Mirror box with 1 candela of luminous intensity light source

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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
 

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  • #2
Ibix
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What are your thoughts?

I'd want to think about the shape of the box, the shape of the light source, and how I was planning to make a measurement before answering.
 
  • #3
jbriggs444
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Think about units of measurement too. If you want to know how much light is in the box, the "candela" is not the right unit. Luminous intensity is an amount of light per unit time per unit solid angle. It is a good unit to characterize a steady source. It is not a good way to measure the amount of stuff in a container.
 
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  • #4
hutchphd
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It is a good unit to characterize a steady source
It is really only a good unit to describe a steady point source. I hate being picky but for me anything else in photometry/luminosity leads to misunderstanding. It is assuredly not the right unit for the box.
(1)You need to find the power emitted by the source
(2)The power leakage through the walls
(3) Balance input and output
You need to choose a geometry for the box. Remember the archetypal physics joke: "assume a spherical chicken"
 
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  • #5
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surely some wiz-kid can work this out in a jiffy, 1 candela (x) 99.9% =?
 
  • #6
jbriggs444
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surely some wiz-kid can work this out in a jiffy, 1 candela (x) 99.9% =?
That sounds like the aggregate luminous intensity from the first reflection from the walls - if the inside surface of the walls could be reasonably treated as a point source.
 
  • #7
Vanadium 50
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It is not a good way to measure the amount of stuff in a container.
surely some wiz-kid can work this out in a jiffy,

Presumably you are reading the replies. Are you just not believing them?
 
  • #8
hutchphd
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Yes the physics can be worked out and you can help. It is not clear to me what the question is. A candela is not "inside" anything. The reflection from a spherical concentric .999 mirror will focus most reflected light back to the original point source and then through the focus.
What prompts the question?
So at steady state what would one see from outside?
What would be the distribution of light inside?
 
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  • #9
sophiecentaur
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This problem could view viewed like a simpler model of a transmission line with a source at one end an a 'short circuit' the other (reflection coefficient of 0.99). If the source has a frequency that resonates with the line then the Energy in the line will be 1/0.01 of the power input per cycles. Total energy would be 100 times the Energy per cycle(?), when the rate of energy in would be the rate of energy lost.
That's a bit tongue in cheek but it could be a way into the problem. If we're dealing with light then the diameter of the box would determine the number of reflections per second. That way, a big box would build up more total energy but the density would be the same(?).
 
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  • #10
tech99
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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
May I answer the question as a radio engineer - after all, light and radio waves are made of the same stuff.

Let's assume the 1 candela source is located inside the box. It is possible that the box will be resonant at the frequency of the source if it happens that a whole number of waves can exactly fit in the box. In such a case, light bounces around several times before the amplitude is diminished to 37%, and if we count the number of waves until that happens it gives us a figure called the Q, or Quality Factor, of the resonator. A bigger box will give more waves and a higher Q, and better reflecting mirrors will also give a higher Q. The following web site explains about Q factor quite well:-

https://www.rp-photonics.com/q_factor.html

The standing waves inside the resonator contain very large stored energy, and if the Q is, say, a million, then a 1 Watt source would create (1/2pi) Megajoules of stored energy. From the point of view of an observer in the box, this will be spread over the cross sectional area of the box, so might not represent a very high intensity measured in, say, Watts/sq metre, because the original source itself might have been very small and intense.

If, on the other hand, the box does not happen to be resonant at the frequency of the source, it will be very difficult to launch energy into it, because reflected waves will oppose the process of radiation. In radio engineering we will see a large reactance (inductance or capacitance) presented to the source, which will tend to stop the flow of current in the source. The acceleration of electrons in the source, required for radiation, will be hindered.
For a case where the source has a broad spectrum, we might expect narrow frequency bands where high intensity is found, and of course these will be coloured.
 
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  • #11
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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

Do you have the official correct answer?

Assuming that the light source inside the box uniformly emits 1 candela of luminous intensity in all directions (the total luminous flux emitted is ##~4\pi~## lumens), and the box itself does not absorb any light energy to convert it into heat, then the box should be still a 4 pi lumens light source viewed from the outside.

However, if 99.9% of energy is reflected, is it possible to measure total ##~4000\pi~## lumens from the inner wall of the box ?
 
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  • #12
hutchphd
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However, if 99.9% of energy is reflected, is it possible to measure total 4000π lumens from the inner wall of the box ?
For a sphere of radius of any radius the ingoing integrated flux will be ##(999)4\pi## because of the leakage I believe.

The original OP had no official geometry specified so we are as official as it gets I think. For any other geometry (except diffusely rough surface) I think the interior result will be nonuniform in a very complicated way. But the integrated result external is still energy conserving.
 
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  • #13
sophiecentaur
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If, on the other hand, the box does not happen to be resonant at the frequency of the source,
With the dimensions that are implied (a 'box' and light ) - and the likely tolerances for the geometry, I reckon the idea of specific resonance frequencies is probably not relevant. Matching the source could probably be treated as 'wide band'. (Or a very fine structured comb)
For a radio Engineer, like you and me, the problems of matching in and out of the system immediately come to mind but that could probably take a back seat in the light of the other other practicalities.
 
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  • #14
tech99
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I don't think anyone has mentioned that candela is a photometric unit of *power*; a 1 cd source continuously emits power- the enclosed volume will have an increasing energy density as time goes on. I'm not sure it makes sense to speak of measuring luminous intensity (or radiant intensity, the radiometric unit) within a reflective cavity. At least, I am not familiar with quantifying 'intensity' (as opposed to energy) within such a cavity.
If the box acts as a resonator (unlikely, as mentioned in #14), then the box will "slowly fill up" with energy, like a tank, but will approach a limit. When that happens the source is entirely supplying the losses. When the source is switched off, the stored energy will be slowly delivered to the lossy items in the cavity, including the source itself.
 
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  • #15
sophiecentaur
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If the box acts as a resonator (unlikely, as mentioned in #14), then the box will "slowly fill up" with energy, like a tank, but will approach a limit.
The term Q factor comes to mind, although this usually refers to a resonant system. But it would also apply to an insulated box with a heater in it. The final temperature inside the box would be reached when Power input by the heater is equal to the power loss through the walls of the box etc. etc.
 
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  • #16
hutchphd
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I don't think anyone has mentioned that candela is a photometric unit of *power*
?? (first five entries) I think it was noticed.

the enclosed volume will have an increasing energy density as time goes on
That transient will be measured in nanoseconds. The assumption implicit is that the sides are either reflective or transmissive. The steady state has been pretty accurately described from a number of more and less appropriate analogies for the simple spherical case.
Personally I don't wish to explore all of the arcane units for light flux (the fundamental unit is named for a candle !!)
We seem to have lost the OP long ago. Ah well.
 
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  • #17
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It is not clear to me what the question is.
basically i was just trying to work out how much more light there would be due to all the reflections
 
  • #18
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as a radio engineer - after all, light and radio waves are made of the same stuff.
For a radio Engineer

actually i was trying to figure this out for RF, i only resorted to the visible light example because i could not find anyone to answer it in RF, and i figure it is the same, also i was asking in broad band, you read my mind.
so could you tell me your estimation of a scenario with a 3W antenna transmitting a broadband white noise from 50MHz to 900MHz inside an aluminium room 3m wide, how many 3W equivalents would be in there due to the thousands of reflections?
that is say that from the 3W point source antenna there would be 0.003W that would strike a measuring antenna directly at 2m distance, but counting and adding all the reflections of the RF power how much would be incident on the antenna? could it reach as high as 0.8W or maybe 1W?
 
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  • #19
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I don't think anyone has mentioned that candela is a photometric unit of *power*; a 1 cd source continuously emits power- the enclosed volume will have an increasing energy density as time goes on. I'm not sure it makes sense to speak of measuring luminous intensity (or radiant intensity, the radiometric unit) within a reflective cavity. At least, I am not familiar with quantifying 'intensity' (as opposed to energy) within such a cavity.
i would prefer the answer in energy or percentage actually. watts, or Jules, as in if the light source had 1W or 1kJ
 
  • #20
hutchphd
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that is say that from the 3W point source antenna there would be 0.003W that would strike a measuring antenna directly at 2m distance,
Sorry but I have no idea what you are saying here.
The answer to your general question will depend upon a host of things you have not specified, including how the antenna is fed and shape and material of the cage.
 
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  • #21
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Sorry but I have no idea what you are saying here.
The answer to your general question will depend upon a host of things you have not specified, including how the antenna is fed and shape and material of the cage.
yes, as in the mirror original question the walls would be aluminium in this RF mirror equivalent, it would also be a box (cuboid shape room). the feed, please ignore all that detail, just abstract thought estimation, as the actual answer would be impossible to work out and would have to be measured.
 
  • #22
sophiecentaur
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broadband white noise from 50MHz to 900MHz inside an aluminium room 3m wide
Could you just assume a surface Impedance of about 1mΩ and a free space Impedance of 377Ω? That is a ratio of 3770, which would be an indication of the 'Q' of the box. That assumes that the alumninium surface is continuous. Seams and a door would probably increase the effective surface Impedance by a factor of at least ten (??) . So could we say that the power flows inside the room would be perhaps 400 times the input transmitter power, once equilibrium was established ( as @hutchphd notes, the time for this could involve a few hundred transits of the waves, which would take just a few microseconds - but a timescale that you could easily measure)
The spectrum of the signal buzzing about inside the box would be strongly 'coloured'.
 
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  • #23
jbriggs444
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i would prefer the answer in energy or percentage actually. watts, or Jules, as in if the light source had 1W or 1kJ
The dimensions of the box enter in.

Suppose that the box is a sphere that is one light-nanosecond (approximately 30 cm or one foot) in radius with a point source at its center. Then in two nanoseconds the emitted light will have had time to go to the reflective walls and back.

The reflected light behaves as though it emerged from the same point source and adds to the original source.

We ignore the possibility that the original source absorbs the reflected incident light. In practice, an incandescent filament absorbs visible light [has to -- second law of thermodynamics] so this is a questionable assumption.

Suppose that we work in watts and have a 1 watt source. For the first two nanoseconds, we just have the 1 watt source. Then for the next two nanoseconds we have an additional 0.999 watt source and so on. Summing the infinite series, ##\sum_{i=0}^{\infty}k^n = \frac{1}{1-k}##. Effectively we have a 1000 watt source.

But that analysis is just a bit over-simplified. It tells us about the "outbound illumination" in the interior of the container but ignores the half of the trip that the light takes back. The "inbound illumination" starts at .999 watts and decays. The sum of that geometric series is 999 watts.

How much light is in the enclosure? We have one nanosecond's worth of outbound light in flight and one nanosecond's worth of inbound light in flight at any given time. 1000 watts times one nanosecond plus 999 watts times one nanosecond = 0.001999 joules.

If one doubles the size of the enclosure, the enclosed energy at equilibrium doubles.
 
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  • #24
tech99
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actually i was trying to figure this out for RF, i only resorted to the visible light example because i could not find anyone to answer it in RF, and i figure it is the same, also i was asking in broad band, you read my mind.
so could you tell me your estimation of a scenario with a 3W antenna transmitting a broadband white noise from 50MHz to 900MHz inside an aluminium room 3m wide, how many 3W equivalents would be in there due to the thousands of reflections?
that is say that from the 3W point source antenna there would be 0.003W that would strike a measuring antenna directly at 2m distance, but counting and adding all the reflections of the RF power how much would be incident on the antenna? could it reach as high as 0.8W or maybe 1W?
This problem seems equivalent to a generator feeding a long transmission line and a load which can be moved along the line.
Case (a) - already discussed. If the load is absent and the line is resonant at the source frequency, oscillation will slowly build up in the line. There will be large energy stored in the standing waves and once the store is full, the entire output of the source will be delivered to the line losses.
Case (b) - already discussed. If the load is absent and the line is not resonant at the source frequency. The source will not be able to deliver very much power to the line because it is presented with a high reactance, which hinders the flow of current in the source. For a mirror box, we assume that it behaves just as a mirror, preventing radiation. In such a case we still have standing waves close to the mirrors, where the intensity may be quadrupled.
Case (c) - load placed at a position along a resonant line. At certain positions the output of the source can be delivered to a resistive load, such as a receiving antenna, with only small losses.
Case (d) - load placed along a line having no resonances. If the load is at the far end of the line it is capable of receiving nearly all the generator output. However, in other positions, if there are no resonances, the source will be unable to deliver all its output to the load, irrespective of the position, as the source will have reactance in series with its output.
Therefore for the general case of a transmitter and a receiver in a non resonant reflective box, I think case (d) will apply. The transmitter will be unable to deliver all its output to the receiver as it will be subjected to reflected energy, which will place reactance in series with its output.
 
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  • #25
tech99
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To answer your final question, in a free space environment the maximum power the receiving antenna could receive is half the transmitter power, but in the reflective environment I would expect this to be reduced due to the reflected energy arriving back at the transmitter, which will constitute mismatch and place reactance in series with the antenna radiation resistance.
 
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