Minimizing Reflection of Light: Coating Glass with n = 1.44

dragonrider
Messages
27
Reaction score
0
A material with an index of refraction of 1.44 is used to coat glass. The index of refraction of glass is 1.5. What is the minimum thickness of the coating that will minimize the reflection of light with a wavelength of 4680 A? Answer in units of µm.

I know that the thickness for this is t = λ/4ncoating

But I am not sure on how to find thickness when ncoating = 1.63
 
Last edited:
Physics news on Phys.org
The problem is with reflections at surfaces. You will get a different reflected wave when light reflects off of a higher index of refraction than on a lower index of refraction. I'll let you figure out what the difference is. It should be in your book. Also, I'm assuming that you know how you got your answer for t above when n_{coating}=1.44. If not then you'd better figure that out too.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top