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Colour phenomena: the role of absorptionMUSE Workshop

WCPE Istanbul, July 1-6, 2012

Published by the MUSE group (More Understanding with Simple Experiments) , Physics Education Division (PED) of the European Physical Society (EPS) http://education.epsdivisions.org/muse/

Activity 11

Recalling the classical rules of colour phenomena

Consider the set-up with a ping-pong ball, also shown in Figure 1.

Figure 1. The “colour mixer” used in this workshop (designed by Gorazd Planinsic 2004, see Appendix)

Explore the effect of switching on separately each switch:How would you explain what you see,-with switch 1 on?-with switch 2 on?-with switch 3 on?It is up to you how you choose switch numbers but once you choose them, make sure you are using them consistently throughout this activity.Explain precisely the role of the ping-pong ball. In particular, explain why ping-pong ball appears to be white in day light. Explore the effect of switching on two switches on at the time:How would you explain what you see,-with switch 1 and 2 on?-with switch 2 and 3 on?-with switch 3 and 1 on?Explain precisely the role of the ping pong ball.Switch on all the three switches:

1 The MUSE group (G. Planinsic, E. Sassi, L. Viennot) takes responsibility for the content of this paper (July 2012). The intellectual property remains with the author Laurence Viennot.

http://education.epsdivisions.org/muse/



Explain what are the similarities and differences with respect to the case when the ping pong ball (with all LEDs switches off) is simply seen in day light.Consider the first sheet of rules (I) (handed out to the participants at this moment).

Write on the headline of the sheet the type of colour mixing that is involved in this experiment.

Colour Mixing – rules, sheet I

.................... colour mixing

Separating the various radiations that constitute "white" light gives a "spectrum".The spectrum of visible white light ranges from =400 nm to =700 nm.(wavelength; measured in a vacuum; 1 nm = 10-9 m).The spectrum is here schematically divided into three thirds.

Coloured lights with a spectrum corresponding to a third of the spectrum of white light in long wavelengths appears red

in intermediate wavelengths appears green

in short wavelengths appears blue

Combining these three lights, in various proportions, results in a large range of colours and can also give white light. Combining two of these lights in the correct proportion respectively gives a light that is

- yellow, if green light and red light are added

- cyan, if green light and blue light are added

- magenta, if red light and blue light are added



Activity 2

Working with pigments

Pigments absorb some parts of the spectrum of white light.

Consider the second set of rules (II) (handed out to the participants at this moment) shown in Figure 2.

Figure 2. The classical rules of the subtractive role of pigments, also valid with filters



Figure 3. Coloured letters to be illuminated by different coloured lights in a dark room (Chauvet 1999).

Imagine we place the image shown in Figure 3 (the letters printed with different pigments: red, green, blue yellow, magenta and cyan) at the rear inside the dark cardboard box.Predict and explain what you will see, in the absence of ambient light (your face being placed at the entrance of the box), if the letters are illuminated -by a red light (from the ping-pong mixer situated inside the box);

-by a green light (from the ping-pong mixer situated inside the box);

-by a blue light (from the ping-pong mixer situated inside the box).

Perform this experiment and discuss its outcomes. In particular:-check what (which letters) you can see with more or less ambient light (sticking more or less the cardboard “doors” of the box again your cheeks), and no other lighting;

-investigate what you can see when (resp.) red, green, blue, all lights are illuminating the letters.



Activity 3

Working with a laser pointer

Place the sheet with colored rectangles (Figure 4) inside the dark box, at the rear.

Figure 4. Sheet of paper with white, black, and differently coloured (red, green, blue, yellow, cyan and magenta) areas

Predict, explain, discuss what you will see if each of the different zones of this sheet is successively illuminated by a red laser pointer (633nm).

Impact zone

Predict what will you see when the zone is illuminated by the laser pointer?

Explain your reasoning

White

Black

Green

Blue

Red

Yellow

Magenta

Cyan



Performing the experiment

From the entrance of the dark box (not strictly closed), illuminate each zone of the sheet with the laser pointer.Describe what you see. Comment and discuss the outcome.

Impact zone

Describe what you observed when the zone was illuminated by the laser pointer

Based on observation try to explain the outcome of the experiment

White

Black

Green

Blue

Red

Yellow

Magenta

Cyan

Based on observation can you think of a general rule which applies to all cases?



Activity 4

Working with the reflection curves

Figure 5. Reflection curves of the blue (left), green (middle) and red (right) pigments used in this experiment. The reflection factors for a white zone of the sheet (curve in black) and a black zone of the sheet (curve in light grey) are also shown. A 633 nm laser beam (red vertical line) is situated with respect to the scale of wavelengths. Credit: G. Planinsic.



Figure 6. Reflection curves of cyan (left), yellow (middle) and magenta (right) pigments used in this experiment. The reflection factors for a white zone of the sheet (curve in black) and a black zone of the sheet (curve in light grey) are also shown. A 633 nm laser beam (red vertical line) is situated with respect to the scale of wavelengths. Credit: G. Planinsic.

Consider the preceding experiment with a laser beam.What can be concluded from the drawing of a vertical line at the position = 633nm?

Discuss the role of the intensity of incident light in the previous experiments. Hint: look at the graphs shown in figure 5 and 6.

Consider the different outcomes of experiments with, respectively, the ping-pong ball and the laser beam. Try to explain the differences between the two cases



Why this experiment?This experiment has been designed on the basis of a content analysis, with special attention to what contradicts some teaching rituals and/or some well-know common ideas about colour phenomena. The first item below is more of the “ritual” type, the second refers mainly to common ideas.-“All or nothing”

Concerning light and colour, some prototypical statements contradict what can daily be observed by students. “Black objects do not reflect any light” is one of these inadequate statements that can be found in some teaching materials 2. In fact, the impact of a red laser beam remains visible whatever the colour of the surface on which the light is sent might be. Such ritual comments are based on an oversimplified view of the real world involving an ‘all or nothing’ reasoning process, whereas light should be considered in a quantitative way.

In the same way, the rules of additive mixing are classically stated in an “all or nothing” style. They provide for instance the outcome of two beams of coloured lights superimposed on a white screen (e.g. “with red plus green, you get yellow”) correspondingly, the outcome of sending a coloured light beam on a filter or a pigment may be given by a rule of subtractive mixing such as “with red plus yellow, you get red”. But such a reduced stating of the rules does not permit a proper interpretation of what students can observe in daily life. This is not only for practical reasons (for instance an omnipresent ambient light) but also because of the need to consider both the composition of the light sent and its intensity.

The experiment presented in this workshop is intended to help students pass from an ‘all or nothing’ way of reasoning to another involving ‘more or less’ terms, in order to make the visibility of the impact of a laser beam on any coloured surface intelligible. So doing, we intend to facilitate the comprehension of absorption in the context of colour phenomena, a concept that has proved to be very difficult (Chauvet, 2006). -The very status of colour.

Colour is often understood as being an intrinsic property of an object: you would see it - say red - because it is red. In the same logic, the rules of color mixing can be often misunderstood for the outcome of a mixing of paints.

Depending on the context, it may be useful to underline a very basic aspect: colour is a perceptive reaction of the visual system to a light entering the eye. It is therefore essential that, before any other concern, (that is: in the very first part of the workshop, attendees reach a sound comprehension of the role of the ping-pong ball or other diffusing agents (pigments). When illuminated in white light (day light), the ball is seen white because its envelope ensures a diffuse transmission of the whole “visible” part of the spectrum of this light. When illuminated with a coloured light of one (or two) third(s) of spectrum, the envelope diffusively transmits a light of approximately same composition, therefore it is seen of same colour.

By contrast, a coloured filter or a pigment transmits (reflects) a light the composition of which may be different from that of the incident light. The action of such an object on the incident light is summed up by spectral absorption/reflection curves. These curves indicate, for each part of the spectrum, a factor (<1) by which the power of the incident light is multiplied: they have a multiplicative meaning. When it comes to interpreting the reflection or transmission curves, and the link with the situation proposed (laser beam), it is often observed that students have much difficulty in accepting the following statement: a millionth of a big quantity can be more than a tenth of a small one (Viennot & de Hosson 2012) .

When two filters are superimposed, or when two pigments are blended, two multiplicative curves intervene in transforming the incident light. This means to focus on the multiplicative aspect of what is somewhat improperly named subtractive synthesis.

References

Chauvet, F.1999. STTIS Project, Colour sequence University " Denis Diderot ", LDAR (Laboratoire de didactique André Revuz); and STTIS (Science Teacher Training in an Information Society) web sites: (retrieved 1.7.2010): http://www.lar.univ-paris-diderot.fr/sttis_p7/color_sequence/page_mere.htm or http://crecim.uab.cat/websttis/index.htmlPlaninsic, G. 2004. Color Light mixer for every student, The Physics Teacher, 42,138-142Planinsic , G. & Viennot, L. 2010. Shadows: stories of light. http://education.epsdivisions.org/muse/example-shadows-documents/SHADOWS_stories_of_light.pdfViennot, L. & de Hosson, C. 2012. Beyond a Dichotomic Approach, The Case of Colour Phenomena, International Journal of Physics Education, 34:9, 1315-1336.

2 http://maths-sciences.fr/documents/quatrieme/sources-de-lumiere-4eme.pdf, http://www.sciences92.ac-versailles.fr/spip/spip.php?article27. Links verified on December 8th of 2011.

http://crecim.uab.cat/websttis/index.html

http://www.lar.univ-paris-diderot.fr/sttis_p7/color_sequence/page_mere.htm

http://www.lar.univ-paris-diderot.fr/sttis_p7/color_sequence/page_mere.htm

http://www.sciences92.ac-versailles.fr/spip/spip.php?article27

http://www.sciences92.ac-versailles.fr/spip/spip.php?article27

http://maths-sciences.fr/documents/quatrieme/sources-de-lumiere-4eme.pdf



AppendixDo it yourself: how to make a LEDs colour mixer Schema of the electronic circuit of the LED colour mixer is shown in Fig. 4. The values of the fixed and variable resistors may vary depending on the voltage of the battery and type of the LEDs. The value of the resistors should be chosen so that the current through each LED does not exceed the maximal forward current as specified by the LED manufacturer (normally 20mA) at the lowest setting of the variable resistor.

Figure 7: Schema of the electronic circuit of the LED colour mixer

LEDs in our colour mixers are taken from a flexible LED strip that utilizes square LEDs without lens. In case you cannot obtain such LEDs you can make yourself an equally good point-like light source by modifying conventional LEDs as described below.

Most commercially available LEDs have an epoxy drop lens above the PN junction — the light-emitting element. Unlike point light source such LEDs emit light in a limited cone. In addition, imperfections in lens design cause the angular distribution of light intensity to be non-uniform. With a simple modification the LED can be converted into a light source that is a good approximation of a point light source. Start with LEDs that are encased in clear plastic. Using a hacksaw carefully saw off the part of the LED body that makes the lens, as shown in the figure below.

Figure 8: Transforming conventional LED into a point light source: (1-2) saw off part of the LED that makes the lens, (3) brush the sawed surface with fine sandpaper and (4) finally with a white toothpaste.

Make sure that you do not cut too close to the PN junction (1 mm away is OK). Remove scratches in sawed surface by brushing and polishing. Use fine sandpaper starting from lower grades (such as grade 600) and finishing with higher grades (grade 1200). After this you have to polish the surface with an abrasive polishing paste (a white toothpaste also works well) until the surface looks clearly transparent.

Reference: Planinsic, 2004. Color light mixer for every student. Physics Education, 42, 138-142.Planinsic, G. & Viennot, L. 2010. Shadows: stories of light http://education.epsdivisions.org/muse/example-shadows-documents/SHADOWS_stories_of_light.pdf/view

http://education.epsdivisions.org/muse/example-shadows-documents/SHADOWS_stories_of_light.pdf/view



Activity 1’: for students

Recalling the classical rules of colour phenomena

Consider the set-up with a ping-pong ball, also shown in Figure 1.

Figure 1. The “colour mixer” used in this workshop (designed by Gorazd Planinsic 2004, see Appendix)

How would you explain that you see the ping-pong ball white? Explain.The ping pong ball diffusively reflects the white light it receives from outside, without notably transforming the spectral composition of this light. The observer’s eye receives a light of same composition as day light.

Would the ping-pong ball be seen white when illuminated with any light (including no light at all)? Explain.No, the light it reflects depends on the received light. Lights of a given colour will be reflected without spectral transformation. Therefore, the ball will reflect a light of same colour as the incident light. Illuminated in red, it will be seen red. In a totally dark room, with no incident light, the ball will be invisible.

Explore the effect of switching on separately each switch:How would you explain the appearance of the ping-pong ball with a given (single) switch on?When the incident light is from inside (a LED), there is diffuse transmission. The reasoning is the same as in the preceding item, changing “reflects” for “transmits”.

Explain precisely the role of the ping-pong ball. See above

When we see the envelope of the ball, do we receive some light from it? What can we say of the spectrum of this light?Yes, when we see the envelope of the ball, we receive some light from it. This light has the same spectral composition as the light received by this envelope. The observer’s eye receives a light of same composition as that emitted by the LED



In particular, explain the connection between your argument and the fact that this ping-pong ball is white in day light. The envelope of the ball can diffusively transmit or reflect any spectral band inside that of white light and, in day light, our eye receives a light of spectral composition close to white light.Same questions with two switches on at a time-with switch 1 and 2 on;-with switch 2 and 3 on;-with switch 3 and 1 on.

Concerning the light received by the observer’s eye, all the answers above hold.Concerning the visual impression this light generates, the functioning of visual system involves three types of receptors, or “cones”, respectively sensitive to red, blue and green. Nervous connections and further processing by the brain result in the rules summed up in sheet I (below).

Switch on the three switches:Three thirds of spectrum of white light make a spectrum of white light.

What are the similarities/possible differences with respect to the case when the ping-pong ball is simply seen in day light?It might be that the three lamps do not reproduce exactly the composition of day light. In fact, the visual impression of white is compatible with spectra of different compositions, provided these spectra cover a wide part of the spectral interval of day light

Consider the set of rules in sheet (I) Ask any question you feel necessary to understand the link between these rules and what you have just seen.

Write on the headline of sheet I the type of colour mixing that is involved in this experiment. Explain.Additive colour mixing of lights. Each lamp emits about a third of the spectrum of white light.



Colour Mixing – rules, sheet I

.................... colour mixing

Separating the various radiations that constitute "white" light gives a "spectrum".The spectrum of visible white light ranges from =400 nm to =700 nm.(wavelength; measured in a vacuum; 1 nm = 10-9 m).The spectrum is here schematically divided into three thirds.Coloured lights with a spectrum corresponding to a third of the spectrum of white light in long wavelengths appears red

in intermediate wavelengths appears green

in short wavelengths appears blue

Combining these three lights, in various proportions, results in a large range of colours and can also give white light. Combining two of these lights in the correct proportion respectively gives a light that is

- yellow, if green light and red light are added

- cyan, if green light and blue light are added

- magenta, if red light and blue light are added

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