Question

An airplane is made invisible to radar by coating it with a 5.00 -mm-thick layer of an antireflective polymer with the index of refraction \(n=1.50 .\) What is the wavelength of radar waves for which the plane is made invisible?

Short answer

Answer: The wavelength of radar waves for which the airplane is made invisible is 15.0 mm.

Step-by-step solution

Write down the given values and the formula for destructive interference

We are given the thickness of the polymer coating, \(t = 5.00\,\text{mm} = 5.00\times10^{-3}\,\text{m}\), and the index of refraction of the polymer, \(n = 1.50\). The condition for destructive interference is given by: $$(2nt = m\lambda)$$ where our goal is to solve for the wavelength of the radar waves, \(\lambda\).

Rearrange the formula to solve for \(\lambda\)

To find the wavelength of the radar waves, we need to isolate \(\lambda\) in the interference formula. We can do this by dividing both sides of the formula by \(m\): $$\lambda = \frac{2nt}{m}$$

Find the minimum value of \(m\) that satisfies the condition for destructive interference

Since \(m\) is an integer and we want to get the smallest wavelength that causes destructive interference, we will use the smallest integer value that still satisfies the condition, which is \(m=1\). This will ensure that the radar waves with this wavelength experience destructive interference, making the airplane invisible.

Calculate the wavelength of radar waves for which the plane is made invisible

Using the value of \(m=1\), we can now calculate the smallest wavelength of radar waves that experience destructive interference upon reflecting at the interface: $$\lambda = \frac{2nt}{m} = \frac{2(1.50)(5.00\times10^{-3}\,\text{m})}{1} = 1.50\times10^{-2}\,\text{m}$$ So, the wavelength of radar waves for which the plane is made invisible is \(\lambda = 1.50\times10^{-2}\,\text{m}\) or \(15.0\,\text{mm}\).

Key concepts

Antireflective Coating

Antireflective coatings are designed to reduce reflection from surfaces. They work mainly through a principle called destructive interference. By applying a thin layer of material with a certain thickness and refractive index, unwanted reflections can be minimized.

The special layer is typically engineered so that the light reflecting off the top surface of the coating and the light reflecting off the surface beneath the coating cancel each other out. This happens when the two waves are out of phase by half a wavelength and effectively destroy each other, creating "destructive interference."

• This concept is essential in many applications ranging from eyeglasses to camera lenses.
• By manipulating the thickness and refractive index, the coatings can be tuned to specific wavelengths, such as radar.

In our exercise, the 5.00 mm thick polymer helps make the airplane "invisible" to radar by creating destructive interference for radar waves with a 15.0 mm wavelength.

Index of Refraction

The index of refraction, often symbolized by \(n\), is a fundamental property of materials that describes how light travels through them. It quantifies how much light slows down and bends when passing through the material compared to air. This property influences the amount of bending of light rays at the interface of two different media.

The formula for destructive interference involves the index of refraction, symbolized by \(n\). A material's index affects the thickness needed for antireflective coatings to work correctly.

• The index is crucial in calculations for any type of coating.
• The higher the index, the slower the light travels through the material.

In our problem, the polymer has an index of refraction of 1.50, indicating the speed reduction of light when entering it. This information was crucial for designing the coating to interfere destructively with radar waves.

Radar Invisibility

Radar invisibility is achieved when an object cannot be detected by radar waves, generally through clever engineering of surfaces to disrupt or negate signals. Antireflective coatings play a key role in this process in some stealth vehicles.

By covering a plane with a specific polymer, constructive radar wave reflections are minimized, achieving "invisibility." The coating exploits the principle of destructive interference so that the reflected radar signal is significantly reduced or even cancelled.

• This makes radar invisibility an excellent example of applying optical principles to practical problems.
• Stealth technology is vital in modern military applications.

Achieving radar invisibility involves meticulous calculation of index of refraction and thickness to ensure the highest degree of interference at key radar frequencies. In our exercise, this means calculating the wavelength of radar for which the plane becomes 'invisible' because of its antireflective properties.