Optics, Minimizing Reflection with Refraction Index Numbers HELP?

In summary, the conversation is about finding the minimum thickness of a coating on a lens to minimize reflection of 401 nm light. The index of refraction of the coating material is 1.31 and the index of the glass is 1.56. The person seeking help is asking for an equation or guidance on how to approach the problem. Two potential equations are given, one for constructive interference and one for destructive interference, with a variable representing the phase change at each surface. The thickness of the coating is denoted as "d" and the refractive index of the film as "n2".
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Homework Statement


What is the minimum thickness of coating which should be placed on a lens in order to minimize reflection of 401 nm light? The index of refraction of the coating material is 1.31 and the index of the glass is 1.56.


Homework Equations


I haven't been able to attempt the problem because the book we use for optics has nothing covering this information. If you don't mind giving me an equation to start with or some guidance on how to start the problem I \'d greatly appreciate it. Our teacher was out of town so we were left to try and figure this out on by ourselves... :/ Thanks for anything you can help me with!


The Attempt at a Solution

 
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  • #2
I believe that there are two potential equations that could help you. I am by no means great at this (since we also are going over it, and i have a test tomorrow) but i think the equations given to me may help.

{ } + (2dn2)/λ = m this is for constructive
{ } + (2dn2)/λ = m + 0.5 this is for destructive

the { } a variable: Add 1/2 if phase change occurs @ 2nd surface, Subtract 1/2 if phase change occurs at 1st surface.

d = thickness
n2 = refractive index of film

hopefully this helps you. if you have any questions about anything, let me know. also the n2 is n(subscript)2.
 
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Related to Optics, Minimizing Reflection with Refraction Index Numbers HELP?

1. What is the purpose of minimizing reflection with refraction index numbers?

The purpose of minimizing reflection with refraction index numbers is to reduce the amount of light that is lost due to reflections at the interface between two materials. This can improve the efficiency and clarity of optical systems, such as lenses and mirrors.

2. How do refraction index numbers affect reflection?

The refraction index is a measure of how much a material bends light. When light passes through a material with a higher refraction index, such as glass, some of the light is reflected at the interface between the material and air. By adjusting the refraction index of the material, the amount of light reflected can be minimized.

3. Can all materials have their refraction index adjusted?

No, not all materials have adjustable refraction index numbers. The refraction index is a property of the material itself and is determined by its chemical composition. However, some materials, such as glasses and plastics, can have their refraction index altered by adding certain chemicals or by changing their physical structure.

4. How do you calculate the refraction index of a material?

The refraction index of a material can be calculated by measuring the angle at which light is bent when passing through the material and comparing it to the angle at which light would be bent if it were passing through air. This is known as Snell's law: n1sinθ1 = n2sinθ2, where n1 and n2 are the refraction indices of the two materials and θ1 and θ2 are the angles of incidence and refraction, respectively.

5. Can refraction index numbers be used to completely eliminate reflection?

No, it is not possible to completely eliminate reflection with refraction index numbers. Some amount of reflection will always occur at the interface between two materials. However, by carefully selecting and adjusting the refraction indices of materials, reflection can be minimized to a very low level.

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