Why is a particular wavelength favored by transmission through a thin film?

Question

A sheet of glass is coated with a $500\text{ nm}$ thick layer of oil ($n=1.42$). For what visible wavelengths of light do the reflected waves interfere
a) constructively?
b) destructively?

c). What is the color of reflected light? What is the color of transmitted light?

Here are the solutions:

21.60. Model: Reflection is maximized for constructive interference of the two reflected waves, but minimized for destructive interference.

Solve: (a) Constructive interference of the reflected waves occurs for wavelengths given by EQuation 21.32: $$\lambda_m = \frac{2nd}{m} = \frac{2(1.42)(500\text{ nm})}{m} = \frac{(1420\text{ nm})}{m}$$ Thus, $\lambda_1 = 1420\text{ nm}$, $\lambda_2 = \frac{1}{2}(1420\text{ nm}) = 710\text{ nm}$, $\lambda_3 = 473\text{ nm}$, $\lambda_4 = 355\text{ nm}$, ... Only the wavelength of $473\text{ nm}$ is in the visible range.

(b) For destructive interference of the reflected waves, $$\lambda = \frac{2nd}{m - \frac{1}{2}} = \frac{2(1.42)(500\text{ nm})}{m - \frac{1}{2}} = \frac{1420\text{ nm}}{m - \frac{1}{2}}$$ Thus, $\lambda_1 = 2\times 1420\text{ nm} = 2840\text{ nm}$, $\lambda_2 = \frac{2}{3}(1420\text{ nm}) = 947\text{ nm}$, $\lambda_3 = 568\text{ nm}$, $\lambda_4 = 406\text{ nm}$, ... The wavelengths of $406\text{ nm}$ and $568\text{ nm}$ are in the visible range.

(c) Beyond the limits $430\text{ nm}$ and $690\text{ nm}$ the eye's sensitivity drops to about 1 percent of its maximum value. The reflected light is enhanced in blue ($473\text{ nm}$). The transmitted light at mostly $568\text{ nm}$ will be yellowish green.

Now, here is the part that bothers me. C) does not make sense to me. The reflected light is enhanced in blue is correct in my head. But why would the transmitted light be mostly at 568 nm?

If you send white light through this, all waves are reflected AND transmitted. The transmission does not ensue phase changes. But the reflection does. So, if you look only at those transmitted waves, why would only 568 nm be favored?

Answer 1

Imagine you have a monochromator that you can tune to give you whatever wavelength you like. Now send light at a wavelength of 568 nm towards the interface.

What has been calculated is that destructive interference takes place, so that less of the incident radiation is reflected. Let's take an extreme case where the reflection from the first and second interfaces give reflected light of similar amplitude (but out of phase). This can be achieved by selecting an appropriate value for the refractive index of the film. In this case NO light is reflected.

So where does the energy of the incoming electromagnetic waves go? It must all be in the transmitted wave.

Now if you send light of a different wavelength, some of that is reflected and some of it is transmitted. i.e. not all of it is transmitted as in the case of light at 568 nm.

The net effect will be a greenish tinge to the transmitted light.

Edit: The OP is puzzled that the destructive interference between the reflected waves leads to enhanced transmission through the film compared with a situation that some reflection occurs. The easiest way to think about this definitely is conservation of energy. If (i) there is no reflected component at one wavelength, but (ii) there is at another, and if the Poynting vectors of the incident EM waves are the same in each case, then the power present in the transmitted EM wave must be enhanced in case (i) simply to conserve energy - to be exact, for normal incidence, the flux of the transmitted waves must equal the flux of the incident waves in case (i).

Let's take the case of normal incidence and (i) where we have managed to arrange that the amplitude of the reflected wave at the 2nd interface (and then transmitted back through the first interface) equals the amplitude of the reflected wave at the first interface. This is achieved by setting the refractive index of the oil film to be somewhere between that of the air and glass layers. In this case the summed E-field of the net reflected EM wave is zero - perfect cancellation. But we know from Faraday's law that the component of the E-field parallel to the plane of the interface must be the same immediately on either side of the interface. So the transmitted wave E-field at the first interface must have the same amplitude as the incident wave E-field.

In case (ii) we have a similarly thick film, but the wavelength is changed so we do not get destructive interference, but the amplitudes of the waves reflected at the first interface and second interface (then transmitted back through the first interface) are still equal. In this case the net reflected wave amplitude will be non-zero and have some phase shift that results in the total E-field on the air side of the interface being less than the E-field of the incident wave alone. Thus the transmitted wave also has an amplitude that is smaller than that of the incident wave.

Now you have to go through a similar consideration inside the film, but there is the additional complexity that some of the light reflected from the 2nd interface is also internally reflected off the first interface back into the oil (and so on...)! All this has to be solved in a set of simultaneous equations. However the basic fact is that in case (i) the amplitude of the EM wave transmitted through the first interface is larger than in case (ii). This in turn leads to a larger transmitted wave through the 2nd interface because the ratio of transmitted to incident light at this second interface depends only on their respective refractive indices and hence is the same in both cases.

Hopefully I have convinced you that it is easier to think about where the energy goes...