Optics, Minimizing Reflection with Refraction Index Numbers HELP?

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To minimize reflection of 401 nm light on a lens, the minimum thickness of the coating can be calculated using the refractive indices of the coating (1.31) and the glass (1.56). The relevant equations for constructive and destructive interference are provided, which involve the thickness of the coating, the refractive index, and the wavelength of light. A phase change at the boundaries may affect the calculations, requiring adjustments in the equations. The discussion emphasizes the need for guidance due to a lack of resources in the textbook. Understanding these principles is crucial for solving the problem effectively.
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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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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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