# Transmitted Intensity

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.

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

Homework Helper
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.

Homework Helper
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}
[/itex]

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?

Last edited:
Staff Emeritus
Homework Helper
Another approximation is that air's refractive index is close to 1.

Homework Helper
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.

Staff Emeritus
Homework Helper
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.

Homework Helper
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.

Staff Emeritus