everyday-lifeheatreflectionvisible-light
Aluminum foil is said to be not absorbing light at all.
It reflects light. So, does it mean that a more shiny aluminum foil will reflect more light and thus make the room more cooler as compared to less shiny foil?
If yes, then where does unreflected light go in case of less shiny aluminum foil?
Does less shiny and more shiny actually make a difference? Why?
In order to keep whole room cool will it help if I stick the foil to the outside of wooden door (exposed to the sunlight) too?
Wrong:
"since infrared waves have a shorter wavelength"
Infrared has longer wavelength than visible and visible longer wavelength than ultraviolet .
White is a term for visible light mixed wavelengths. In the plot you can see that almost half of the sun's radiated energy arrives as visible light. The white buildings reflect this visible light which otherwise,impinging on the surfaces would be absorbed and turned into infrared by the interactions, adding to the arriving infrared.
What is absorbed and what is reflected depends on the chemical bonds of the surfaces, whether the incoming radiation can excite molecular states of the materials. Infrared is in frequencies/wavelengths of the black body radiation of bodies in the temperature ranges comfortable for the human body, so they easily raise the vibrational and rotational levels of solids and liquids and the kinetic energy of gasses.
37C curve seen here practically all in infrared, and lower temperatures more so.
Thus white paint will not reflect infrared as efficiently as visible, a large part of infrared will be absorbed as also some part of visible will scatter at the surface and degrade to infrared. Infrared can be reflected by metal mirrors, from the collective fields in metals . If you put aluminum foil in front of a heater you are sheltered from most of the heat which is reflected, but some of it is absorbed as can be seen by touching the foil.
If I expose an object to EM radiation only from the infrared spectrum, will it only reflect back infrared?
Yes, but most of it will be absorbed ( except by mirror metal surfaces) because the materials have the receptors for these wavelengths. This is due to the fact that larger wavelengths have photons with less energy which cannot excite higher energy levels.The energy of the photons goes as h*c/lamda where h is plancks constant, lamda the wavelength and c the velocity of light.
Is this true for other types of EM radiation?
No.Visible and ultraviolet by scatterings degrade their energy down to infrared frequencies, depending on the material.
Is it possible to make an object that looks white and absorbs a lot of infrared radiation?
Usually most of the infrared will be absorbed except by mirror metal surfaces.
If an object reflects most of the EM radiation that it receives of a particular wavelength λ, will it also reflect most of the radiation it receives of wavelengths less than λ (and absorb most of the radiation of wavelengths larger than λ)? Is this why objects that reflect most visible light (and hence look white) also reflect most infrared radiation (since infrared waves have a shorter wavelength)?
There is no such rule. It depends on the material and its chemical bonds.
Your inability to see the dust until you narrow the slit has nothing to do with the narrowness of the beam but instead the dynamic range of light that your eye can see at one time.
A bit of searching turns up reports of a contrast ratio for you eye at one time as between 100:1 and 1000:1. This means if you're in a room with a range of brightness greater than about 100 to 1 the brightest things will all be washed out as white and the darkest things will all be essentially black. This is obvious in photos that are "backlit" like this one:
These horses aren't black but because the ratio of the bright light to the dark horses exceeds the dynamic range of the camera the sky is washed out white and the horses are in silhouette.
Your eye can adjust over time to a huge range but it can't utilize the whole range all at once.
In the case of dust reflecting light, if you allow a lot of light into the room the relative brightness between the small amount of light the dust is reflecting and the rest of the illuminated room prevent you from seeing the dust.
This is fundamental to signal processing. Why can't you hear a whisper in a noisy room? The noise of the crowd obscures the whisper. The difference between the signal you're trying to pick up and the background noise is called the signal-to-noise ratio. In the case of dust, the light let into the room is scattered and reflected in the room and causes the room to be illuminated. This is the noise that obscures the signal from light reflected off of the dust.
Best Answer
First, there's no perfect reflector nor absorber. In fact - even Aluminium does absorb some radiation (by which it gets heated, can be noticed at incident high frequency radiation). One more thing is that aluminium foils are designed in a way to reflect light.
Here's the Wiki article quote...
The shiny and not-shiny surfaces are totally a favor of production technology (credit goes to the rollers). Now, to the "why" question.
As a physics parameter, we use reflectivity to address the shininess. As we can see, the reflectivity is quite high for the bright surface, compared to dull one. The unreflected light (as you say) can go anywhere. It can go inside the aluminium foil (i.e) it's absorbed and hence the 12% & 20% loss...
Response to comment (based on edit): That's a nice
strangeidea. With some perfection (I mean, there should be very less allowance of any sorta radiation inside), the room will be relatively cooler. But, in reality (where we can't expect idealistic things), there will always be some radiation inside. But, it keeps the room warmer compared to the outside. But, always be careful when playing with such things because, any sort of harmful radiation (if any- what about a heater or even an electric iron?) inside the room will be reflected back to you by the matte side - which can be very harmful...