get a drink and maybe even some change. Money comes in lumps of different sizes, just like light, and the machine is like a molecule that doesn't change unless you give it enough energy. The analogy isn't quite right, because you can put 8 dimes in the machine and get something; the molecule is a defective drink machine that forgets that you have already put coins in it.
Having experience with developing black and white photos, the lighter shades on the enlarger indicate the photo will be too dark and the darker shades indicate the photo will be too light. We think there is a correlation between the two processes.
Joe's comment:
You might have been using a red safelight. It used to be that black and white film didn't see red (not enough energy in those lumps), but they make film better all the time.
Question number two is the one we were least comfortable with. We put the red object on the white card using a strong light source. Looking through the spectroscope we discovered that we had two visible spectrums. The complete spectrum was visible with a white light. The pigment in the red object reflected the red wavelength while absorbing the other wavelengths.
Joe's comment:
That's exactly what I hoped you would see. If a red object turned white light into red light (instead of absorbing the not-red part), a really bright red object might be brighter in the red part than the white light that is illuminating it.
Harder is to look at the spectrum of a yellow object and recognize it.
There is another problem you should be aware of, that will come up if you look at a green stop light -- all the wavelengths are there to some extent, and the eye is very good at ignoring the differences in brightness. So green looks very much like white. That's why we told you to compare them side-by-side.
(Central Pest Control)
> Diffraction Gratings > > Discussion questions > 1. The red light is narrow and all the colors of the spectrum are > visible. The green and blue light spectrum have no yellow mostly blue and > green. If a black and white photograph was taken the absorption spectrum > lines would still be evident. >
Joe's comment:
It depends on which red light you are looking at. If we make red light with the red filter, I find that the blue and green a strongly suppressed; the green does let through some of all wavelengths, but the violet and red ends of the spectrum are faint; the blue filter takes out most of the red. The really interesting question (which gets asked again, later) is whether you can tell what color a light is from its spectrum. It isn't easy, because the eye has a way of ignoring brightness of light -- so most light has the entire spectrum present, and the fact that some parts are very dim gets overlooked and is even hard to see when you know you are looking for this.
> > 2. The red light is seen as red through the spectroscope. It does not > contain the higher wavelengths of visible light. The white light contains > the entire visible spectrum so all colors are seen. When the white and red > object are looked seen at the same time the spectroscope showed just the red > spectrum. This is because the white light does contain the red wavelength. >
Joe's comment:
OK.
(Cordia Lions)
1. What does the spectrum of a very red light look like? How does it differ from the spectrum of a blue light or a green light? Would a black and white photograph of the spectra look different? (Less sure about-- the black and white photograph?) I did each of the light activites in the book with my 5th and 6th Grade Class. We used the frosted light and looked throght the red, blue, and green filters. We found that the red filter removes all other colors except for the red. The blue and the green filters were less effective. Some of the colors were removed, but not all.
Joe's comment:
Unfortunately the blue and green filters are not as good as the red one. Using a double blue and a double green gets rid of more of the "wrong" parts of the spectrum, but make the "right" part somewhat less bright.
In the case of the blue filter, I think the problem is that there isn't that much blue light present in incandescent light, and so a really good blue filter would not make very bright light. So they don't put in as much blue colorant as they might, letting red and green leak through. The eye is very good as seeing dim light (and ignoring the fact that it is dim, too!) so that even a little red and green getting through fills in the spectrum seen through the diffraction grating.
Making a red filter or a blue filter is easier than making a green filter. Think of the spectrum written out as Red Orange Yellow Green Blue Indigo Violet. The red filter absorbs everything to the right of Red, and the blue filter absorbs everything to the left of Blue. But the green filter has to obey two rules -- it absorbs everything to the left and to the right of green, and yet be fairly transparent to green, so that we get a lot of light through. We could make the window wider, and let in some Yellow and some Blue, but the problem would be the same: we have to get two things right. But it turns out that getting even one thing right isn't very easy
I would imagine that a black and white photograph of the spectra would show a difference in the colors, in that colors with higher frequencies would appear daker, while colors with lower frequencies would appear lighter.
Joe's comment:
I had in mind something simple: assume the black-and-white film sees all wavelengths, and records bright lights of any color as white and darkness (of any color!) as dark. Then the spectrum would be recorded as a pattern of shades of gray, and the spectrum of red light would be light gray on one side and darker gray or black over most of the range, while blue light would be light gray on the other side, and so on. We would still be able to tell what color a light was, from the black-and-white picture of its spectrum.
Your answer is more interesting, because you are proposing that the response of the film to different colors is different. This does happen; in particular, the first films that were invented couldn't see red light at all. This was also true of the early kinds of duplication, and people learned to make notes about corrections in light blue, because it was invisible on the copies. Modern film (even black and white) and modern photocopiers are much more equitable, and see all colors much the same. In a way it is too bad, because the physics of why the film and photocopiers could not see red is very interesting.
3. A diffraction grating has very closely spaced grooves on it. The color
effects you get from a compact disk are also diffraction, because there are
closely spaced rows of little holes, recording the data. Compare the
spectrum made by a diffraction grating to the spectrum made by reflecting
light from a compact disk: how are they alike, and how are they different?
(More sure about)
Both the compact disk and the diffraction grating produced a spectrum. In
answering this question, I am unsure about the spectrum of which light
source to compare the spectra of the disk to. In class, we compared the
spectra produced by the florescent lighting in the class room, the spectra
of a frosted lightbulb, the spectra of a clear lightbulb, and the spectra
produced by looking out the window at the natural light produced by the sun.
We found that each type of light source produced a different spectra with
different colors, as well as different lengths for each band of color. We
also looked at the spectra created by a compact disk. We decided as a class
that the light produced by looking through the spectometer at florescent
light is most like the spectra produced by the compact disk. Both of these
sources displayed the colors of red, orange, yellow, green, blue, and violet
and also showed bands of each colors msot similar.
Joe's comment:
OK. There wasn't a right answer here -- I just wanted you to look at them.
Here is something to think about. There are many ways to make colors, and several ways to turn white light into a spectrum. The rainbow made by actual rain differs from the spectrum seen through the diffraction grating in that rain makes mainly one colored arc, while a diffraction grating has several -- if you look at a white (W) object, you see spectra RYGB like this:
R Y G B R Y G B RYGB W BGYR B G Y R B G Y R
There sometimes is a second rainbow, but its colors are in the reverse order: starting from the inside of the bow, the colors go
BGYR RYGB where the second set is much more dim than the first. So the rainbow is not a diffraction grating. Now we can ask, does the compact disk resemble the diffraction grating or the rainbow? I think you will find that if you slowly tilt the CD more and more, you will go through the spectrum ... and then again ... and then again, and each spectrum is in the same order (unless you go through the mirror reflection place, which is the W at the center).
Guest reviewer
> 1. Part of this queston seemed hardest. > The red light causes the spectrum to look all red. The red takes away all of > the other colors but the red. With the blue and the green all of the colors > are there only slightly differetly. For instance with the blue light the > yellow seems very narrow and the red seems not as vibrant. The hard part is > what would happen with a black and white photo of the spectra? Would it look > different? I do not know. I imagine you could see shade variations with > the black and white photo of the blue and green but the red may look like > all the same color which it is. >
Joe's comment:
The blue and green color filters are not as good as the red one, and let some light of the wrong colors leak through. Comparing no filter, one blue filter and a double blue filter will reveal this best. Then the eye is a pretty bad brightness detector -- it is designed to ignore variations in brightness, so that tigers in dappled sunlight will be readily recognized. So the spectra look much the same, even though the brightness of the red and green light making it through the blue filter is really greatly decreased. But if you compare them side by side (by covering only the top half of the diffraction grating, for example) the difference is more visible.
The black and white photograph will record the brightness and dimness. So the photograph of the spectrum of a white light will be a gray rectangle showing all the directions in which you see light of any color; the spectrum of a red light will be a gray rectangle that is smaller -- it covers only the outside third of the white light rectangle. The photograph of the blue light will be a full=length rectangle that is darker gray on the outside (red) end. (I guess I'll have to figure out how to take these pictures).
> > 3. This one seemed easiest. > On the CD the colors seemed more brilliant than the colors on the > diffraction grating. They seemed to be the same colors in the same order on > both the CD and the DG. On the CD, if you held the CD just right, the > colors went around in a circle as opposed to the difraction grating where > the colors went in a line. The yellow band on the CD is very narrow. I could > also make the spectrum appear as a more of a straight line on the CD and > also make a certain color very wide. sometimes the CD spectrum rminded me > more of the spectrum formed on a bubble. >
Joe's comment: This is probably not a very good question, since there is no particular point to be made. But your description is very good; the basic point is that the colors appear in the same order (including the phenomenon that there is a white reflection in the middle, with spectra to either side of it, and you go through the rainbow several times as you keep tilting the CD more and more), suggesting that this (and the bubble spectrum) has the same origin as the diffraction grating spectrum. (In contrast, a prism spectrum is different: there is only one copy of the spectrum, no white version, and the color that is deflected the most is blue instead of red).