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What is the thinnest film of a coating with \(n=1.42\) on glass \((n=1.52)\) for which destructive interference of the red component \((650 \mathrm{nm})\) of an incident white light beam in air can take place by reflection?

Short Answer

Expert verified
The thinnest film of such a coating for destructive interference to occur is 229 nm.

Step by step solution

01

Analyze the nature of interference

Given that destructive interference takes place, the path difference must be \( (2n - 1) \lambda /2 \) or an odd multiple of half-wavelengths where n is the refractive index of the film. Here, we are asked to find the thinnest film, which means we should take the smallest odd number, that is 1.
02

Find the Wavelength of Light Inside the Film

The given wavelength is for light in air or vacuum. The wavelength of light inside the film is given by \( \lambda' = \lambda / n = 650 nm / 1.42 \). We calculate this to be approximately 458 nm.
03

Calculate the Thickness of the Film

Now, calculating the thickness of the film using the formula for destructive interference \( t = (2n - 1) \lambda' /2 \). Substituting the values we get \( t = (2.1-1) 458 nm / 2 \) which simplifies to \( t = 229 nm \).

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Key Concepts

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

Thin Film Interference
Thin film interference is an extraordinary phenomenon that occurs when light waves reflected by the upper and lower boundaries of a thin film, like a soap bubble or a layer of oil, interact with each other. This interaction leads to interference, which can either enhance or diminish the intensity of the reflected light.
When the thickness of a thin film is in the range of the wavelength of light, the light waves can interfere in a constructive or destructive manner. Constructive interference leads to brighter, more intense colors, whereas destructive interference may cause the colors to appear dimmer or even disappear. This happens because:
  • Constructive interference occurs when the path difference between the light reflected from the top and bottom surfaces of the film is an integer multiple of the wavelength.
  • Destructive interference occurs when this path difference is an odd multiple of half-wavelengths, resulting in the cancellation of some wavelengths.
A practical application of thin film interference is in anti-reflective coatings, where specific destructive interference is utilized to reduce glare and increase the transmission of light through lenses.
Refractive Index
The refractive index is a crucial concept that helps us understand how light propagates through different mediums. It's a dimensionless number that indicates how much the speed of light is reduced within a substance compared to vacuum. The formula to find the refractive index, usually denoted as "n," is:\[ n = \frac{c}{v} \]where:
  • \( c \) is the speed of light in a vacuum, typically about \( 3 \times 10^8 \) meters per second.
  • \( v \) is the speed of light in the material.
A material with a higher refractive index implies that light travels slower in that material. In the scenario of thin film interference, the refractive index is crucial for calculating the wavelength of light within the film.
For example, a film of refractive index 1.42 implies that light travels slower by a factor of 1.42 compared to a vacuum. This change in speed results in a change in the wavelength, which directly affects how we perceive light's interference when reflecting from thin films.
Wavelength of Light
The wavelength of light is another fundamental concept in understanding thin film interference. It is the distance between successive crests of a wave and determines the color of light that we see. Normally, when light travels from one medium to another, its speed and consequently its wavelength change, although its frequency remains constant.
In the case of our problem, the given in vacuum or air wavelength is 650 nm, representing the red component of white light. When this light enters a medium with a refractive index (such as our film with n=1.42), the wavelength inside the film becomes:\[ \lambda' = \frac{\lambda}{n} \]Thus, for our red component:\[ \lambda' = \frac{650 \mathrm{nm}}{1.42} \approx 458 \mathrm{nm} \]This shorter wavelength is crucial in determining the optical path difference leading to interference.
This transformation in wavelength affects the conditions for interference and is used to find the minimum thickness for destructive interference in a thin film. Knowing the behavior of light's wavelength in different media, aids in predicting and manipulating light to achieve desired effects, such as those found in optics and photonics technologies.

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Most popular questions from this chapter

Coherent light of frequency \(6.32 \times 10^{14} \mathrm{~Hz}\) passes through two thin slits and falls on a screen \(85.0 \mathrm{~cm}\) away. You observe that the third bright fringe occurs at \(\pm 3.11 \mathrm{~cm}\) on either side of the central bright fringe. (a) How far apart are the two slits? (b) At what distance from the central bright fringe will the third dark fringe occur?

In a two-slit interference pattern, the intensity at the peak of the central maximum is \(I_{0}\). (a) At a point in the pattern where the phase difference between the waves from the two slits is \(60.0^{\circ}\), what is the intensity? (b) What is the path difference for \(480 \mathrm{nm}\) light from the two slits at a point where the phase difference is \(60.0^{\circ} ?\)

Newton's rings are visible when a planoconvex lens is placed on a flat glass surface. For a particular lens with an index of refraction of \(n=1.50\) and a glass plate with an index of \(n=1.80,\) the diameter of the third bright ring is \(0.640 \mathrm{~mm}\). If water \((n=1.33)\) now fills the space between the lens and the glass plate, what is the new diameter of this ring? Assume the radius of curvature of the lens is much greater than the wavelength of the light.

Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eye. If the lenses are medium flint glass of refractive index 1.62 and the coating is fluorite of refractive index \(1.432,\) (a) what minimum thickness of film is needed on the lenses to cancel light of wavelength \(550 \mathrm{nm}\) reflected toward the eye at normal incidence? (b) Will any other wavelengths of visible light be cancelled or enhanced in the reflected light?

A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index \(1.50 .\) At room temperature \(\left(20.0^{\circ} \mathrm{C}\right)\), this film is just thick enough for light with wavelength \(582.4 \mathrm{nm}\) reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to \(170^{\circ} \mathrm{C},\) you find that the film cancels reflected light with wavelength \(588.5 \mathrm{nm} .\) What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.

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