How Is Light Intensity Transmitted Through a Glass Slab Calculated?

[Math Jeans]

Homework Statement

Show that for normally incident light, the intensity transmmitted through a glass slab with an index of refraction of n is approximately given by:

$$I_T = I_0 \cdot [\frac{4 \cdot n}{(n + 1) ^ 2}]^ 2$$

Homework Equations

Reflected intensity is given by:

$$I_R = I_0 \cdot [\frac{n_2 - n_1}{n_2 + n_1}] ^ 2$$

The Attempt at a Solution

I've attempted to solve this numerous times, however, the equation that I always end up with is different than the one shown.

$$I_T = I_0 \cdot [\frac{4 \cdot n}{(n + 1) ^ 2}]$$

 

[alphysicist]

Hi Math Jeans,

What approach did you use? What is the relationship of $I_R$ and $I_T$? It seemed to give the right answer (once the identification of $n_2 \to n$ and $n_1 \to 1$ is made, of course).

 

[Math Jeans]

Well. My approach was to use I_T=I_0-I_R, and simply use the formulas from that point. But like I said, I don't know how to turn that into an approximation.

 

[alphysicist]

I believe the approximation they refer to is the approximation they make in deriving your expression for $I_R$ in the first place (they assume that $\mu\approx\mu_0$ to derive it).

However, what I think you are not taking into account is that the light that is transmitted through the glass slab has to be transmitted through two surfaces; each surface transmits the quantity you found:

[tex] I_T = I_0 \frac{4 n}{(n+1)^2}<br /> [/itex]<br /> <br /> if an intensity of $I_0$ hits it. (Assuming the glass is in air, of course.) The total fraction of transmitted light passing through both surfaces is the expression they give. Do you get that result?[/tex]

 

[Redbelly98]

Another approximation is that air's refractive index is close to 1.

 

[alphysicist]

After I posted I realized that yet another approximation is the assumption that all of the light that passes through the first surface makes it to the other surface; depending on the glass and the light frequency this could be a very bad assumption.

 

[Redbelly98]

I think it's a reasonable assumption, or at least what is required to get the answer given in post #1.

Absorption or scattering losses in a good quality optical glass would indeed be negligible.

 

[alphysicist]

I think it's a reasonable assumption, or at least what is required to get the answer given in post #1.

Well, all of the assumptions/approximations are required to get the given answer.

Absorption or scattering losses in a good quality optical glass would indeed be negligible.

If you don't mind exploring this a bit further--I think your statement here is a bit circular in its logic. Absorption or scattering losses being negligible is what we mean by high quality glass. When we say "good quality optical glass" that means two things: the glass has a very low extinction coefficient for a range of frequencies due to its makeup/manufacturing; and also that we are using frequencies of light for which it is very transparent (in that range for which the coefficients are low).


(Of course I am considering UV light as "light", which is a matter of definition of which an infinite amount of argument could be had!)

My point is that setting $n_{\rm air}\approx 1$ is almost "universally" good in the sense that I believe it would be very unusual to have a case in which the error in making this approximation would be large.

However, I think it would be easy to find types of glass and/or frequencies of light that would give a huge error in the given formula. We have to specify the glass and specify the light frequency--namely, the high quality optical glass you mention in your post and probably visible light (to cover the majority of glasses).

(I'm also not sure about setting $\mu_{\rm glass}\approx\mu_0$; I don't know if there are common types of glass for which this is a bad approximation. I don't think there would be.)



But I have to say I am definitely no expert in the properties of glass! If you think what I have written is wrong I'd appreciate you letting me know.

 

[Redbelly98]

No, what you are saying is quite reasonable. Having spent many years working in optics labs, I am used to thinking of quality optical glass as commonplace and so I just sort of take it for granted.