CP A thin uniform film of refractive index 1.750 is placed...

CP A thin uniform film of refractive index 1.750 is placed on a sheet of glass of refractive index 1.50 . At room temperature $\left(20.0^{\circ} \mathrm{C}\right)$, this film is just thick enough for light with wavelength $582.4 \mathrm{nm}$ reflected off the top of the film to be cancelled by light reflected from the top of the glass. After the glass is placed in an oven and slowly heated to $170^{\circ} \mathrm{C},$ you find that the film cancels reflected light with wavelength $588.5 \mathrm{nm}$. What is the coefficient of linear expansion of the film? (Ignore any changes in the refractive index of the film due to the temperature change.)

Key Concepts

Coefficient of Linear Expansion

The coefficient of linear expansion quantifies how much a material's length (or thickness) changes per unit temperature change. It is a fundamental parameter in understanding how temperature variations can modify the physical dimensions of a film, which in turn affects its optical properties, particularly in interference applications.

Thermal Expansion

Thermal expansion refers to the change in a material's dimensions in response to temperature changes. In the context of thin films, an increase in temperature can lead to an increase in film thickness. This change affects the optical path difference and thus alters the interference condition for reflected light.

Thin-Film Interference

Thin-film interference is the phenomenon where light waves reflecting off the boundaries of a thin layer interfere with each other, either constructively or destructively. The effect is determined by the film's thickness, refractive index, and any phase shifts encountered upon reflection, which leads to the colorful patterns seen in soap bubbles or anti-reflective coatings.

Optical Path Difference

Optical path difference (OPD) is the difference in the distance light travels within a medium, taking into account the medium's refractive index. In thin films, the OPD between light reflected from the top surface and that from the bottom surface governs the interference condition—whether the reflected waves will cancel out or enhance each other.

Phase Change on Reflection

When light reflects off a boundary leading to a medium with a higher refractive index, it undergoes a phase change of ? (half a wavelength). This phase shift is critical in thin-film interference calculations because it alters the effective optical path difference between the reflected beams, thereby influencing the interference outcome.

Transcript

00:01 Hi there, so for this problem, we have a thin, uniform field of refractive index that is given, and that index, remember that, well, thing this was labeled that as n, is equal to 1 .75, and it's placed on a sheet of glass of refractive index that is equal to 1 .5.

00:33 Then the question for this problem is about a room temperature that is at a temperature of 20 celsius degrees.

00:48 This film is just thick enough for light with a wavelength.

00:52 So the wavelength given for this is equal to 582 .4 nanometers.

01:01 And after the glasses placed in an oven and slowly heated to a value of 170 sills degrees, then we will obtain that you find that the film comes off the reflected life with a wavelength that is equal to 588 .5 nanometers.

01:19 So the question is, what is the coefficient of linear expansion of the film? so with that said, the first thing that we need to do is it considered the interference between raised reflect? from the upper and lower surfaces of the film to relate the thickness of the film to the wavelengths, for which there is destructive interference.

01:44 The thermal expansion of the film changes the thickness of the film with the temperature changes.

01:50 Now, for this film, on this glass, there is a net of the wavelength, divided by two, phase change due to the reflection, and the condition for destructive interference is that 2 times the thickness equals to the order n times the wavelength divided by the index of refraption, where in this case, n is the value that we are given for this.

02:31 Now, the smallest non -zero thickness is given by the expression, where the thickness is just simply the wavelength, divided by two times.

02:40 The index of refraction.

02:42 So let's calculate this at the two temperatures that we are given...

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