Guest

how a non-reflecting film reduces reflection?derive the expression for thickness of non-reflecting films?

how a non-reflecting film reduces reflection?derive the expression for thickness of non-reflecting films?

Grade:

1 Answers

Ashwin Sinha
520 Points
13 years ago

Dear Sathya Vinod,

Anti-Reflection Coatings

Thin film anti-reflection coatings greatly reduce the light loss in multi-element lenses by making use of phase changes and the dependence of the reflectivity on index of refraction. A single quarter-wavelength coating of optimum index can eliminate reflection at one wavelength. Multi-layer coatings can reduce the loss over the visible spectrum.

The idea behind anti-reflection coatings is that the creation of a double interface by means of a thin film gives you two reflected waves. If these waves are out of phase, they partially or totally cancel. If the coating is a quarter wavelength thickness and the coating has an index of refraction less that the glass it is coating then the two reflections are 180 degrees out of phase.

Thin films Example calculation
Index

Reflection concepts
 













Multi-Layer Anti-Reflection Coatings

A single layer anti-reflection coating can be made non-reflective only at one wavelength, usually at the middle of the visible. Multiple layers are more effective over the entire visible spectrum.


 













Single Layer Antireflection Coating

A single layer anti-reflection coating can be made non-reflective only at one wavelength, usually at the middle of the visible. Multiple layers are more effective over the entire visible spectrum.

Single layer antireflection coatings are generally calculated for a midrange wavelength like 550 nm (green). With the assumption of a coating thickness of a quarter wavelength in the medium, the reflection can be calculated by using the normal incidence reflection coefficients.

For a coating of index
on glass of index
the amplitude of reflection from the first surface (air-to-coating) is
=
The amplitude of reflection off the back surface (coating-to-glass) is
=
Since these reflections are 180° out of phase, the resulting reflected intensity (square of amplitude) is
 %
Without the coating, the reflected intensity would be  %.
The transmitted intensity is then %
compared to % for an uncoated surface.
The reflected intensities at the extreme wavelengths of the visible spectrum are:
Reflected intensity at 400 nm:  %
Reflected intensity at 700 nm:  %
Reflected intensity at nm =  %

You may have noticed that the lenses of your camera or binoculars have a purplish tint to them when viewed at a glancing angle like the illustrations below. This is a result of optimizing the antireflection coating for the midrange wavelengths, the greens. More of the reflected light is in the red and blue extremes of the visible spectrum. Analyzed from the point of view of the CIE chromaticity diagram, combining red and blue puts you on the "line of purples". Even with the simplerNewton color circle, you can see that emphasizing red and blue will give you somethin akin to magenta.


 


 

 

 

Please feel free to post as many doubts on our discussion forum as you can.
If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly.
We are all IITians and here to help you in your IIT JEE  & AIEEE preparation.

All the best…………
 
Regards,
Askiitians Expert
Ashwin

Plz. approve my answer by clicking 'Yes' given below, if you loved it......... Plz. Plz. Plz. don't forget to do it.......WinkSmileCool

Think You Can Provide A Better Answer ?

ASK QUESTION

Get your questions answered by the expert for free