How do I see things of the bright room, being in the dark room?

Question

Visible light emitted or reflected from the objects around us provides information about the world.

If I sit in a dark room, and see the bright room outside, I am able to see all the objects of that bright room. In other words, it mean light emitted or reflected from those objects of the bright room are entering into the dark room. Shouldn't that make my room bright? But, my room is dark and I am able to see those objects of bright room, both are contrary?

Edit: If we consider another situation like the one in the picture below, lets assume that we move as far from the street light, that even in the presence of street light the place remains as dark as it was before. I believe even then we can see the street light, won't we?

If we agree that we still can see the lamp, it mean that we are able to see the lamp, without the ray from lamp hitting our eye? Won't it be contrary? Edit: Lets consider a laser beam in the dark room (in the picture room is not totally dark, but assume a dark room). Look the laser in the same angle as shown in the picture, as laser remains narrow over long distance, I hope we can assume light is not spreading in other directions. But I hope we can see it even though light is not spreading and reaching me, how is it possible?

Answer 1

In the situation you are describing, the only thing one can take away is that the room you are in is brighter than it would be if the bright room were not there. It does not mean that your room has to be as bright as the bright room, just that your room is brighter than it would be otherwise.

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For your new situation with the street lamp:

If you move very far away, but still within sight of the lamp, the light from the lamp or nearby bright objects will still reach your dark room. Your dark room will not be as dark as it was before, even if you judge it to be just as dark using only your eye. Since light is reaching the dark room, that light will cause the room to be a bit brighter.

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Using the idea of intensity of light:

An intensity of zero at some point in space means there is no electromagnetic radiation; there is no light. If you stand in a place where the intensity of visible light is actually equal to zero, there is no light there. You won't be able to see anything with your eyes or any other visible light detector.

Make sure you're not confusing zero intensity with a very small intensity. These are very different ideas.

It might help if you take a particle-view of light. Think about the photons leaving the lamp and entering your eye. If they enter your eye, and you move away, they're going to hit the walls of the room instead.

Answer 2

There exists coherent light and incoherent light

The laser is an example of coherent light, i.e. it can be described by a wave with a known amplitude and phase .

Amplitude is connected with the power that the beam distributes. In lasers this is very concentrated . Amplitude is also connected with the number of photons in the beam .

The street lamp is an example of incoherent light: individual atomic excitations emit photons incoherently so there exists no wave whose phases are known.

You also completely ignore diffusion of light. Diffusion of light is what gives us twilight when the sun goes down . Photons from the light hit on air atoms and change directions, no longer pointing at the source, as infinitesimal reflections.

Now in your last entry, the laser, the reason that one can see the green light is because of diffusion of the laser light in the air of the room. A small part of its photons hit the molecules of air and reach not only our eyes, but the whole surrounding ambiance, reflecting also from the walls etc. If the laser were in vacuum you would not see the beam unless you crossed it ( careful of your eye, though that would be the least of your worries if you were in vacuum :) )

Diffusion also accounts for the light in a dark room next to a lit room even if the source does not shine into the room . Light diffuses in the air enters the dark room hits the walls and furniture and "lights dimly" the dark. If the street lamp were in vacuum the light would illuminate only in the direct optical rays from the source and shadows would be sharp and black.

The further away one is from a light source the smaller the intensity of the beam, which in direct optical ray falls as 1/r^2. For diffusion it is more complicated because each rescatter from the molecules becomes a source and each reflection of light also becomes a source, still intensity falls rapidy with distance, that is why the dark room is mainly dark and the laser green line is clearly defined: the scatterings are few and mainly next to the beam, and do not markedly reduce the intensity of the laser beam.

Answer 3

I will try to answer your question as simply as I can but involving some biology (only a very basic thing). Now when do you see an object? Only when the light of that object hits your eyes and your eye detects it. When you look at an object, electrical signals travel via the optic nerves to an area in your brain called the thalamus. This then sends the information to the visual cortex, where it is examined in detail. Different parts of the visual cortex simultaneously process the colour, shape, movement and depth of the object. Other parts of the cortex put this information together to give you a complete picture of the object. Now you might think I am going off topic but it is just to build up a basic concept so that you do not get confused.

Now for the technical bit, say that the light from the lamp goes directly into your eye and nowhere else. Then, the room you are viewing from (i.e. the dark room) will not become any brighter than it initially is. You have to understand that only a fraction of the light rays emitted by an object hit your eye and are sufficient to create a visual picture. The rest hit the walls of your room and then hit your eye making the room seem a bit brighter than you originally think it is. If your room is closer to the light source then it will be more bright as the light loses less energy when it travels over a shorter distance. But if your room is far away, then the light from the source won't be enough for your room to become very bright. Though there will be a very little change in the brightness, you will not be able to make out the difference but the light reaching your eye will be enough to create a visual picture.

To conclude, in a room the light loses energy when it is reflected from wall to wall but when it directly hits your eye, it has sufficient energy to create a clear visual picture. It goes the same for viewing the objects of a bright room.

To give an example, take the example of the stars. You are viewing the stars from the window of your room. You are still able to see its light as it directly hits your eye, but the energy of the light isn't enough that it is able to light the room as most it gets absorbed as soon as it hits the wall of your room. The example of stars includes many other technical things but it would be best to not get into them. I hope this answers your question and clears your misinterpretation.

Edit after laser question See, with lasers there are two cases. One is invisible laser which you see only after it is reflected by an object. The one shown in the picture is visible in the sense that you can see the path of the laser. the phenomenon is known as Rayleigh Scattering. For more information please refer to the wiki page of the article on Rayleigh Scattering to understand the concept in detail.

http://en.wikipedia.org/wiki/Rayleigh_scattering

It is basically the reflection of particles from the dust and fog and other particles in the air. I hope this answers your question.

Answer 4

I don't know if you already understood it, but here is a simple explanation.

Take a torchlight, look at it directly. You'll be blinded. Now, shine a nearby object, you will see the bright, but it won't hurt.

Repeat the experiment with a very dim torchlight. If you look directly you will notice the light, but if you shine an object, you won't see anything.

Replace the torchlight with the light room and the nerby object with the dark room. Done!

Answer 5

your room is dark because either there is not a lot of light coming into it or because all objects in your dark room absorb the light, don't reflect it back. so the light coming from the bright room does not reflect from the walls and other objects of your room, hence it's dark.

in the case of the street light, it's the same. light coming from the bulb is not reflected off the air, it goes through it. notice how the guy and the road on the picture are relatively bright, these objects are reflecting light in all direction, including in the directions to your eyes. imagine that the light coming from the bulb is split into many rays. a few of them hit your eyes, making the bulb look bright. a few of the rays hit the road and the guy, they're reflected all over, and some of them hit your eyes - therefore the guy and the road look brighter than the air, but not as bright as the bulb itself.

Answer 6

Hello GODPARTICLE, Answer for your question lies in reflection and intensity properties of light. Here is a good article which can "throw light on your question".

http://www.cs.colorado.edu/~mcbryan/5229.03/mail/97.htm

Now in that article you can see figure under 1.Dull surfaces. In that figure you can replace V which is viewer with something in your dark room say cup, bottle , pencil etc... and now you can figure out why you can not see those things as bright as it those things were in bright room. As you can read in that article "Dull surfaces reflect scattered light equally in all directions". So bottle or table in your dark room scatters light from point source in all direction. Now as far as I think what you are trying to ask is I am standing in dark part of your edited picture (street light) and one bottle is besides me. Now I can see all the things like a cup resides in brighter part of your street light picture and I can see it and I can also see point source (here street light) but why I can not see bottle which besides me in dark area??

The answer lies as I have already said as follow:

In your situation(street light) the bottle is in place of V in figure of the given link and you are seeing bottle by light which is reflected two times from point source(street light). In real situation like yours bottle is not in place of v. But light that has been reflected already off lots of surfaces and then reaches to bottle and each time intensity is decreased before reaching to our eyes so we can not see bottle.

Answer 7

I am converting my comment into an answer with some mine amendment.
You only see most of the light entering in the room which directly falls upon your eyes. On the other hand when this light coming from outside falls upon the objects placed inside the dark room then most of it is absorbed and the remaining is scattered all around, hence you see do not see the things in the dark room well.
Amendment:
In the picture you have shown we do not see the road because most of the light falling on the road is absorbed and the remaining is scattered. On the other hand we can see the street light because light emitted from it directly falls on our eyes.
To check my argument practically place a mirror on the road at appropriate angle. The mirror will shine. Now remove this mirror you will see the dark road. This implies that most of the light falling on road is absorbed.
Edit:
@GODPARTICLE "...we are assumed to stand in such a place where light from the street lamp won't reach..". This is only possible if there is an obstacle in the path(this does not seem to be the case in your picture). We can see the light is the proof that light is reaching there. You can also put a photodetecter to prove if you don't believe your eyes. As I said in my answer: most of the light which falls on the surface of road undergoes diffuse reflection and remaining is absorbed.
If you place a plane mirror on the road then the light will not undergo diffuse refelection. A detectable light will fall on your eyes. This proves that light doesn't only falls on your eyes, it does fall on whole of the road. If there is no mirror placed then most of the light is absorbed and the remaining is scattered in all directions which causes an appearance of dark.

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Edit-2
You say that:
"We stand at a place at which the intensity of light reaching our eye becomes zero s.t. there is no obstacle in the way of light."
This statement can be disproved.
First method: Assume the street light emit spherical wavefronts of light. The intensity of light at any distance $d$ from the source(street light). The intensity(power per unit area) of the light at $d$ is $$I = \frac{P}{A} = \frac{P}{4 \pi d^2}. $$ So at any finite distance from the street light the intensity doesn't become $0$.
If we further assume that the air absorb the light even then at any distance $d$ the intensity of light will not become $0$.
I do not know how to calculate the intensity of light for a spherical wavefront including absorbtion. I know for a plane wavefront. If the light emitted from a source is a plane wavefront then at any distance $d$ from the source the intensity of the light will become $$I = I_{0} \, e^{-\alpha \, d}$$, where $\alpha$ is the absorption cofficient. So at any finite distance $d$ the intensity of light will not become zero.

Second method: In this method I will disprove your argument by Proof by contradiction. A proof by contradiction is always the best proof of all.
To disprove the statement "Light reaching at a distance $d$ has $0$ intensity" I will prove the statement "Light reaching at at a distance $d$ does not has $0$ intensity." I do an experiment in which I stand in your dark room and see the street light. I use a convex lens and allows the light reaching in the dark room to fall on a small amount of water for a particular amount of time $T$ then I calculate the rise in the temprature of water. Knowing the specific heat of water I can calculate the energy of incident light. I know the cross-section area of my lens.
The intensity is equal to (Energy calculated/cross-section area)/T. Since the intensity comes out to be a positive quantity I conclude:
"If I can see the light then it has a finite intensity."
Now I again rearrange my experiment. This time I sit in a room in which I do not see the street light. I again calculates the intensity with my apparatus. I repeat this experiment many a times. The result of my experiments are:
" When I do not see the light Sometimes the intensity I measures comes out to be positive. Sometimes the intensity I measures comes out to be zero."
So from my experiment I conclude:
"If I do not see the light then the intensity sometimes comes out zero and sometimes positive"
This implies:
"In any case if I see the light the intensity of the light is not $0$."
This result disproves your argument: "...as the intensity of light is zero, even in that situation I believe that I can see the street lamp."

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Hope this helps.
Regards.