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Chapter 34: Problem 61

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

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

Step by step solution

01

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, \(\lambda\).
02

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 \(m\): $$\lambda = \frac{2nt}{m}$$
03

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.
04

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 \(15.0\,\text{mm}\).

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