Exploring Visible Spectrum of Light: Seeking Advice

In summary, the size that each color has within the spectrum corresponds to there intensity. The largest band (red) therefore means that red light is the best for absorbing light.
  • #1
barooonscape
6
0
im doing an extended essay at school and i am loooking for help to adjust/refine my ideas.

basically I am investigating properties of the visible spectrum of light.
one idea is that the size that each colour has within the spectrum green, large, yellow, small
corresponds to there intensity?\

any help?
 
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  • #2
I assume that by "intensity" you mean that we perceive some colours as brighter than others, even if they have the same intensity.
What do you mean by the "size" that each colour has? How are you going to make this rigorous?
 
  • #3
The problem of color is not so much a physics problem, but a biology problem. It would probably be useful to look up how the eye senses color and the brain processes color. Here's a link to start you off:

http://en.wikipedia.org/wiki/Color#Color_in_the_eye
 
  • #4
primarily i was investigating the rate of absorbition of heat through different colours.
and i found that the closer the colours are to the infa red side of the spectrum(red orange yellow) the more heat absorbes, and i deduced this is becasue infa red is in fact heat anf therefore colours nearer to red get absorb more

however this was a bit bog standard for my essay.
therefore i looked at the visible spectrum and i saw that certain bands of colour wher larger. see :http://www.gamonline.com/catalog/colortheory/spectrum.gif

i was wondering is there any follow ups to the first part of my essay that would be interesting and get the markers attention. for example does the largest band (red) therefore mean that red light is the best for absorbing light. honestly I am a bit lost.

and ty for the quick replies

my nexyt
 
  • #5
Intensity is related to the number of photons (light particles) hitting your eye in a given amount of time per unit area and is not related to the range of wavelengths for a color. To see this simply note that the intensity for even a single color can vary greatly (for example, a "bright red shirt" vs. a "dull red shirt" are both red).

The range of wavelengths the eye perceives as being a certain color is determined by the properties of the cone cells for that color in the eye. This is as far as I can see a Biology question, and I would search for things such as rhodopsin, cone cells, etc. to get my material if I were interested in this topic.
 
  • #6
The widths of the colors bands in the spectrum are determined by the eye. There are no physical bands in the spectrum of the white light. It is a continuous spectrum, with no bands. The eye receptor (and the brain) associates some "colors" (or responses) to various wavelengths. It has to do with spectral sensitivities of the cell receptors in the retina, I guess.

As for pressure of light, this depends on both intensity and wavelength. Shorter wavelengths carries more momentum (p=h/lambda) per photon so for same number of photons will "put more pressure". But I don't think his question was about light pressure.
 
  • #7
While perception by the human eye of a particular color is part of the equation, in essence, it’s still the wavelength of light that actually determines the color we will perceive. For instance, wavelengths that are detected as green by the human eye are detected as green for ALL human eyes (except those whom may be colorblind or have other color disorders).

The amplitude of the wavelength determines intensity of the light energy. The greater the amplitude of the wavelength, the greater will be the photon emission therefore a perception of greater brilliance.

Visible light ranges from approximately 400,000,000 MHz to 700,000,000 MHz (approx 750 nm to 428 nm wavelengths respectively).

Wavelength is derived as follows:

c / f = w

c (constant of light-speed, 299,792,458 m/s) / frequency (in Hz) = wavelength (in meters)

therefore,

299,792,458 m/s / 400,000,000,000,000 Hz = .000000749481 meters (essentially 750 nano-meters)

The visible light spectrum (as split in ascending frequency order by a prism) begins with Red then continues with Orange, Yellow, Green, Blue, and Idigo (the highest viewable light frequency).

Infrared (invisible heat energy) is below visible red frequencies and ultraviolet (invisible energy responsible for causing sunburn) is above indigo.

I hope you found this helpful, barooonscape.
 
  • #8
Thank you Gnosis/barooonscape.

You told me a lot. That most EM spectrums are backward, red should be on the left because it has the lower 400,000,000 MHz and indigo the higher at 700,000,000 MHz (waves per second). The longer wavelength, lower frequency regions are located on the far left of the spectrum and the shorter wavelength, higher frequency regions are on the far right.

Second, yellow, the brightest color is less in proton emission to keep the EM scale ascending evenly. At equal brilliance yellow would have the highest proton emission and exert the most pressure on the Earth or any object. Adjusting the yellow to a high intensity would cause it to have the highest proton emission of any color, right?

Because magenta is not seen in an EM chart but is higher on the chart approaching X-rays with shorter wavelengths and higher frequencies it would have an even higher pressure and proton emission with more inertia and momentum. Combining the physically unseen magenta wavelength and yellow make red, could red have most radiation pressure on a reflecting surface? Or would it be green, combining the visible blue and visible yellow light?

I hope my post isn't considered obnoxious and removed by anyone before I get help.
 
  • #9
Don Jusko said:
Thank you Gnosis/barooonscape.

You told me a lot. That most EM spectrums are backward, red should be on the left because it has the lower 400,000,000 MHz and indigo the higher at 700,000,000 MHz (waves per second). The longer wavelength, lower frequency regions are located on the far left of the spectrum and the shorter wavelength, higher frequency regions are on the far right...

Yes, the higher frequency with its associated shorter wavelength makes it seem sort of backwards, and for that reason, it is often a source of confusion/simple error. It’s merely the nature of frequency verses its "cycle duration time". As frequencies increase, their cycle durations must become shorter in time duration in order to produce more cycles per second hence, a higher frequency. In turn, this causes the frequency’s wavelength to become shorter in its physical length (typically measured in meters).

Here’s another angle to view this topic to perhaps shed even greater comprehension…

Let’s examine a 60 Hz frequency. Its cycle time is .016666 seconds. This is derived by:

1 second / 60 Hz = .0166666 second cycle time

Now ask yourself the following question:

“Given .016666 seconds of travel time (the time required to complete a single cycle at 60 Hz), how far could light travel at its light-speed constant of 299,792,458 m/s?”

Answer: 299,792,458 m/s x .0166666 second cycle time = 4,996,341.105 meters

Therefore, the wavelength of a 60 Hz frequency is incredibly long at 4,996,341.105 meters, just under 5 million meters long. Holy Toledo!

Let’s continue this progression…

Now assume the frequency is increased significantly to 1 MHz (therefore, 1,000,000 Hz). A full cycle of a 1 MHz frequency is .000001 seconds and derived by:

1 second / 1,000,000 Hz = .000001 second cycle time

Now ask yourself the same question using the cycle time for 1 MHz:

“Given .000001 seconds of travel time (the time required to complete a single cycle at 1 MHz), how far could light travel at its light-speed constant of 299,792,458 m/s?”

Answer: 299,792,458 m/s x .000001 second cycle time = 299.792458 meters

Now evident by the comparison, as frequency increases, wavelength becomes shorter, as it must. Naturally, this same process continues as frequencies continue to increase.

Now let’s assume a light frequency of 400,000,000,000,000 Hz (roughly the visible red light range). A single full cycle of this ultra high frequency is a mere .0000000000000025 seconds!

1 second / 400,000,000,000,000 Hz = .0000000000000025 second cycle time

Now ask yourself the same question using the cycle time for 400,000,000 MHz:

“Given a mere .0000000000000025 seconds of travel time (the time required to complete a single cycle at 400,000,000 MHz), how far could light travel at its light-speed constant of 299,792,458 m/s?”

With such an incredibly short cycle duration time, light won’t be given a chance to travel very far therefore, the wavelength of this frequency is going to be quite short.

Answer: 299,792,458 m/s x .0000000000000025 second cycle time = .000000749 meters (749 nm)

Summarizing:

The shorter the “cycle duration time” of a frequency, the shorter will be the time afforded for travel at light-speed, which reduces the distance traversed per that cycle duration time thereby shortening the wavelength associated with the frequency.
 
  • #10
Thank you Gnosis, you are very clear in your explanation. The hight and depth of one wave changes by frequency but not the speed at which it travels.

If a light emerges from a single point scattering 360 degrees x 360 degrees in all directions is it diminished by the distance between each originating wave as the distance increases? If the number of waves emitted from that single point is finite. Say the size of the emitting point is 3-D, 1x1x1, how has the density of the waves decreased with distance away from the starting point? Does a one sided 1x1 emitting surface have a given number of starting waves? If all the emitting waves are focused into a parallel emission equal the size of the original 1x1 source, at what rate does the light diminish over distance in a vacuum?

Assuming a 3-D light can be focused to one degree, how would the intensity of that degree be expressed?

What frequency wavelength exerts the most pressure? I want to get this beam of light down to one powerful frequency color that can move objects.
 
  • #11
Hi
Be careful: You are using the word 'color' (colour where I come from!) incorrectly. Light is not coloured.
* Light is a wave in the electromagnetic field
* Colour is a sensation that light elicits in human beings
* Many different sets of electromagnetic stimulations can induce the same colour sensation (a phenomenon known as metamerism)

So: With that clarification in mind: good luck with your investigations

M
 
  • #12
Hi dylanesque,
Oh, that was good, thanks. What would be your thought on the most pressure giving wave in the EM field?
 
  • #13
Don Jusko said:
Hi dylanesque,
Oh, that was good, thanks. What would be your thought on the most pressure giving wave in the EM field?

Given that the visible spectrum of light frequencies are essentially in a fairly narrow range of frequencies, per identical “instrument measured” amplitudes (not “perceived” amplitudes due to the eye’s greater sensitivity to yellowish-green wavelengths), I’m not given cause to presume that one light frequency would move an object better than another light frequency. However, since I haven’t specifically tested this process, I’m not speaking with a conclusive knowledge concerning this aspect.

I will say however; just as the frequencies immediately below visible red cause a distinctive heating effect and just as the frequencies immediately above the indigo range causes sunburn, there could potentially exist a complimentary physics component that would allow one frequency of light to move an object better than another. Here’s where tests would need to be conducted and measurements taken and compared. In any case, it would likely require a great deal of energy to move an object of any significance.
 
  • #14
H Gnosis,

Thank you. Perhaps I'm on the wrong tract. I know light's pressure can be measured, that's what got me started, but I might be looking at the wrong part of the spectrum. Maybe the invisible magenta and Ultraviolet wavelengths can move an object with less energy required then the Electromagnetic visible light.
 
  • #15
Don Jusko said:
I know light's pressure can be measured, that's what got me started, but I might be looking at the wrong part of the spectrum. Maybe the invisible magenta and Ultraviolet wavelengths can move an object with less energy required then the Electromagnetic visible light.
It's difficult to understand what you are getting at with your questions about light pressure. Realize that the pressure exerted by a beam of light depends on its intensity, not its frequency. (The momentum per photon depends on frequency, of course.) A beam of red light of great intensity will exert more pressure upon absorption than a beam of blue light of weaker intensity.
 
  • #16
Yup. It is a property of light that energy and momentum are proportional: pc = E.
That's not true for massive things: E = p^2/2m in classical physics, and E^2 = m^2 c^4 + p^2 c^2 in SR.

But for light, there' s proportionality. That means that if you want to transfer a certain amount of momentum with light, that you will have to spend a certain amount of energy, independently of whether this momentum comes into small bunches or big bunches.

With particles, this is not true, for instance. If in Newtonian physics, you want to transfer a lot of momentum with not much energy, then you have all interest in taking MANY SMALL bunches (many slow bullets instead of one fast one). Also, you have all interest in taking MASSIVE bullets. So MANY, SLOW, MASSIVE bullets will transfer a certain amount of momentum with much less energy than one single fast and light bullet.

But for light, the ratio is constant. It is c.
 
  • #17
Hi Doc Al, thank you, I understand what you have said.

I understand that reflection receives twice the pressure then absorption.
"Maybe magenta and Ultraviolet can move an object with less energy required then visible light"
That clear glass will filter out UV rays so this test is not impossible using sun light.
That an object can be lifted and be moved by a tight laser beam.

Hi Vanesh,
I couldn't understand what you were saying. That energy and momentum are proportional. It sounds simple but I can't put it all together, I've lost the concept.
 
  • #18
barooonscape said:
im doing an extended essay at school and i am loooking for help to adjust/refine my ideas.

basically I am investigating properties of the visible spectrum of light.
one idea is that the size that each colour has within the spectrum green, large, yellow, small
corresponds to there intensity?\

any help?

no, you discovered the hot water bro wow.
 
  • #19
Don Jusko said:
Hi Vanesh,
I couldn't understand what you were saying. That energy and momentum are proportional. It sounds simple but I can't put it all together, I've lost the concept.

Hi Vanesh,
Ok, I think I have it.
The higher frequency has more energy and momentum of protons.
More of a lower frequency with a longer wavelength uses less energy and would have the same momentum.
A lot of Alfa waves can move matter more easily then the same amount of energy used in ultraviolet waves.
 

1. What is the visible spectrum of light?

The visible spectrum of light is the portion of the electromagnetic spectrum that is visible to the human eye. It ranges from approximately 400 to 700 nanometers and includes all the colors of the rainbow: red, orange, yellow, green, blue, and violet.

2. How can I explore the visible spectrum of light?

There are several ways to explore the visible spectrum of light. One way is to use a prism to separate white light into its component colors. Another way is to use a spectroscope, which uses a diffraction grating to separate light into its different wavelengths. You can also use filters, such as colored glasses or gels, to isolate specific colors of light.

3. Why is it important to study the visible spectrum of light?

Studying the visible spectrum of light can help us understand the properties of light and how it interacts with matter. It also has practical applications in fields such as optics, photography, and color theory. Additionally, studying the visible spectrum can lead to a deeper understanding of the natural world and the phenomena that occur within it.

4. What are some examples of objects that emit visible light?

Some examples of objects that emit visible light include the sun, stars, light bulbs, candles, and fire. Any object that is hot enough to emit thermal radiation will also emit visible light. This includes objects like lava, hot metal, and the human body.

5. Can we see all colors within the visible spectrum?

No, we cannot see all the colors within the visible spectrum. Our eyes are only sensitive to a narrow range of wavelengths, so we are limited to seeing the colors within that range. Additionally, some animals can see colors that are outside of our visible spectrum, such as ultraviolet and infrared light.

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