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A possible means for making an airplane invisible to radar is to coat the plane with an antireflective polymer. If radar waves have a wavelength of \(3.00 \mathrm{~cm}\) and the index of refraction of the polymer is \(n=1.50\), how thick would you make the coating?

Short Answer

Expert verified
The thickness of the coating should be 1.00 cm.

Step by step solution

01

Understand the Phenomenon

The problem is using the phenomena of destructive interference. For destructive interference to occur, the path difference must be half the wavelength, or using the formula, layer thickness (t) is equal to \( \frac{wavelength}{2n} \) where wavelength is the wavelength of the light and n is the refractive index.
02

Substituting and Solving

Substitute the given wavelength (3.00 cm) and refractive index (1.50) to calculate the layer thickness. So we get \[ t = \frac{3.00 cm}{2*1.50} = 1.00 cm\]

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Key Concepts

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

Destructive Interference
Destructive interference happens when two waves meet in such a way that their crests and troughs are aligned to cancel each other out. This is especially useful in physics for reducing reflection or making something like a coating practically invisible to certain wavelengths. In the context of antireflective coatings, destructive interference involves a thin film of material on a surface causing reflected waves to interfere with each other in a destructive way.

For this cancellation to happen, the path difference between light waves must be equal to half of the wavelength they are handling. Using the equation: \[t = \frac{\text{wavelength}}{2n}\] where **t** is the thickness of the antireflective coating, and **n** is the refractive index of the coating material. By ensuring the thickness of the coating meets this requirement, the reflected waves cancel each other out, significantly reducing the reflection.
Wavelength
The wavelength of a wave is the distance between two consecutive crests or troughs of the wave. It's a critical factor when designing antireflective coatings because it determines the thickness needed for effective interference. In optics, and specifically when dealing with radar waves, calculating the correct wavelength of light is essential for achieving the desired effect.

In our example, we are working with radar waves having a wavelength of 3.00 cm. This specific length is because radar operates at a specific frequency. Using the wavelength in the formula, we find the required thickness of a polymer coating that would cause destructive interference, ensuring that the radar waves are significantly diminished or nullified as they reflect from the surface. By tailoring the thickness proportionally to the wavelength, the coating can significantly reduce radar visibility.
Refractive Index
The refractive index is a measure of how much light bends, or refracts, as it passes from one medium to another. It indicates how fast light travels through a material compared to how fast it travels in a vacuum. The refractive index is denoted by **n** and is a dimensionless number that plays a crucial role in designing antireflective coatings.

In the exercise at hand, the refractive index of the polymer is given as 1.50. This means light will travel 1.50 times slower in the polymer compared to a vacuum. When dealing with antireflective coatings, given the wavelength of the light, the refractive index helps determine how thick the coating should be for maximizing destructive interference.

Adjusting the thickness of the coating according to the refractive index is essential because it controls the phase change when light reflects off the surface. By precisely tuning the thickness using the refractive index, materials engineers can create coatings that effectively decouple the reflection and transmission of waves, leading to applications like stealth technology in aviation.

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Most popular questions from this chapter

Two rectangular optically flat plates \((n=1.52)\) are in contact along one end and are separated along the other end by a \(2.00-\mu m-\) thick spacer (Fig. P24.24). The top plate is illuminated by monochromatic light of wavelength \(546.1 \mathrm{~nm} .\) Calculate the number of dark parallel bands crossing the top plate (including the dark band at zero thickness along the edge of contact between the plates).

The two speakers are placed \(35.0 \mathrm{~cm}\) apart. A single oscillator makes the speakers vibrate in phase at a frequency of \(2.00 \mathrm{kH}_{z}\). At what angles, measured from the perpendicular bisector of the line joining the speakers, would a distant observer hear maximum sound intensity? Minimum sound intensity? (Take the speed of sound to be \(340 \mathrm{~m} / \mathrm{s} .)\)

A light beam is incident on a piece of fused quartz \((n=\) 1.458) at the Brewster's angle. Find the (a) value of Brewster's angle and (b) the angle of refraction for the transmitted ray.

Light with a wavelength in vacuum of \(546.1 \mathrm{~nm}\) falls perpendicularly on a biological specimen that is \(1.000 \mu \mathrm{m}\) thick. The light splits into two beams polarized at right angles, for which the indices of refraction are \(1.320\) and \(1.333\), respectively. (a) Calculate the wavelength of each component of the light while it is traversing the specimen. (b) Calculate the phase difference between the two beams when they emerge from the specimen.

Light of wavelength \(5.30 \times 10^{2} \mathrm{~nm}\) illuminates a pair of slits separated by \(0.300 \mathrm{~mm}\). If a screen is placed \(2.00 \mathrm{~m}\) from the slits, determine the distance between the first and second dark fringes.

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