A layer of transparent plastic (n=1.61) on glass (n=1.52) looks dark when viewed in reflected light whose wavelength is 589 nanometers in vacuum. Find the two smallest possible nonzero values for the thickness of the layer.
Answers are 183 nm and 366 but I don't understand it. I got 91.46 and 183. I realize I divided by two twice but why would you only divide by two once since it's destructive interference and you get 2 pathlengths from the standard equation.
Sorry I got 91.46 and 274.379. You were supposed to add 1/2 wavelength for the second value but why can't you add 1 whole wavelength especially since destructive interference is 1/2 wavelengths, 3/2 wavelengths, 5/2...etc.?
To understand why the thickness of the layer needs to be divided by two only once when calculating the two smallest possible nonzero values, let's go through the calculation and reasoning step by step.
When light reflects off a thin film, such as the layer of transparent plastic on glass in this case, interference can occur between the reflected waves from the top and bottom surfaces. Depending on the phase difference between these waves, constructive or destructive interference can happen.
For destructive interference to occur, the path length difference between the two waves should be equal to an odd multiple of half the wavelength. Mathematically, this is expressed as:
2t = (2n1 - 1)λ / 2
Where:
- t is the thickness of the layer
- n1 is the refractive index of the transparent plastic
- λ is the wavelength of light in vacuum
Now we can substitute the given values:
- n1 = 1.61 (refractive index of the transparent plastic)
- λ = 589 nm (wavelength of light in vacuum)
Let's solve this equation to find the two smallest possible nonzero values for t.
First, divide both sides of the equation by 2:
t = (2n1 - 1)λ / 4
Next, substitute the provided values:
t = (2 * 1.61 - 1) * 589 nm / 4
t = (3.22 - 1) * 589 nm / 4
t = 2.22 * 589 nm / 4
t ≈ 325 nm
So, one possible nonzero value for the thickness of the layer is approximately 325 nm.
Now, let's calculate the second possible nonzero value. Since we need destructive interference, there can also be a path length difference of λ/2 (half the wavelength) in addition to the odd multiple of λ/2. In this case, we multiply the odd multiple by 2:
t = (2 * 2n1 - 1)λ / 4
Substituting the values again:
t = (2 * 2 * 1.61 - 1) * 589 nm / 4
t = (6.44 - 1) * 589 nm / 4
t = 5.44 * 589 nm / 4
t ≈ 793 nm
So, the second possible nonzero value for the thickness of the layer is approximately 793 nm.
Therefore, the two smallest possible nonzero values for the thickness of the layer are approximately 325 nm and 793 nm.