Why are umbrellas specifically in black? Of course we do have colored ones, but black is the majority. Is there any scientific reason behind it?
[Physics] Why are umbrellas black
radiationthermal-radiationvisible-light
Related Solutions
This is similar to the reasons one way mirrors work.
If you look through a black net then no light is reflected from the net so the eye sees only the light coming from the objects on the far side of the net. The amount of the external light that reaches you is reduced, but the brain is pretty good at reconstructing images from only partial data, so the view looks unchanged.
If you look through a white net then the eye receives a mixture of the light reflected from the net and the light from outside transmitted through it. If the room you are in is dark and the outside is bright, then the amount of light reflected from the net is small compared to the transmitted light and you still don't see the net. However if the room is light and the outside dark then the light reflected from the net swamps the light transmitted and you only see the net. In between you'll see both the net and the view.
A black body need not be of black color. Black body means a body which ABSORBS all wavelengths completely.
This gives rise to two conclusions:
- A black body doesn't reflect any light. A mirror can't be a black body because when you shine a light beam on a mirror, it doesn't absorb it completely, because it reflects some wavelengths.
- A black body can't be a transparent object. Because when you shine a beam of light on a transparent object, it doesn't absorb it, instead it allows some wavelengths to go through it.
Now, consider a situation where you have a black body in the sunshine. According to our definition, it will absorb all the wavelengths coming from the sun onto it. Saying that it absorbs those wavelengths, it means that it is absorbing their energy. According to Kirchhoff's law, a body having some finite temperature will emit ALL wavelengths of radiation, but the wavelengths themselves can be of different intensities like red is being emitted more intensely than the blue one. The wavelength which is emitted most intensely will be the color that body and I will call that wavelength "max intense wavelength". Our black body lying in the sunshine will also emit radiation of all kinds of wavelengths with different intensities and there would be a wavelength which would be emitted most intensely and that will determine the color of the black-body.
If that max intense wavelength lies in the visible region then the black-body will have some visible color like blue or red.
But when the max intense wavelength doesn't lie in the visible region, because we see only the visible range of the Electromagnetic spectrum, we will see only that wavelength which has maximum intensity in the visible region, though the overall maximum of intensity is out of range of visible light.
When the max intense wavelength is far far away from the visible region then the visible wavelengths will have very low intensities and hence no color is visible and black-body looks black.
For a black-body to look white, it will have to emit wavelengths corresponding to the visible region with nearly equal intensities because white light is composed of visible colors but with EQUAL intensities of all the colors.
Our sun has its max intense wavelength lying in the visible region and human eyes are evolved to see the most intense wavelength which is visible light.
In our situation of black-body absorbing radiation from the sun, it is radiating some energy on its own because of its finite temperature. So it is receiving some power from the sun and emitting some power on its own. When the temperature of black-body becomes constant then it means that power incoming will be equal to power outgoing.
If your black paint absorbs all wavelengths completely then only it is a black-body.
Best Answer
Physics answer
A black body at a temperature T, area A (on either side) will radiate heat by: $$\dot{Q}_{radiated ,downwards}=\sigma A T^4$$. If incident radiations are of power $P$, then $$\dot{Q}_{body}=mc\frac{dT}{dt}=P-2\sigma AT^4$$ (the 2 comes into place because it radiates in both directions)
We can integrate this if we want, but we mainly want to qualitatively analyse it. From this equation, we can see that as time passes, temperature will increase to an equilibrium value ($T_{eq}=(P/2\sigma A)^{1/4}$). I am assuming that $P>2\sigma A T^4$ in this analysis (I've never seen an umbrella that feels hotter than direct sunlight). At this equilibrium value, you will receive heat $\sigma A T^4$, i.e., $\frac{P}{2}$.
For a non-black umbrella, assuming no transmitted heat, we have $e=a=1-r$ (coefficient of emission,absorption and reflection respectively). Out of incident heat power $P$, $aP$ is absorbed. The body radiates heat on either face as $e\sigma AT^4=a\sigma AT^4$. So we have $$\dot{Q}_{body}=mc\frac{dT}{dt}=aP-2a\sigma AT^4$$. From this, we get the same value of equilibrium temperature as for a black body. But, radiated heat power is $eA\sigma T^4$, so it will radiate less heat than the black umbrella onto you (Since both $T=T_{eq}$ are the same).
So paradoxically, we get that a black umbrella is the worst thing to use on a sunny day.
IMHO
Black seems to be the majority because of a cultural bias. I'm assuming you're from India (by looking at your name). Since India was once a British territory, we can trace the use of black umbrellas back to the Brits. The Brits used to use black because black was considered formal attire (Black suits are actually the most uncomfortable from the physics point of view). This must have proliferated to India (Most British traditions have). Umbrellas before the British must've been colorful (See the painting here). If you take a look at Japanese umbrellas, they're all colorful. So black is really not the majority worldwide.