Problem 51 - University Physics with Modern Physics

Question

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.)

Step-by-Step Solution

Answer

The coefficient of linear expansion of the film is approximately \\(2.59 \times 10^{-5} \, ^{\circ}C^{-1}\\).

Step 1: Determine the Initial Thickness of the Film

The condition for destructive interference for a film with wavelengths is given by \(2nt = m\lambda\), where \(n\) is the refractive index, \(t\) is the thickness, \(m\) is an integer, and \(\lambda\) is the wavelength. Initially, with \(n = 1.750\) and \(\lambda_0 = 582.4\,\text{nm}\), the equation is \(2 \times 1.750 \times t = m \times 582.4\). Solving for \(t\), we get \(t = \frac{m \times 582.4}{2 \times 1.750}\).

Step 2: Determine the New Thickness of the Film

After heating to \(170^{\circ}\text{C}\), the destructive interference condition changes because the wavelength in air changes to \(\lambda_1 = 588.5\,\text{nm}\). Using the same condition \(2nt' = m\lambda'\), we have \(2 \times 1.750 \times t' = m \times 588.5\). Solving for \(t'\), we obtain \(t' = \frac{m \times 588.5}{2 \times 1.750}\).

Step 3: Calculate Change in Thickness

Subtract the initial thickness from the changed thickness: \(\Delta t = t' - t = \left(\frac{m \times 588.5}{2 \times 1.750}\right) - \left(\frac{m \times 582.4}{2 \times 1.750}\right) = \frac{m \times (588.5 - 582.4)}{2 \times 1.750}\).

Step 4: Calculate the Coefficient of Linear Expansion

The coefficient of linear expansion, \(\alpha\), is defined by \(\Delta t = \alpha t \Delta T\), where \(\Delta T = 170 - 20 = 150^{\circ}C\). Substituting the values, we have \(\alpha = \frac{\Delta t}{t \times 150}\). Use the change in thickness formula to replace \(\Delta t\) in this equation, and solve for \(\alpha\).

Key Concepts

Refractive Index

The refractive index, often symbolized by the letter 'n', is a dimensionless number that indicates how much light slows down when traveling through a medium compared to its speed in a vacuum. It tells us the bending degree of light as it passes from one medium into another. In terms of formula, the refractive index is expressed as:

• The ratio of the speed of light in vacuum to the speed in the medium: \(n = \frac{c}{v}\), where \(c\) is the speed of light in a vacuum, and \(v\) is the speed of light in the medium.

For example, a refractive index of 1.750 means light travels 1.750 times slower in the film compared to vacuum. Similarly, a refractive index of 1.50 implies that light is slowed less than in the film. Understanding refractive indices is crucial because they directly affect how light waves interact and interfere in films, which leads to phenomena like colored reflections or complete cancellation, known as destructive interference.

Coefficient of Linear Expansion

The coefficient of linear expansion is a critical property when studying how materials behave under temperature changes. Denoted often by \(\alpha\), it describes how much a material expands per unit length for every degree change in temperature. This is particularly relevant for precise applications like coatings or films.

• Linear expansion formula: \(\Delta L = \alpha \cdot L_0 \cdot \Delta T\), where \(\Delta L\) is the change in length, \(L_0\) is the original length, and \(\Delta T\) is the temperature change.

In this particular problem, the film’s thickness changed when it was heated from \(20^{\circ}C\) to \(170^{\circ}C\), causing a shift in the interference pattern—indicating expansion. Calculating the coefficient of linear expansion involves measuring how much the material expanded due to the said temperature change, providing insights into the material's thermal sensitivity.

Destructive Interference

Destructive interference occurs when two or more waves overlap and combine in such a way that they cancel each other. For light waves, this happens when the peaks of one wave align perfectly with the troughs of another, resulting in a net nullification of the wave's amplitude. With respect to thin films, this phenomenon is key to understanding and predicting color patterns and reflectivity changes.

• Condition for destructive interference: A path difference of \((m+\frac{1}{2})\cdot\lambda\), where \(m\) is an integer and \(\lambda\) is the wavelength.

In the described exercise, the light reflected from the top and bottom interfaces of the film experience destructive interference at specific wavelengths. Initially, a wavelength of 582.4 nm was canceled, but as the film heated and expanded, this changed to 588.5 nm, further illuminating the real-world effects of thermal expansion on interference.