How internal reflection affect the performance of the thin film emissive display devices?
Probably many people here could answer if you gave more details.
Just explain a little bit more the context. Is it an exam question? Is it a practical question for an application?
Remind us about what is called a "thin film emissive display".
What do you call an "emissive display".
Why do the "thin film" come into play, what is its role?
Could you give us a picture of this device?
Without these background information I can only suggest you to read about the theory of thin film optics and the role of the refractive index of the film. The final conclusion depends on the background I am missing.
It's just a normal paper work question.
Sometimes you can just forget thin film, and take emissive display into account.
Emissive display is a display which convert electricity directly into light.
Thank you guys~
How can we answer a question about a system without precise description.
The geometry of a thin film is extremely important of course.
But we need to know about the emissive display ("ED") too.
Of course, we could make some assumption to fill in the gaps.
Let's assume this ED is made of a transparent substrate.
Let's assume this substrate is very thick (reflexion on the interface will never come back)
Let's assume inside this substrate there is some light emission.
A long time ago I wanted to use a prism of alumina for a microwave optics setup.
Alumina was convenient for my application because of its high refractive index.
But the problem was that high refractive index would also imply high reflexion losses.
Therefore I re-derived the theory of thin film coating and found the classical result.
If I remember well, the reflexions can be totally supressed by a coating if:
- the refractive index is the square root of the substrate refractive index (assume air index = 1)
- the thickness of the coating has to be chosen properly (1/4 wavelength I think)
If the refractive index of the coating or its thicknes are not as needed, the reflexion losses will increase. From the Maxwell's equation the full details of the reflexion losses can be calculated. Probably you can find the result on wiki.
It's not easy to answer, anyway thank u very much~
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