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A thin layer of magnesium fluoride \((n=1.38)\) is used to (a) What thickness should the coat a flint-glass lens \((n=1.61)\). magnesium fluoride film have if the reflection of \(565-\mathrm{nm}\) light is to be suppressed? Assume that the light is incident at right angles to the film. (b) If it is desired to suppress the reflection of light with a higher frequency, should the coating of magnesium fluoride be made thinner or thicker? Explain.

Short Answer

Expert verified
(a) The thickness should be approximately 102.36 nm. (b) Thinner for higher frequency light.

Step by step solution

01

Identify the Condition for Destructive Interference

To suppress reflection, the film should cause destructive interference. Destructive interference for thin film happens when the optical path difference is half of the wavelength.
02

Use the Formula for Minimum Thickness

For destructive interference, the minimum thickness \( t \) is given by \( t = \frac{ ext{wavelength in film}}{4} \). The wavelength in the film is \( \lambda/n \) where \( \lambda \) is the given wavelength in vacuum and \( n \) is the refractive index of the film.
03

Calculate the Wavelength in the Film

Convert the wavelength of the light to the wavelength in magnesium fluoride: \( \lambda_f = \frac{565\, \text{nm}}{1.38} \approx 409.42 \, \text{nm} \).
04

Calculate the Minimum Thickness

Apply the formula for minimum thickness: \( t = \frac{409.42\, \text{nm}}{4} \approx 102.36 \, \text{nm} \).
05

Analyze Frequency Effect for Thickness

For higher frequency light, the wavelength decreases. To maintain destructive interference, the thickness must be decreased because the optical path length needs to stay in step with the wavelength changes.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Destructive Interference
In the realm of thin film interference, destructive interference is a fascinating phenomenon. It occurs when two waves combine in such a way that they cancel each other out. This can happen with light waves in a thin film, such as a coating on a lens. For destructive interference to happen in our scenario, the optical path difference, which is basically the extra distance one wave travels over another, should be equal to half the wavelength of the light.
This ensures that the waves are perfectly out of phase, meaning the crest of one aligns with the trough of the other, leading to cancellation. By suppressing reflection, this phenomena plays a crucial role in antireflective coatings, making the lenses less shiny and improving the clarity of lenses or glasses.
Optical Path Difference
The concept of optical path difference is key to understanding thin film interference. It is the effective difference in distance that light travels when passing through a medium, compared to traveling the same path in a vacuum. When light hits a thin film, part of it reflects off the top surface, while another part continues through, reflecting off the bottom surface of the film.
When these two reflections meet again, they may either enhance or cancel each other, depending on the optical path difference.
  • If the path difference is a multiple of the wavelength, they will constructively interfere, reinforcing the light.
  • If it's a half-multiple of the wavelength, they destructively interfere, cancelling out the light.
This is how a thin film can produce beautiful colors or reduce unwanted reflections, by manipulating the optical path difference.
Refractive Index
Refractive index is a fundamental property of materials that affects how light travels through them. It's defined as the ratio of the speed of light in vacuum to the speed of light in the material. For thin films, the refractive index determines how much the light will bend or slow down when it passes through the material.
This bending alters the wavelength of the light while it's inside the film, which is a central aspect when calculating the conditions for interference.
In our example, magnesium fluoride has a refractive index ( ) of 1.38, meaning light travels slower through it compared to a vacuum by that factor. This refractive index directly influences the thickness that the film needs to be for achieving destructive interference for a given wavelength of light.
Minimum Thickness Calculation
Calculating the minimum thickness of a thin film for destructive interference is a practical application of the concepts we've discussed. It involves determining the smallest thickness at which a thin film will cause specific wavelengths of light to destructively interfere. The formula often used is:
\[ t = \frac{\text{wavelength in film}}{4} \]where the wavelength in the film is found by dividing the wavelength in vacuum by the refractive index of the film.
Using this formula ensures that the optical path difference is half of the wavelength, achieving destructive interference.
If you need to adjust the film for different frequencies of light (higher frequency means shorter wavelengths), you'd correspondingly decrease the thickness of the film to maintain effective interference, as demonstrated in the original problem.

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Most popular questions from this chapter

A screen is placed \(1.00 \mathrm{m}\) behind a single slit. The central maximum in the resulting diffraction pattern on the screen is \(1.60 \mathrm{cm}\) wide- that is, the two first-order diffraction minima are separated by \(1.60 \mathrm{cm}\). What is the distance between the two second-order minima?

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