Transmitted Intensity

• Math Jeans
In summary: But you are right, the statement "good quality optical glass" does have a specific meaning in this context and it is important to make that clear. And yes, there are definitely types of glass and frequencies of light that would not give accurate results with the given formula. Thank you for pointing that out!In summary, the intensity transmitted through a glass slab with an index of refraction of n, when normally incident light is used, is approximately given by the formula I_T = I_0 * [4 * n / (n + 1)^2]^2. This is derived from the formula for reflected intensity, which is given by I_R = I_0 * [(n_2 - n_1) /

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.

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

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).

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.

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?

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

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.

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.

Redbelly98 said:
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.

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.

What is transmitted intensity?

Transmitted intensity is a measure of the amount of light or radiation that passes through a material or medium.

How is transmitted intensity calculated?

Transmitted intensity is calculated by taking the ratio of the transmitted intensity to the incident intensity, and then multiplying by 100 to get a percentage.

What factors affect transmitted intensity?

The factors that affect transmitted intensity include the properties of the material (such as its thickness and composition), the wavelength of the incident light, and the angle of incidence.

What is the relationship between transmitted intensity and absorbance?

Transmitted intensity and absorbance have an inverse relationship, meaning that as one increases, the other decreases. This is because as more light is absorbed by a material, less is transmitted through it.

Why is transmitted intensity important in science?

Transmitted intensity is important in science because it allows us to understand how light or radiation behaves when it passes through a material. This can help us determine the properties of the material and how it interacts with different wavelengths of light.