Problem 45 - 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 (20.0\(^\circ\)C), this film is just thick enough for light with wavelength 582.4 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\(^\circ\)C, you find that the film cancels reflected light with wavelength 588.5 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 \( 6.87 \times 10^{-6} \text{/°C} \).

Step 1: Understand the Condition for Destructive Interference

To achieve destructive interference, the optical path difference must be an odd multiple of half the wavelength: \( 2n_{film}d = (m+\frac{1}{2}) \lambda \). Here \( n_{film} = 1.750 \), and \( \lambda \) represents the wavelengths (582.4 nm and 588.5 nm) at temperatures 20.0° C and 170° C. "\( m \)" is an integer that remains constant between temperatures since only the film length changes between those conditions, and thus the interference frequency remains the same.

Step 2: Relate Film Thickness to Temperature-Dependent Wavelengths

Calculate the thickness condition at 20.0° C: \( d = \frac{(m+\frac{1}{2}) \times 582.4 \text{ nm}}{2 \times 1.750} \). Similarly, at 170° C: \( d_{170} = \frac{(m+\frac{1}{2}) \times 588.5 \text{ nm}}{2 \times 1.750} \). Since m is constant for both equations, their ratio will give the dilation due to thermal expansion.

Step 3: Calculate Change in Thickness

By relating the two thicknesses, we have:\[ \frac{d_{170}}{d} = \frac{588.5 \text{ nm}}{582.4 \text{ nm}} = 1 + \alpha(T_{170} - T_{20}) \] where \( \alpha \) is the coefficient of linear expansion, \( T_{170} = 170 \) and \( T_{20} = 20 \) (both in degrees Celsius).

Step 4: Solve for the Coefficient of Linear Expansion \( \alpha \)

We know:\[ 1 + \alpha(170 - 20) = \frac{588.5}{582.4} \]Thus,\[ \alpha = \frac{\frac{588.5}{582.4} - 1}{150} \] Calculate \( \alpha \) using the change in thickness ratio.

Key Concepts

Refractive Index

When light travels through different media, its speed and direction change. This change is measured by a property called the refractive index (or index of refraction). The refractive index, denoted as \( n \), is a measure of how much the speed of light is reduced inside a medium compared to its speed in a vacuum. In this exercise, we are dealing with a thin film of refractive index 1.750 placed on glass with a refractive index of 1.50.

• A higher refractive index means that light moves slower through the medium.
• The speed reduction influences how light waves interfere with each other.

It is also crucial to understand that refractive indices can change with temperature, but in this problem, any such changes are ignored.

Wavelength Dependency

The interference pattern of light is highly dependent on its wavelength. Wavelength, which is the distance between successive peaks of a light wave, is a fundamental characteristic of light that affects how it interacts with materials. In this exercise, we observe that different wavelengths (582.4 nm and 588.5 nm) are involved at different temperatures.

• At room temperature (20.0°C), the wavelength involved in the destructive interference is 582.4 nm.
• Upon heating to 170°C, the wavelength at which destructive interference occurs changes to 588.5 nm.

The wavelength dependency is due to the fact that the optical path difference must match specific conditions for interference, which varies with the wavelength of light used.

Thermal Expansion

Materials expand when heated; this is known as thermal expansion. It's a common response to increasing temperature and is quantified in terms of the coefficient of linear expansion, \( \alpha \). This coefficient gives us a measure of how much a material's dimensions increase with temperature.

• In this exercise, the film expands as the temperature increases from 20°C to 170°C.
• Due to expansion, the film's thickness changes, altering the conditions for destructive interference.

The calculation involves comparing the film thickness at two temperatures to determine \( \alpha \), using the known change in interference wavelengths.

Destructive Interference

Destructive interference occurs when two light waves meet in such a way that they cancel each other out. For destructive interference to happen in reflected light, the optical path difference must be an odd multiple of half the wavelength. This condition ensures that the crests of one wave align with the troughs of another, leading to cancellation.

• In the context of the film, when light reflects off the top and bottom surfaces, the corresponding waves interfere.
• At room temperature, 582.4 nm light experiences destructive interference. After heating, this shifts to 588.5 nm.

This shift in wavelength is a critical indicator used to measure the thermal expansion of the film, as the change signifies a modification in the optical path contributing to the interference.