Glass Thickness for Light Absorption: 90%, 99%, 99.9%

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A 0.5 m thick slab of glass absorbs 50% of light, and the discussion revolves around calculating the thickness required for higher absorption rates of 90%, 99%, and 99.9%. The primary equation for light absorption discussed is I(z) = e^(-αz)I(0), where z is the thickness and α is the absorption coefficient. Participants explore various formulas and concepts, including the relationship between intensity and thickness, and the integration of differential equations to derive the absorption expression. The conversation emphasizes the importance of understanding the absorption coefficient and its impact on the required thickness for different absorption levels. Ultimately, the original poster gains clarity on how to approach their homework questions.
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Homework Statement


A slab of glass 0.5 m thick absorbs 50% of light passing through it. Compute how thick of a slab of identical glass you would need that:
- absorbs 90% of light
- absorbs 99% of light
- absorbs 99.9% of light

Homework Equations


I spent an hour looking for formulas in my textbook and couldn't find any, which is the primary reason I'm having trouble.

The Attempt at a Solution


N/A
 
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The simplest mathematical formulation to describe light absorption is through the equation ##I(z) = e^{-\alpha z} I(0)##, where ##z## the distance traveled and ##\alpha## an absorption coefficient.
 
I typed light absorption as a function of glass thickness into the Google search engine and it came up on the first hit.
 
blue_leaf77 said:
The simplest mathematical formulation to describe light absorption is through the equation ##I(z) = e^{-\alpha z} I(0)##, where ##z## the distance traveled and ##\alpha## an absorption coefficient.
So the quantity that we are trying to measure here is intensity? I do see a formula in my textbook saying I(z) = I0e-τ(x). Could that be it?

EDIT: This actually appears to be a formula for intensity through a medium of gas. Nevermind.
 
Astro Student said:
I(z) = I0e-τ(x)
Shouldn't the exponent be a function of z?
 
blue_leaf77 said:
Shouldn't the exponent be a function of z?
Yes, I mistyped it. Both I and tau should be functions of x according to my textbook. I'm still not finding a formula for absorption of light by a solid anywhere in my textbook, only for gases.
 
If you can't find it in your book, you should look for it somewhere else, shouldn't you?
 
Another formula in my textbook says ΔI/Δ = -(nSΔx)σ/S = -nσΔx. σ here is the cross-section, Δx the thickness, and n the number density. For the sake of this problem, the number density and cross-section would remain constant with Δx as the only independent variable to change ΔI/I. Would this be an acceptable solution to this problem?
 
Astro Student said:
ΔI/Δ = -(nSΔx)σ/S = -nσΔx.
This is accurate only when the thickness and/or the absorption coefficient are small. Look what happen if you make the deltas infinitesimal (##\Delta \rightarrow d##),
$$
\frac{dI}{I} = n\sigma dx
$$
and then integrate both sides. What expression will you get?
 
  • #10
Astro Student said:
A slab of glass 0.5 m thick absorbs 50% of light passing through it.

What would two of these slabs do?

I spent an hour looking for formulas in my textbook and couldn't find any, which is the primary reason I'm having trouble.

Perhaps you underestimate your own powers of reason.
 
  • #11
blue_leaf77 said:
This is accurate only when the thickness and/or the absorption coefficient are small. Look what happen if you make the deltas infinitesimal (##\Delta \rightarrow d##),
$$
\frac{dI}{I} = n\sigma dx
$$
and then integrate both sides. What expression will you get?
lnI = -nσ dx + C1.
If I use the knowledge that at x = 0 I = I0 we can use differential equations (I worked them out on a sheet of paper) to get C1 = lnI0 and then I(x) = I0e-nσx.

I think...
 
  • #12
Astro Student said:
lnI = -nσ dx + C1.
If I use the knowledge that at x = 0 I = I0 we can use differential equations (I worked them out on a sheet of paper) to get C1 = lnI0 and then I(x) = I0e-nσx.

I think...

Yes, that's right, and how does it compare to the equation in comment #2?
 
  • #13
blue_leaf77 said:
Yes, that's right, and how does it compare to the equation in comment #2?
It's looks similar. In the equation in comment 2 I would assume number density and cross section were combined into a single constant.
 
  • #14
So, what is still halting you to answer your homework questions?
 
  • #15
blue_leaf77 said:
So, what is still halting you to answer your homework questions?
Nothing anymore. Thank you very much!
 
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