Reflection of light with known refractive index.

[Silversonic]

This is for my 1st year undergrad course but it feels quite basic and not worthy of the 'Advanced Physics' section. Yet there's one thing I'm confused about - I have the answer, just something I don't understand.

Homework Statement

Light falls normally on a glass surface. What fraction of the incident light intensity is reflected if the refractive index n of the glass is 1.5?

Homework Equations

Now R/I = (1-n)/(1+n)

R is the amplitude of the reflective wave and I is the amplitude of the incident wave, R/I is of course their ratio i.e. the fraction of light reflected back. (n is the refractive index)

The Attempt at a Solution

I have the answer but I have no idea why it makes sense. It says

Intensity of reflected light = ((1-n)/(1+n))^2 = ((1-1.5)/(1+1.5))^2 = 0.04.

Now why is it squared? I'd understand if it wasn't but for some reason it is. It's basically saying (R/I)^2 which I can't comprehend. Is this something really obvious? It's quite late at night so maybe I've missed something.

 

[soothsayer]

What you have there isn't a formula for the intensity of the reflected light, its a value for the fraction of light intensity that is reflected, which is why it's dimensionless. The square comes from the fact that the reflected light is not, all other things being equal, as intense as the incident light, it loses energy in the reflection, I think.

Also, "Advanced Physics" usually deals with stuff you'd find in upper division college courses, for future reference.

 

[Stargazer19385]

Light falls normally on a glass surface. ...

Intensity of reflected light = ((1-n)/(1+n))^2

R is the amplitude of the reflective wave and I is the amplitude of the incident wave, R/I is of course their ratio i.e. the fraction of light reflected back. (n is the refractive index)

Thank you for that equation. I'm ray tracing a lens system and need to predict how bad the reflections will be. What is the equation for the intensity of the reflection if the light is not normal to the surface? Is there an equation that is a function of the angle of incidence? Thanks.

And of course I realize that n+1 should be replaced with n1+n2 for glass to glass surfaces.

 

[Stargazer19385]

One other question:
I want to coat an acrylic lens with MgF. Acrylic n = 1.5, and MgF n = 1.38. The reflection between the two will be negligible, 0.0006. The reflection from air to MgF is about 0.025, which is better than 0.04 of air/acrylic. The problem is that Acrylic has a much higher thermal expansion coefficient than glass, and I'm worried the MgF will crack or bubble off on a very cold or hot car day. So here is my question: If I put a MgF layer a few atoms thick on the Acrylic, will that avoid the thermal expansion issue? And will the light even recognize a layer so thin? I think it does for metal plating on Mylar sheets and sun glasses, which are about a few atoms thick. This would not be an issue with glass lenses, but I want to save weight.

 

[Stargazer19385]

And one more question:

I read that multi coatings transmit over 99 percent of light. I don't know how that is possible with the leading surface reflecting at least 2 percent. I realize if the lenses are high index, and you make the film the right thickness, you can send an antiwave that cancels the first reflective wave. But wouldn't that mean you still lost both rays? Or does their zero sum somehow change it to as if both rays did not get reflected and instead made it through? Is it only correct that films can reduce reflections, or is it also true that they greatly increase transmission to 99.7% in some cases?

I thought I read somewhere that antireflective films are not as effective on lenses that have lower indexes of refraction. That makes sense to me, though I still wonder if there are techniques for such lenses as well.

 

[ehild]

That is a rather complicated topics. You should read about Optics of thin films, books of Knittel or Vasicek, for example. They are old books, but very detailed. And you have to know the basic principles of Wave Optics.
Yes, the reflectance depends on the angle of incidence and the polarization of the incident light.
In case of a surface film, interference is involved and it influences the reflectance/transmittance. In case of dielectrics, the sum of the transmittance and reflectance is very nearly 1.
You can get very low reflectance (about zero) with a quarter-lambda coating at normal incidence, at a specified wavelength.
The reflectance can be tailored by multiple-layer coatings.

ehild

 

[Stargazer19385]

Thanks for the leads. Unfortunately my local library only has novels and kids science books, so I'll have to wait until I'm back in a university to use their library, or else spend a lot on books from Amazon. For now I'll try to master the optics chapters in my college physics book. I don't plan to build the lens for another two years, though I just enjoy designing it for now. I want a fully hyperbolic eyepiece for a 90 degree apparent field of view, which no binoculars have.

For now, I mostly just want to know if it is possible to coat a 1.5 index lens in a way to get the reflectance down well below 1%, or if the 2.5% is the only practical goal to accept without paying a fortune.

For the black side of the lens, which I plan to make flat, I will use moth eye texture film for a nearly smooth index transition, resulting in almost 0% reflectance. I think it was Sharp that says they can nano print the film cheaply and will be using it on their TV screens.

 

[ehild]

I can not help you with the lens, but for an antireflecting coating they can use a material with refracting index n=sqrt(n_üveg) and thickness equal to one quarter wavelength, at both sides. That ensures practically zero reflectance at the chosen wavelength, at normal incidence, and also works well for angles less than 20°. For glass, it requires refractive index N=1.225 for the coating and about 120 nm thickness. That low index of refraction can be achieved with layers of nano-particles, or with a porous film. The fabrication is easy and cheap. The attached transmittance spectrum is that of a microscope slide covered with porous silica films on both sides, made by one of my students. The reflection loss is 1-transmittance, about 0.5% at 0.75 μm wavelength.

The actual equation for the reflectance are a bit complicated, so wait till you reach there with your studies.

Thin layer coating is one of the simplest way to reduce reflectance, but there are a couple of other methods. You wrote about moth eye, that is one. A graded-index film reduces reflectance very effectively.

Read http://en.wikipedia.org/wiki/Anti-reflective_coating.

To design wide-angle and broad-band antireflection coatings is a challenge for the researchers. Go ahead!

ehild

 

[Stargazer19385]

Thanks for the reply and graph. Now that I know you have that experience, I'm more interested in asking about the new techniques than in the standard coatings. It is very encouraging that your student was able to cheaply coat a slide with nano particles, although the average reflectance over the whole visual spectrum still looks like 4%: 10% reflected at 400 nm, and 0.5% at the best wavelength. That is an improvement if it is the total for both sides. That is the problem with the light wave cancellation being film thickness dependent. Ideally constructed moth texture should be broad band and broad angle. I take it researchers still have not figured out how to make ideal moth texture inexpensively.

How good is current moth eye texture technology?

Can moth eye texture be applied to a curved surface? In one common news article I read, Sharp uses a nano printer to make moth films for their TV screens. I suspect I'd be able to buy sheets of flat texture cheaply, but unless it stretches nicely, I don't know how to apply it to a curved surface.

I suspect the nano particles are kind of like moth texture, but without as continuous and complete of a graduation. The porous membrane also uses a similar method, but with the coating elements connected before they are attached to the substrate. I also read that even Sharp is not printing ideal moth eye texture, but rather something that looks very similar to the porous texture description, as a cheaper approximation to the sharper moth spike texture.

 

[Stargazer19385]

I already read about the single coatings before coming here, and the square root being the best index for a single film, paired with 1/4 wave length, and that no material has a low enough index unless the substrate has a 1.9 or higher index (or you use those nano particles). My lack of understanding is with multicoats, non-normal angles, and the feasibility of modern methods such as fabricating moth eye texture.

 

[ehild]

Moth-eye structure makes the refractive index graded, gradually decreasing it from bottom to top. Such graded index films smooth out the interference waves of reflectance arising from a coating film, and reduce the reflectance. To reach minimum reflectance in the broadest range, the refractive index has to vary from that of the substrate to that of the surrounding air (1) slowly. The layer has to be thick, to ensure that the change of the refractive index along a wavelength distance is small.

A thick film can cause absorption loss. Also, the moth eye structure can cause light scattering, which reduces transmittance, too. There is no recipe for all.

ehild

 

[Stargazer19385]

I was a math major, not a physics major. I never will take any optics classes, but I have taken many math classes past calculus. Where can I find information on how to calculate the intensity of reflections given the indexes of refraction and angle of incidence? I doubt very much it would be too complex for me.

 

[ehild]

I would suggest Optics of thin films by Knittl or by Vasicek.

You need Maxwell equations, plane-wave solutions of Maxwell equation, boundary conditions at an interface, and you can derive the reflectance and transmittance equations by yourself for both modes, parallel and perpendicular to the plane of incidence.
For a single interface, google "Fresnel equations"

ehild

 

[Stargazer19385]

I would suggest Optics of thin films by Knittl or by Vasicek.

You need Maxwell equations, plane-wave solutions of Maxwell equation, boundary conditions at an interface, and you can derive the reflectance and transmittance equations by yourself for both modes, parallel and perpendicular to the plane of incidence.
For a single interface, google "Fresnel equations"

ehild

Thank you.

There were some Maxwell equations in one of my chemical engineering classes, and some boundary condition differential equations and partial differential equations in some of my other classes. I'm rusty, but I'm sure I'll figure it out again with all those key words and terms you just gave me to search for.

 

[Stargazer19385]

http://en.wikipedia.org/wiki/Fresnel_equations

Looks like I got most of the information I need right in the graphs on Wikipedia. It won't be exact for other indexes, but 1.5 is the one for acrylic, which I want to use, so this works for me.

Looks like the big risk is internal reflection before the light goes from glass back to air. I think this is why many eyepieces have more than one lens, so they can share the angles and keep the incidence below 30 degrees. This is probably why the 82 degree Naglers have 3 achromats, curved right to meet the light as head on as possible each time they bend it. And that is why they say with a plano convex lens, the plano face should face the objective.

I also used Snell's law and some trig and found that chromatic aberration is directly proportional to aperture, but my calculations did not find fast scopes to be any worse than slow ones. So that is another puzzle to figure out later. I based my calculations on the linear displacement on the retina. I also found that a higher index of refraction reduces chromatic aberration, if only the Abbe number stayed the same. With the lower Abbe number of high index glass, my calculations said it is an even trade off.

On other forms, I read that chromatic aberration is proportional to aperture cubed divided by focal length squared. I don't know why my use of Snell's law and trig contradicted that. My calculation even makes sense because the lower spread of the slow scope still reaches the same final spread because of the longer distance. When it reaches the center line, the spread will have the same later width, which will project onto the retina as the same wide spread.