A 420nm layer of plastic (n = 1.22) is placed on top of a piece of glass ( n = 1

Question

A 420nm layer of plastic (n = 1.22) is placed on top of a piece of glass ( n = 1.45). Determine the various visible wavelengths that have maximum constructive and destructive interference for reflection for light shining vertically onto the system. (Visible light in air has a range of wavelengths from about 400nm to about 700nm.)

A 420nm layer of plastic (n = 1.22) is placed on top of a piece of glass ( n = 1.45). Determine the various visible wavelengths that have maximum constructive and destructive interference for reflection for light shining vertically onto the system. (Visible light in air has a range of wavelengths from about 400nm to about 700nm.)

Explanation / Answer

For maximum we will apply 2nt = m(wavelength)

We will need trial and error to find the wavelengths that are in the visible specturm with different m values

2(1.22)(420 X 10^-9) = 1(wavelength)

wavelength = 1025 nm which is NOT A VALID SOLUTION, so try m = 2

2(1.22)(420 X 10^-9) = 2(wavelength)

wavelength = 512.4 nm (VALID ANSWER)

Now for m = 3

2(1.22)(420 X 10^-9) = 3(wavelength)

wavelength = 341.6 nm (NOT A VALID ANSWER)

So the only answer for maximum is 521.4 nm

For the minimum, same idea, but the formula changes slightly

2nt = (m + .5)(wavelength)

The value of 1 for m gives a valid answer

2(1.22)(420 X 10^-9) = 1.5(wavelength)

wavelength = 683.2 nm (VALID ANSWER)

For m = 2

2(1.22)(420 X 10^-9) = 2.5(wavelength)

wavelength = 409.9 nm (VALID ANSWER)

The only two answers for minimum are 683.2 nm and 409.9 nm

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