Mirror box with 1 candela of luminous intensity light source

[genekuli]

The only thing that resonance achieves is matching to whatever load you place in the box.

say that, on an rx antenna, that a 1W tx source might induce 1mW. But inside a RF reflective (metal) enclosure, this same scenario might induce 1W on an rx antenna? or what would it be like in that kind of example?

 

[anorlunda]

say that, on an rx antenna, that 1W tx source might induce 1mW. But inside a RF reflective (metal) enclosure, this same scenario might induce 1W on an rx antenna? or what would it be like in that kind of example?

I think you started at the right place. You simplified down to a hollow sphere, and didn't say anything about the source of radiation. Adding complexities will make it much harder to calculate and won't bring any more principles, and maybe not any more useful details.

What are you trying to accomplish with these questions?

 

[sophiecentaur]

this same scenario might induce 1W on an rx antenna?

The cavity is effectively just a transformer. The position of the transmitting antenna and the receiving antenna will affect the coupling. In just the right positions, you could expect all the Tx power to go into the Rx when both terminals have the right Impedances. Think of a waveguide.
I've already made the point that this problem has not been defined properly so there is no proper answer. This is even worse when you want to 'measure' the Energy inside the room.

 

[hutchphd]

OMG- obviously, when the light is turned off the energy density decays rapidly. That's not what the OP was asking

I was answering the related question of building up to steady state when the light is turned on. Same process: charge the capacitor/ discharge the capacitor.
I have no idea what question you were answering.

 

[sophiecentaur]

I obviously always leave a room too quickly to notice that.

 

[Shane Kennedy]

Summary:: Mirror box (reflects 99.9%) with 1 candela of luminous intensity light source, how many candelas of luminous intensity will be inside?

considering that a simple mirror may reflect 99.9% of the visible light

Candela is not light output, it is intensity in a particular direction

 

[genekuli]

Candela is not light output, it is intensity in a particular direction

yes that is correct, my question refers to intensity in a particular direction, basically; for an observer inside the box, how many candles would they see, relative to 1 candela? i think it would be around 1000 candelas perhaps.

 

[jbriggs444]

yes that is correct, my question refers to intensity in a particular direction, basically; for an observer inside the box, how many candles would they see, relative to 1 candela? i think it would be around 1000 candelas perhaps.

For an idealized spherical cavity with near-perfect mirrors, the illumination will follow an inverse square law.

For specular reflection in an irregular cavity, I believe that you are correct. The incoming illumination on a point receiver will be approximately 1000 candelas, independent of position within the cavity.

 

[alan123hk]

yes that is correct, my question refers to intensity in a particular direction, basically; for an observer inside the box, how many candles would they see, relative to 1 candela? i think it would be around 1000 candelas perhaps.

I agree with you. As I mentioned in the previous post #11, assuming that the box itself does not absorb any light energy to convert it into heat, and considering that the shape of the box is not necessarily spherical, the illuminance in different directions may be different, but the total lumens should still be 4000π, which means that there is an average 1000 candles in all different directions. I believe this is the theoretical value observed on the inner wall of the box when the system reaches a steady state.

 

[genekuli]

i know this abstract physics question would have already been theorized hundred of years ago. it is a classic abstract physics question akin to the light on a train at light speed.
the issue with this one is i can't find it anywhere, i think it probably has its own wiki page because mirrors and candles have been around a long time

 

[hutchphd]

which means that there is an average 1000 candles in all different directions. I believe this is the theoretical value observed on the inner wall of the box when the system reaches a steady state.

This is absurd. So if I am in a perfectly mirrored room and I enter it with a single candle I will be immediately flooded with light from all directions like Kleig lights! Somebody tell the electric company.

 

[alan123hk]

This is absurd. So if I am in a perfectly mirrored room and I enter it with a single candle I will be immediately flooded with light from all directions like Kleig lights! Somebody tell the electric company.

Sorry I don't understand what you mean at all.

 

[genekuli]

This is absurd. So if I am in a perfectly mirrored room and I enter it with a single candle I will be immediately flooded with light from all directions like Kleig lights! Somebody tell the electric company.

what do you think would happen, you know how when a mirror is facing another mirror they reflect each other reflections till it reaches too dull to see in a 1000 or so reflections.

 

[hutchphd]

but the total lumens should still be 4000π, which means that there is an average 1000 candles in all different directions.

what do you think would happen, you know how when a mirror is facing another mirror they reflect each other reflections till it reaches too dull to see in a 1000 or so reflections.

In the case of the parallel mirrors you will not see 1000 reflections even for a .999 mirror. Each reflected candle will be optically twice as far away and so the perceived brightness will diminish rapidly. You need to understand what "one candela=one lumen per steradian" actually means. It is an emissive measure for a point source.
The result will in fact depend greatly on the geometry. I suggested the spherical box because then you can look only at exactly the radial part and in fact if you look at original sphere from outside, through the filter, you would just see a single regular candle.
Lumens and Candela are not useful descriptors of a volume, because they are inherently related to flux.

 

[genekuli]

sorry, i am not having luck explaining the simplicity of this abstract question.
the observer's perspective is on the inside. averaging it out, how many reflections would he see with his one candle giving one candela into his eye from the one candle? it would have to be around 1000 candelas right? (ignoring the practicalities like peripheral vision restrictions and such)

 

[hutchphd]

I understand your question. If the mirrors were perfectly reflective and one meter apart, the 1000th reflection would show a candle 1 km distant. The question is not how good are your eyes. The point is the reflections diminish rapidly because of geometry.

 

[genekuli]

I understand your question. If the mirrors were perfectly reflective and one meter apart, the 1000th reflection would show a candle 1 km distant. The question is not how good are your eyes. The point is the reflections diminish rapidly because of geometry.

yes and I'm trying to work out how many candelas worth of light into the eye there would be (on average, from all angles)

 

[hutchphd]

candelas worth of light into the eye there would be (on average, from all angles)

This statement simply makes no sense.
The answer will be quite specific and involve the size and share of the box and your eye ( whose acceptance angle depends upon focus and iris size.)
So my answer is "ask an appropriate question"

 

[alan123hk]

sorry, i am not having luck explaining the simplicity of this abstract question.

Sometimes, when people ask questions, they may not have set all the conditions perfectly and accurately. We can only do our best to do this.

But I think the description you posted in #1 already has enough basic guidance. When answering questions, just like me, we can add our own assumptions.

 

[genekuli]

This statement simply makes no sense.
The answer will be quite specific and involve the size and share of the box and your eye ( whose acceptance angle depends upon focus and iris size.)
So my answer is "ask an appropriate question"

maybe replace eye in my question, with a point. how many candelas are incident on a point in the box

 

[hutchphd]

I'm happy that you think there is enough in #1, so have at it. I cannot even start.

 

[Ibix]

maybe replace eye in my question, with a point. how many candelas are incident on a point in the box

There is zero energy flux through a point since it has zero spatial extent.

If you mean "through a small region" then the answer depends on how big a region, where it is in the cavity, and what shape the cavity is. If you're interested in an actual practical scenario then you need to think about the absorbance of whatever sensor you have in the small region. You seem curiously reluctant to answer these questions, which makes me think you want a general answer. There isn't one.

If you are interested in cavities made from parallel plates then there are relatively straightforward tricks you can try, but it still depends on the answers to the questions above.

Finally, if you are actually asking about microwaves then the question is a lot more complicated since the waves are comparable in size to the cavity and simple analysis of radiated power from point sources just won't do.

 

[alan123hk]

yes and I'm trying to work out how many candelas worth of light into the eye there would be (on average, from all angles)

maybe replace eye in my question, with a point. how many candelas are incident on a point in the box

In order to focus the discussion and reduce possible misunderstandings, maybe we should try to use only the terms of luminous flux (lm), illuminance (lux, lm/m²) and luminous intensity (cd, lm/sr), they all have strict definitions in Photometry. The luminous flux (lm) in a light source is defined by the integral of the inner product of the luminous efficiency function and the spectral power distribution.

Of course, the simplest assumption is that the light source is located at the center of the sphere. In this case, we can expect that in a steady state, the illuminance and luminous intensity will be equal everywhere on the surface of the sphere. If the light source is not in the center of the sphere, the brightness distribution on the surface of the sphere will be uneven even in a steady state.

Maybe we shouldn't even mention the light meter, because this may cause another problem, namely the accuracy of the light meter, the reflectivity of the sensor surface of the light meter is different, which affects the overall balance and so on.

 

[hutchphd]

In light (ha-ha) of the OP statement,

actually i was trying to figure this out for RF

I would suggest that we abandon all arcane photometric terminology and strictly speak in radiometric terms. There will be much less baggage.
The OP was attempting to simplify the situation by opening Pandora's box.
As near as I can tell he wants to heat things in a microwave.

.

 

[genekuli]

in RF, say that, on an receiving antenna, that a 1W RF transmitter source might induce 1mW. But inside a RF reflective (metal) enclosure, this same scenario might induce more than the 1mW on a receiving antenna because of the duplication of apparent radiation sources of the reflections? maybe it might induce near 1W?

 

[hutchphd]

As has been said several times the answer depends upon the details. State a specific question: shape size material wavelength. Specific. This is physics not a debating society

 

[genekuli]

it would be RF broadband 50MHz - 900MHz
in a 3m cuboid, 0.5 metal aluminium walls

 

[hutchphd]

How is it fed? The wavelength then is .3 m to 6m. This is a waveguide problem and will change drastically over this band of frequencies. It may be very complicated. What precisely do you want to know?

 

[genekuli]

a 3m waveguide can do up to the 6m λ, so it will reflect all the frequencies concerned .
on an receiving antenna, that a 1W RF transmitter source might induce 1mW. But inside a RF reflective (metal) enclosure, this same scenario might induce more than the 1mW on a receiving antenna because of the duplication of apparent radiation sources of the reflections? maybe it might induce near 1W?

 

[hutchphd]

It might. Of course it might not. Its complicated. See previous. I'm done.

 

[Eric Bretschneider]

I am amazed at the number of attempted answers - why?
The question is flawed and can't be answered. As has been pointed out, candela are a measure of luminous intensity. More specifically, 1 candela = 1 lumen/ steradian.

The OP did not specify the solid angle over which the light source has a luminous intensity of 1 candela. A laser might emit over a solid angle of 1E-6 steradian, a uniform point source that emitted over all angles would emit over 4 pi steradians.

So in terms of luminous flux you could have anywhere from about 1E-6 to 12.566 lumens.

The real problem is that candela is a measure for a light source. You have a sealed box that doesn't let light out, so by definition it can't be a light source.

From the information the OP provided you can't even determine how many lumens are inside the box, so asking about luminous intensity is non-sensical. Ask a better question if you want an answer. Better yet, do a search on "integrating sphere equation." At least that would tell you how to ask the question (hint you have to specify the flux of the source). More importantly it would let you calculate the answer.

BTW: since lumens are a psychophysical unit you couldn't even determine the radiant energy inside the box.

 

[genekuli]

I am amazed at the number of attempted answers - why?

I rephrased the question with regards to RF antennas allowing the details that have so far been unspecified

 

[Vanadium 50]

I am amazed at the number of attempted answers - why?

People are trying to help. So far the OP wants to know how many candelas or maybe watts or maybe joules there are in an unspecified system involving light or maybe microwaves. People are guessing, but is it really any wonder we haven't guessed right yet? The OP is unhappy that the "wiz kids" haven't answered his question, but we're not mind readers. (Some of us, however, are very good guessers)

The OP is also starting multiple related theads - three or maybe four - asking similar questions. It's pretty clear he has a question in mind, but he doesn't want to come right out and ask it. It likely involves microwaves, but he's asking about light. I'm not sure why he thinks this secrecy and dancing around the real question will be helpful.

 

[hutchphd]

Better yet, do a search on "integrating sphere equation."

Yes. This will likely be wildly inaccurate for the situation described but it is definitive and provides "scientific formula" which seems to be sufficient to the OP. In fact the following does a lovely job on the integrating sphere:

https://www.labsphere.com/site/assets/files/2551/integrating_sphere_theory_apps_tech_guide.pdf

.

 

[genekuli]

People are trying to help. So far the OP wants to know how many candelas or maybe watts or maybe joules there are in an unspecified system involving light or maybe microwaves. People are guessing, but is it really any wonder we haven't guessed right yet? The OP is unhappy that the "wiz kids" haven't answered his question, but we're not mind readers. (Some of us, however, are very good guessers)

The OP is also starting multiple related theads - three or maybe four - asking similar questions. It's pretty clear he has a question in mind, but he doesn't want to come right out and ask it. It likely involves microwaves, but he's asking about light. I'm not sure why he thinks this secrecy and dancing around the real question will be helpful.

sorry new at this forum thing, the decorum is unfamiliar but this is more accurately what I'm trying to ask