Thin Films with consideration to Amplitude

[cblack]

Homework Statement

Consider the case of a film of thickness [tex]\lambda[/tex]/4 with an index nf where nf<ng=1.5 that is applied to a glass surface. For minimum reflection the reflected rays r1 and r2 must cancel each other out. Since r1 and r2 both reflect fast to slow, they are in phase on reflection, but with r2 traveling [tex]\lambda[/tex]/2 farther than r1 they will be 180 degrees out of phase when r2 exits the film and superpositions with r1. For maximum cancellation, the intensities (ie. amplitude) must be as close to equal as possible.

calculate nf for minimum reflection.

Homework Equations

nf=[tex]\lambda[/tex]air/[tex]\lambda[/tex]coating

The Attempt at a Solution

I don't understand how the amplitude would change after passing through the film or how you would even work that into an equation. I don't have any equations that involve amplitude and it was never discussed in the text. I have no idea how to approach this problem or what equations should be used.

I assumed that [tex]\lambda[/tex]air= 550nm since it's a midrange wavelength of visible light, but that's as far as I can go.

 

[anathema]

Please help!
Thank you for posting your question. I would like to help you understand the problem and provide you with some guidance on how to approach it.

First, let's review what we know about the situation. We have a film with a thickness of λ/4 and an index of refraction of nf, which is less than the index of the glass surface ng=1.5. We also know that for minimum reflection, the reflected rays r1 and r2 must cancel each other out. This means that the amplitude of r1 must be equal to the amplitude of r2, but they must be out of phase by 180 degrees.

Now, let's think about what happens to the amplitude of the reflected rays as they pass through the film. When light passes through a medium with a different index of refraction, its amplitude changes. This phenomenon is known as refraction. In this case, both r1 and r2 are reflecting from fast to slow mediums, so their amplitudes will decrease after passing through the film.

To calculate the new amplitudes, we can use the following equation:

A2=A1(n1/n2)

Where A1 is the original amplitude, A2 is the new amplitude, n1 is the index of refraction of the first medium (in this case, air), and n2 is the index of refraction of the second medium (in this case, the film). Since we are only interested in the relative amplitudes of r1 and r2, we can ignore the constant factor of (n1/n2) and focus on the ratio of their amplitudes.

Now, let's consider the distance that r2 travels in the film. Since the film has a thickness of λ/4, r2 will travel a distance of λ/2 in the film. This means that when r2 exits the film and superpositions with r1, they will be 180 degrees out of phase, as mentioned earlier.

Next, we can use the principle of superposition to determine the total amplitude of the reflected rays after they have superpositioned. This can be done by adding the individual amplitudes of r1 and r2:

A_total=A1+A2

However, since we know that r1 and r2 are out of phase by 180 degrees, we need to use the following equation to calculate the total amplitude:

A_total=A1-A2

Now,

 

[Moetasim]

I would approach this problem by first understanding the concept of thin films and how they interact with light. Thin films are layers of material that are much thinner than the wavelength of light, and they can cause interference and reflection of light. In this case, we are dealing with a film of thickness \lambda/4, which means that the film is one-quarter of the wavelength of light.

Next, I would consider the index of refraction of the film, which is given as nf. This index is a measure of how much the speed of light changes when passing through the film compared to air (ng). In this case, nf is smaller than ng, which means that the film is a slower medium for light.

To calculate nf for minimum reflection, we can use the equation nf=\lambdaair/\lambdacoating. Since we know that the film thickness is \lambda/4, we can substitute this value into the equation and solve for nf. This will give us the index of refraction of the film that will result in minimum reflection.

However, the question also mentions the concept of amplitude and how it affects the reflection of light. In this case, we need to consider the intensity (amplitude) of the reflected rays r1 and r2. For maximum cancellation, the intensities of these two rays must be as close to equal as possible. This means that we need to find the index of refraction that will result in equal intensities for r1 and r2.

To do this, we can use the concept of Fresnel equations, which describe the reflection and transmission of light at an interface between two media. These equations take into account the amplitude of the light as well as the index of refraction of the two media. By solving for the index of refraction that results in equal intensities using the Fresnel equations, we can find the value of nf for minimum reflection.

In summary, to find nf for minimum reflection, we need to use the equation nf=\lambdaair/\lambdacoating and also consider the concept of amplitude and use Fresnel equations to find the index of refraction that results in equal intensities for the reflected rays. It is important to understand the principles and equations involved in order to approach this problem correctly.