Question

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

Short answer

Answer: The wavelength of the radar waves for which the airplane is made invisible is \(3.00 \times 10^{-2}\mathrm{m}\), or 30.0 mm.

Step-by-step solution

Identify the given information

The thickness of the antireflective polymer is 5.00 mm, and its index of refraction is 1.50. The thickness of the layer and index of refraction are important factors in determining the wavelength of radar waves for which the plane is made invisible.

Define equation for destructive interference

When the reflected waves from the front and back surfaces of the coating cancel each other out, destructive interference occurs. For this to happen, the optical path difference between the two reflecting surfaces must be equal to an odd multiple of half the wavelength \((2nt = (2m + 1)\frac{\lambda}{2})\), where \(t\) is the thickness of the layer, \(n\) is the index of refraction, \(m\) is an integer, and \(\lambda\) is the wavelength of the radar waves.

Solve for the wavelength \(\lambda\)

We can rearrange the destructive interference equation to solve for the wavelength \(\lambda\): \(\lambda = \frac{4nt}{(2m + 1)}\). Since we want the smallest wavelength for which the plane is made invisible, m = 0 (this is the fundamental mode of interference pattern). Now we can plug in the known values of \(n = 1.50\) and \(t = 5.00\mathrm{mm} = 5.00 \times 10^{-3}\mathrm{m}\): \(\lambda = \frac{4(1.50)(5.00 \times 10^{-3}\mathrm{m})}{(2(0) + 1)}\) \(\lambda = \frac{6(5.00 \times 10^{-3}\mathrm{m})}{1} = 3.00 \times 10^{-2}\mathrm{m}\).

Present the final answer

The wavelength of the radar waves for which the plane is made invisible is \(3.00 \times 10^{-2}\mathrm{m}\), or 30.0 mm.