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\(\bullet\) Coating eyeglass lenses. Eyeglass lenses can be coated on the inner surfaces to reduce the reflection of stray light to the eve. 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 nm reflected toward the eye at normal incidence, and (b) will any other wavelengths of visi- ble light be cancelled or enhanced in the reflected light?

Short Answer

Expert verified
(a) Minimum thickness is 96 nm. (b) Other wavelengths may also be canceled/enhanced if they satisfy similar interference conditions for different integers.

Step by step solution

01

Understand Thin Film Interference

When light reflects off the surfaces of a thin film, constructive or destructive interference can occur depending on the thickness of the film and the wavelength of the light. To achieve destructive interference (cancellation of light), the optical path difference must be an odd multiple of half-wavelengths, allowing the reflected wave crests to align with wave troughs.
02

Determine Thickness for Destructive Interference

For the film to cancel a specific wavelength (e.g., 550 nm), the thickness must satisfy the condition for destructive interference. The formula is: \[ 2t n_c = (m + \frac{1}{2})\lambda, \]where \(t\) is the thickness, \(n_c\) is the refractive index of the coating, and \(\lambda\) is the wavelength of the light. For minimum thickness, use \(m = 0\): \[ 2t(1.432) = 0.5 \times 550 \ nm. \]
03

Solve for Thickness

Rearrange the equation to solve for \(t\): \[ t = \frac{0.5 \times 550 \ nm}{2 \times 1.432}. \]Calculate the thickness.
04

Calculation

Substitute the known values into the equation: \[ t = \frac{275}{2 \times 1.432} \ nm = 95.97 \ nm. \]Thus, the minimum thickness of the fluorite coating required to cancel 550 nm light is approximately 96 nm.
05

Determine Other Wavelengths

For constructive or destructive interference of other wavelengths, we use the general formula:\[ 2t n_c = m\lambda' = (m + \frac{1}{2})\lambda'. \]Determine if other values of \(\lambda'\) in the visible spectrum (about 400 to 700 nm) satisfy the equation for integer \(m\). If \(m\) can make the equation hold, then these wavelengths could be canceled or enhanced.

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Key Concepts

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

Optical Path Difference
In the realm of optics, the term "optical path difference" (OPD) plays a crucial role, particularly in scenarios involving thin films. Optical path difference is essentially the difference in the paths traveled by two light waves after they reflect off different surfaces. This difference occurs because light can reflect off either the top or the bottom of a thin film, leading these waves to meet either in phase or out of phase.

When dealing with thin films, like the coating on eyeglass lenses, understanding the OPD is essential for predicting whether the resulting interference will be constructive or destructive.
  • A positive optical path difference can lead to waves being out of phase, resulting in destructive interference, where the light is canceled.
  • A zero or even multiple of wavelengths in OPD leads to in-phase waves, causing constructive interference, where the light is enhanced.
The OPD is influenced by multiple factors, including the thickness of the film and the refractive index of the materials involved. When light enters a different medium, its speed changes, thereby affecting the OPD and the interference pattern observed. This change is governed by the principle of refraction, which relies heavily on understanding the refractive index.
Destructive Interference
Destructive interference occurs when two or more waves align such that their peaks and troughs cancel each other out. For eyewear coatings, achieving destructive interference means reducing unwanted reflections. This principle is especially useful for enhancing the clarity of vision through lenses by eliminating specific wavelengths of light, like the 550 nm mentioned in our exercise.

To ensure destructive interference in thin film applications:
  • The optical path difference must be an odd multiple of half the wavelength of the target light, denoted as \( (m + \frac{1}{2})\lambda \).
  • This causes wave crests to align perfectly with wave troughs.
This calculated alignment ensures that when light reflects off the surface of the coating and interacts, it effectively cancels the specific wavelength. Functionally, this makes the lens appear clearer or reduces glare.

When performing these calculations, one must account for the refractive index of the coating material to achieve the correct precise thickness needed for the desired interference outcome.
Refractive Index
The refractive index is a fundamental concept in optics that measures how much light bends, or refracts, as it passes from one material into another. Every material has its own refractive index, denoted as \( n \), which is a dimensionless number representing the ratio of the speed of light in a vacuum to the speed of light in the material.

When light enters a material with a higher refractive index, it slows down and bends towards the normal line. Conversely, when it passes into a material with a lower refractive index, it speeds up and bends away from the normal.
  • Fluorite, for example, has a refractive index of 1.432, indicating it is less optically dense than the eyeglass material, which has a refractive index of 1.62.
The refractive index is crucial in determining the optical path length within a thin film, as it directly affects how much the light slows down in the material.

This ultimately influences the interference pattern observed, making understanding refractive indices imperative for designing coatings for desired optical effects like anti-reflection.
Wavelength Cancellation
Wavelength cancellation is a direct consequence of destructive interference, where specific wavelengths of light are effectively "canceled out" when they interact with thin films. This occurs when waves meet in such a way that their crest-to-trough alignment causes them to negate each other's effects.

This concept is employed to enhance or minimize reflections of particular wavelengths by designing films at precise thicknesses.
  • For instance, in lens coatings, calculating the right thickness means targeting and canceling specific wavelengths that cause unwanted glare.
  • When the thickness of the film results in the optimal optical path difference for a given wavelength, that wavelength undergoes cancellation.
Through careful calculation of the film's thickness and considering the refractive index, certain wavelengths can be canceled or reduced, while others might become more prominent. This effectively allows for greater control over what light penetrates through or reflects off a given surface, offering enhancements in visibility or reduction of visual noise.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Doorway diffraction. Diffraction occurs for all types of waves, including sound waves. Suppose sound of frequency 1250 Hz leaves a room through a \(1.00-\) wide doorway. At which angles relative to the centerline perpendicular to the doorway will someone outside the room hear no sound? Use 344 \(\mathrm{m} / \mathrm{s}\) for the speed of sound in air and assume that the source and listener are both far enough from the doorway for Fraun- hofer diffraction to apply. You can ignore effects of reflections.

\(\bullet\) Electromagnetic waves of wavelength 0.173 nm fall on a crystal surface. As the angle from the plane is gradually increased, starting at \(0^{\circ},\) you find that the first strong interfer- ence maximum occurs when the beam makes an angle of \(22.4^{\circ}\) with the surface of the crystal planes in the Bragg reflection. (a) What is the distance between the crystal planes? (b) At what other angles will interference maxima occur?

\(\bullet\) The walls of a soap bubble have about the same index of refraction as that of plain water, \(n=1.33 .\) There is air both inside and outside the bubble. (a) What wavelength (in air) of visible light is most strongly reflected from a point on a soap bubble where its wall is 290 \(\mathrm{nm}\) thick? To what color does this correspond ( see Figure 23.3\() ?\) (b) Repeat part (a) for a wall thickness of 340 \(\mathrm{nm}\) .

\(\bullet$$\bullet\) Two identical audio speakers connected to the same amplifier produce in-phase sound waves with a single fre- quency that can be varied between 300 and 600 Hz. The speed of sound is 340 \(\mathrm{m} / \mathrm{s} .\) You find that where you are standing, you hear minimum-intensity sound. (a) Explain why you hear minimum-intensity sound. (b) If one of the speakers is moved 39.8 \(\mathrm{cm}\) toward you, the sound you hear has maximum inten- sity. What is the frequency of the sound? (c) How much closer to you from the position in part (b) must the speaker be moved to the next position where you hear maximum intensity?

\(\bullet\) Two satellites at an altitude of 1200 \(\mathrm{km}\) are separated by 28 \(\mathrm{km} .\) If they broadcast 3.6 \(\mathrm{cm}\) microwaves, what minimum receiving-dish diameter is needed to resolve (by Rayleigh's criterion) the two transmissions?

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