Thin-film interference problem

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In summary, the problem involves light of wavelength 470 nm passing through a block of material with a refractive index of 1.30, some of which is reflected and transmitted through air (with a refractive index of 1.00) and then reflected off another block with the same refractive index. The minimum thickness needed for constructive interference is given by t = (m+1/2)*lambda/2n, while for destructive interference it is given by t = m*lambda/2n. The refractive index is taken into account in the wavelength used in the equation. If t is zero, there would be no interference at all.
  • #1
NewtonianAlch
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Homework Statement


Light of wavelength 470 nm passes through a block of material with a refractive index n = 1.30, some of it is reflected off the air-block boundary and some of it is transmitted through air (n=1.00) with thickness t, and is then reflected off another block also with n = 1.30.

What is the minimum thickness, t, for which you will observe:

Constructive interference
Destructive intereference

Homework Equations



t = (m+1/2)*lamda / 2n

t = m*lamda / 2n

The Attempt at a Solution



m would be 0 since it's asking for minimum thickness, therefore destructive interference would always be zero wouldn't it?

I'm a bit confused because another book has the equation:

t = (m + 1/2)*lamda / 2 <- what happened to the refraction index?
 
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  • #2
NewtonianAlch said:
m would be 0 since it's asking for minimum thickness, therefore destructive interference would always be zero wouldn't it?
If t was zero there would be no reflection off the second block, and no interference at all! So you'd not have the setup that the problem specifies.

I'm a bit confused because another book has the equation:

t = (m + 1/2)*lamda / 2 <- what happened to the refraction index?
The refractive index is "bundled" in with the wavelength, lambda; The wavelength used in the equation is the wavelength within the given medium, which depends upon the refractive index.
 
  • #3
gneill said:
If t was zero there would be no reflection off the second block, and no interference at all! So you'd not have the setup that the problem specifies.
Doh! Massive fail.

The refractive index is "bundled" in with the wavelength, lambda; The wavelength used in the equation is the wavelength within the given medium, which depends upon the refractive index.
Ah I see what you mean now.

Thanks a lot!
 

What is thin-film interference?

Thin-film interference is a phenomenon that occurs when light waves reflect off the top and bottom surfaces of a thin film with different refractive indices. This results in the interference of the light waves, causing certain wavelengths to be enhanced and others to be cancelled out.

What causes thin-film interference?

Thin-film interference is caused by the difference in refractive indices between the two materials that make up the thin film. When light passes through these materials, it can be reflected at each interface and the reflected waves can interfere with each other.

What is the most common application of thin-film interference?

The most common application of thin-film interference is in the production of anti-reflective coatings for glasses and camera lenses. By manipulating the thickness and refractive index of the thin film, unwanted reflections can be reduced, resulting in clearer and sharper images.

How does the thickness of the thin film affect interference patterns?

The thickness of the thin film is directly related to the interference pattern. As the thickness increases, the number of interference fringes also increases. This means that thicker films will have a more complex interference pattern with more peaks and valleys.

What is the difference between constructive and destructive interference in thin-film interference?

Constructive interference occurs when the light waves reflected from the top and bottom surfaces of the thin film are in phase, resulting in a stronger amplitude and a bright spot in the interference pattern. Destructive interference, on the other hand, occurs when the waves are out of phase, resulting in a cancellation of light and a dark spot in the interference pattern.

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