Maximum/minimum reflection | Thin-film interference

• IamNotAPhysicist
In summary, the minimum thickness for a transparent film to produce reflection with the smallest angle of incidence is d1=0.1.
IamNotAPhysicist

Homework Statement

A transparent film( n = 1.45) is placed on a glass surface( n = 1.14) and is illuminated by light ( incidence angle α = 60°. What is the minimum thickness d1 of film such that reflection of light with λ is minimized, while if the thickness is d2 then with the same λ the reflection is maximized?

Homework Equations

2dcosβ = (m - .5)λ/n
β = 36.6°
m = 0,1,2,3...
d - film thickness
λ = light wavelength in vacuum (air)

The Attempt at a Solution

I am not good at physics, but I am supposed to do my homework. So, I pointed out that I am dealing with constructive interference and found the equation I showed above(used "Physics for Scientists and Engineers" A.Serway for that).
My first problem is to find λ. I tried to substitute d2 in the equation and then I understood that I don't know what value of m I should use. For example, I used m = 1 and got λ = 648×10^(-9). But in this case, how am I supposed to find d1? If I use the same equaton, the same m and λ I will get d1 = d2. I've got completely confused by these maximum and minimum reflections

I don't seem to understand the question. Is there some kind of relation between d1 and d2? The way I see it, they look like they could be anything. What are their values? What are the values you need to find? Wavelength or distance?

You don't need to find λ. You are supposed to take λ as given, and then find d1 and d2. You answers for d1 and d2 will contain λ, so they will both be functions of λ.

Are you sure about the formula: 2dcosβ = (m - .5)λ/n
If the light ray travels a perpendicular distance d within the film then
wouldn"t the actual ray travel a distance L = d / cosβ ?

IamNotAPhysicist said:
If I use the same equation, the same m and λ I will get d1 = d2. I've got completely confused by these maximum and minimum reflections
You have different equation for interference maximums and minimums. For a maximum, 2dncosβ + λ/2= mλ. For a minimum, the interference is destructive, 2dncosβ + λ/2 = (2m+1)λ/2 ---> 2dncosβ=mλ.

lekh2003
J Hann said:
Are you sure about the formula: 2dcosβ = (m - .5)λ/n
If the light ray travels a perpendicular distance d within the film then
wouldn"t the actual ray travel a distance L = d / cosβ ?
Actually, the ray travels 2d/cosβ in the layer. And the optical path has to be taken into account when calculating interference, which is the geometric path multiplied with the refractive index. The other ray it interferes with, reflected directly at the interface, also travels some distance from their common phase plane AB. The optical path difference between the blue and red rays is 2 n AF -BC.

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ehild said:
Actually, the ray travels 2d/cosβ in the layer. And the optical path has to be taken into account when calculating interference, which is the geometric path multiplied with the refractive index. The other ray it interferes with, reflected directly at the interface, also travels some distance from their common phase plane AB. The optical path difference between the blue and red rays is 2 n AF -BC.
View attachment 214923
Agreed!
The phase difference in the plane wave, as you specified
2 n d / cos β - 2 d tan β sin α does simplify to 2 n d cos β
since sin α = n sin β

1. What is maximum reflection in thin-film interference?

Maximum reflection in thin-film interference occurs when the incident light reflects off the top and bottom surfaces of a thin film at the same phase, resulting in constructive interference. This leads to a bright spot in the reflected light spectrum.

2. How is the maximum reflection point determined in thin-film interference?

The maximum reflection point in thin-film interference is determined by the thickness of the film, the refractive indices of the film and the surrounding medium, and the angle of incidence of the light. These factors must be in the right combination to achieve constructive interference.

3. What is the relationship between maximum reflection and minimum reflection in thin-film interference?

The relationship between maximum and minimum reflection in thin-film interference is based on the principle of interference. Maximum reflection occurs when the reflected waves are in phase, while minimum reflection occurs when the reflected waves are out of phase, resulting in destructive interference.

4. How can the maximum reflection point be shifted in thin-film interference?

The maximum reflection point in thin-film interference can be shifted by changing the thickness of the film or the refractive indices of the film and surrounding medium. This can be achieved by using materials with different properties or by adjusting the angle of incidence of the light.

5. What practical applications does maximum reflection in thin-film interference have?

Maximum reflection in thin-film interference has several practical applications, such as in anti-reflective coatings for glasses and camera lenses, in optical filters for cameras and telescopes, and in the production of colorful patterns on soap bubbles and oil slicks. It is also used in the technology of thin-film solar cells and in the design of thin-film interference devices for measuring the thickness of thin films.

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