Light intensity required to shine a light through an opaque object

[KataruZ98]

TL;DR Summary

I’m interested in a formula useful to calculate the wattage necessary for a light source to have its light passing through a dark body

Pretty much, I was picturing a 1x1x1 meters cube of a dark colored, mostly uniform material (limited porosity, no crevices) with a small chamber in its interior - exactly where the center of mass should be. In it, a body emitting light is activated with the goal to find the intensity necessary for the light beam emitted to be visible from the outside.

Let’s say the opaque body has an absorptivity of X value and both reflectivity and transmissivity are negligible because way lower than X. They’re still nominated as R and T in case they must be used to determine the required intensity, and therefore not ignored. We can also say that all those features regard light in the wavelength of 500nm - equal of course to the one of the light shone inside the body, and after “exiting” the object the light must have an output of 1,500 lumens. Given these conditions, if possible, how can the intensity be mathematically calculated?

 

[malawi_glenn]

Use gamma rays

 

[mfb]

Absorption will be some fraction of the light - mathematically the intensity outside will never be zero, in practice it can be so low that it's effectively zero. This doesn't depend strongly on the light source.

If the absorption is very strong then you can still heat the object until it emits blackbody radiation.

 

[hutchphd]

TL;DR Summary: I’m interested in a formula useful to calculate the wattage necessary for a light source to have its light passing through a dark body

Let’s say the opaque body has an absorptivity of X value and both reflectivity and transmissivity are negligible because way lower than X.

This is not a good start, because it violates conservation of energy. The physics of your situation is that eventually the body will reach a temperature where all of the input energy is radiated by the (black) body. The temperature will depend upon the surface Area (A=6 sq meters), the input power (in Watts) according to Stefan-Boltzmann $$P=A \sigma T^4~~~~~\sigma= 5.67 × 10^{-8} W/⋅m^2.K^4 $$so for a 100W bulb with the cube in the radiative coldness of space this gives a Temperature of around 125Kelvin (please check my arithmetic) This assumes an absorptivity (and therefore emissivity) of unity.

 

[Vanadium 50]

I agree this is a bad start, but for an entirely different reason. Suppose we said "the answer is 3x+2". Now what?

 

[tech99]

If it were a microwave source we would use the Effective Isotropically Radiated Power in Watts, and calculate the intensity at the surface from P/(4 pi R^2) and then apply a correction for the absorption in decibels per metre. I am sure the same is true for light, but the system of units is more difficult.

 

[jbriggs444]

If it were a microwave source we would use the Effective Isotropically Radiated Power in Watts, and calculate the intensity at the surface from P/(4 pi R^2) and then apply a correction for the absorption in decibels per metre. I am sure the same is true for light, but the system of units is more difficult.

Wait, what? How are "microwave" and "light" distinguishable?

And how is $4 \pi r^2$ relevant for the surface area of a cube with sides of length $r$?

 

[tech99]

Wait, what? How are "microwave" and "light" distinguishable?

And how is $4 \pi r^2$ relevant for the surface area of a cube with sides of length $r$?

The intensity will vary over the surface of the cube, so will depend on the radial distance. I thought microwaves and light were both EM waves.

 

[hutchphd]

Part of the problem is that the answer to the original question is "none" No light will shine "through" an opaque object, by definition. If the OP would care to restate the question ln reasonable form, then we can discuss the answer. Otherwise we can circle endlessly..........

 

[jbriggs444]

The intensity will vary over the surface of the cube, so will depend on the radial distance. I thought microwaves and light were both EM waves.

You'd said that the system of units would be different for light than for microwaves. This suggested that you thought that there was some relevant difference.

I see where you were going now with the $4 \pi r^2$ -- going for power per unit source-normal surface area at a particular point on the surface of the cube at radius r from the source. However, you would also need a $\cos \theta$ term in there to use the angle of the cube's surface normal relative to the direction away from the source.

If we are talking about signal attenuation through a medium then there would be logarithmic losses in addition to the inverse square attenuation.

If we are talking black body radiation then we would probably end up doing a finite element analysis and a numerical solution with "cold spots" in the corners and relatively "hotter spots" in the centers of the faces. I'm not good enough to see a way to reduce it to a one dimensional differential equation without a spherical symmetry to exploit.

 

[hutchphd]

If we are talking black body radiation then we would probably end up doing a finite element analysis and a numerical solution with "cold spots" in the corners and relatively "hotter spots" in the centers of the faces.

And of course there are issues of thermal conductivity for an absorbing/radiating cube. Also tangled in this thread are issues of photometry versus luminosity. Not to mention all of the confounding issues of spherical measure vs solid angle measure vs per Area measure. When I was doing this every day and it mattered I always kept a "cheat sheet" on the wall next to my desk. Optics is very old and involuted and wonderful.

 

[KataruZ98]

This is not a good start, because it violates conservation of energy. The physics of your situation is that eventually the body will reach a temperature where all of the input energy is radiated by the (black) body. The temperature will depend upon the surface Area (A=6 sq meters), the input power (in Watts) according to Stefan-Boltzmann $$P=A \sigma T^4~~~~~\sigma= 5.67 × 10^{-8} W/⋅m^2.K^4 $$so for a 100W bulb with the cube in the radiative coldness of space this gives a Temperature of around 125Kelvin (please check my arithmetic) This assumes an absorptivity (and therefore emissivity) of unity.

I see, but the issue I wanted to clear was another - though I may have goofed with the original assumptions. As a more direct question that links to the former I’ve asked, couldn’t a sufficiently intense light pass through an opaque body and be visible to some extent from the outside? I was pretty much depicting a more extreme version of what you get by shining a light behind your hand.

 

[russ_watters]

I see, but the issue I wanted to clear was another - though I may have goofed with the original assumptions. As a more direct question that links to the former I’ve asked, couldn’t a sufficiently intense light pass through an opaque body and be visible to some extent from the outside? I was pretty much depicting a more extreme version of what you get by shining a light behind your hand.

I think people are getting ahead of themselves when discussing the secondary effects, and you've missed the basic logic of the issue you're asking about:

The property of the material you are describing is called "transmissivity". It's a simple proportion of how much light is let through. Tinted windows are sold based on selected values: 5%, 15%, etc. The amount of light let trough is always that proportion of the light shone on it. Shine a 20x brighter than the sun light on a 5% transmissivity window and you get a car interior illuminated exactly as it would be on a sunny day. 20 * 0.05 = 1. Simple math.

Opaque, by definition, is exactly 0% transmissivity. Anything times zero is zero. Maybe there's a loophole where an object we think is opaque really isn't, but that's the only possible "out" available on your question. Some objects may fit that (your hand), but others are known to really, truly be opaque.

 

[KataruZ98]

Opaque, by definition, is exactly 0% transmissivity. Anything times zero is zero. Maybe there's a loophole where an object we think is opaque really isn't, but that's the only possible "out" available on your question.

Oh I see what I did wrong. Yeah, I was thinking to a non-perfectly opaque object - say, 80% of absorptivity instead of 100%. I see also that my question can be easily answered if I know the output light power. Basically, if instead of 1,500lm I said the light visible from the outside had an intensity of 100W, then that of the source inside the cube would be 100/X with X being transmissivity - am I correct?

But what about the thickness of the cube? Shouldn’t that also be accounted for?

 

[russ_watters]

Oh I see what I did wrong. Yeah, I was thinking to a non-perfectly opaque object - say, 80% of absorptivity instead of 100%. I see also that my question can be easily answered if I know the output light power. Basically, if instead of 1,500lm I said the light visible from the outside had an intensity of 100W, then that of the source inside the cube would be 100/X with X being transmissivity - am I correct?

Yes.

But what about the thickness of the cube? Shouldn’t that also be accounted for?

There's probably another name for describing the parameter that way, but it's the same as if you measure/express it as transmissivity per millimeter and multiply by thickness to get total transmissivity.

 

[KataruZ98]

Yes.

There's probably another name for describing the parameter that way, but it's the same as if you measure/express it as transmissivity per millimeter and multiply by thickness to get total transmissivity.

Got it I think. I’d like to research about said transmissivity per mm, so mind me asking about some useful link?

 

[russ_watters]

Got it I think. I’d like to research about said transmissivity per mm, so mind me asking about some useful link?

Sorry, I'm a mechanical engineer not a physicist; I know thermodynamics, but I'm thin on optics except as it supports my astronomy hobby. What I described is mathematically similar to how thermal insulation works. But I don't actually know if it's commonly used that way in optics. For astronomy some filter material comes in sheets, and you can use transmissivity per sheet and layer them on top of each other. It works the same as per mm mathwise. Maybe someone else knows how the subject is truly treated in physics academics.

 

[jbriggs444]

transmissivity per millimeter and multiply by thickness to get total transmissivity.

If a substance has a "transmissivity" of 50% over a width of 1 millimeter then what will be the transmissivity of a thickness of 3 millimeters?

It is not linear. Clearly, it will not be 150 percent. It has to be something less than 50. Maybe we divide by the 3 millimeters? Nope, that is not right either.

It will turn out to be exponential. In the case of three millimeters, think of it as three reductions, each by a factor of 50 percent. That is $0.5^3$.

What you probably want is a rate of absorbtivity per millimeter. And you want a logarithmic scale. For good mathematical reasons, it would conventional to standardize on an absorbtivity of one per millimeter as being enough to get a reduction by a factor of $e$ (Euler's number -- 2.71828).

On this scale, an absorbtivity of 50 percent per millimeter would be rated at $-log(0.5)$ per millimeter. Or approximately = $0.693\ \text{mm}^{-1}$.

For this substance, the fraction transmitted through 31.4 millimeters would be $e^{-31.4 \times 0.693} = 3.5 \times 10^{-10}$ of the original.

[As @russ_watters surely knows well already]

 

[KataruZ98]

On this scale, an absorbtivity of 50 percent per millimeter would be rated at −log(0.5) per millimeter. Or approximately = 0.693 mm−1.

Sorry but isn’t the logarithm on base 10 of 0.5 equal to ~0.301?

 

[jbriggs444]

Sorry but isn’t the logarithm on base 10 of 0.5 equal to ~0.301?

Natural logs. I come from a math background. Despite my name, I use $\log$ to mean a natural log, not a Briggsian log.

In mathematics, the common logarithm is the logarithm with base 10.[1] It is also known as the decadic logarithm and as the decimal logarithm, named after its base, or Briggsian logarithm, after Henry Briggs, an English mathematician who pioneered its use, as well as standard logarithm. Historically, it was known as logarithmus decimalis[2] or logarithmus decadis.[3] It is indicated by log(x),[4] log10 (x),[5] or sometimes Log(x) with a capital L (however, this notation is ambiguous, since it can also mean the complex natural logarithmic multi-valued function). On calculators, it is printed as "log", but mathematicians usually mean natural logarithm (logarithm with base e ≈ 2.71828) rather than common logarithm when they write "log". To mitigate this ambiguity, the ISO 80000 specification recommends that log10 (x) should be written lg(x), and loge (x) should be ln(x).

 

[hutchphd]

Typically and historically Optical Density (OD) is defined as the negative of the logarithm (base 10) of the transmission, where the transmission varies between 0 and 1. But of course the multiplicative/additive relationship still holds for ln (base e).
One other point is important: these measures are typically used to describe transparent materials not translucent ones. The translucent ones (like human dermis) contain "large" scatterers and demand mathematics more akin to diffusion than wave propagation. I think I wrote a short paper on this long ago. So light through your finger is a different thing.

 

[russ_watters]

If a substance has a "transmissivity" of 50% over a width of 1 millimeter then what will be the transmissivity of a thickness of 3 millimeters?

It is not linear. Clearly, it will not be 150 percent. It has to be something less than 50. Maybe we divide by the 3 millimeters? Nope, that is not right either.

It will turn out to be exponential. In the case of three millimeters, think of it as three reductions, each by a factor of 50 percent. That is 0.53....

[As @russ_watters surely knows well already]

Yep, you're right - one of the pitfalls of talking about doing math instead of doing math...especially early in the morning.

It's also more different from heat transfer than I realized. In heat transfer you use rates not percentages, so it's closer to what I described (as long as you use the correct parameter...):

R-value is thermal resistance, u-value is its inverse, thermal conductivity.
Insulation with an R value of 2 / inch layered 10" thick is 2*10=R20.
https://en.wikipedia.org/wiki/R-value_(insulation)

Electrical conductivity too.

 

[hutchphd]

If a substance has a "transmissivity" of 50% over a width of 1 millimeter then what will be the transmissivity of a thickness of 3 millimeters?

This is three filters of $$T=0.5~~~~~~~~T_{total}=T^3=.125$$ The optical densities will add $$OD=-ln2~~~~~~~~ {OD}_{total}= 3OD$$