Problem 32 - University Physics with Modern Physics

Question

Light with wavelength 648 \(\mathrm{nm}\) in air is incident perpendicularly from air on a film 8.76\(\mu \mathrm{m}\) thick and with refractive index \(1.35 .\) Part of the light is reflected from the first surface of the film, and part enters the film and is reflected back at the second surface, where the film is again in contact with air. (a) How many waves are contained along the path of this second part of the light in its round trip through the film? (b) What is the phase difference between these two parts of the light as they leave the film?

Step-by-Step Solution

Answer

a) 36.5 waves, b) Phase difference is \(\pi\).

Step 1: Understanding the Problem

We need to calculate two things: (a) the number of waves contained in the round trip of light through the film, and (b) the phase difference between light reflections at the first and second surfaces.

Step 1: Calculate the Wavelength in the Film

The wavelength of light in a medium is given by the formula: \( \lambda_{film} = \frac{\lambda_{air}}{n} \), where \( \lambda_{air} = 648 \ nm \) and \( n = 1.35 \). Thus, \( \lambda_{film} = \frac{648}{1.35} \ nm \approx 480 \ nm \).

Step 2: Calculate the Path Length of the Round Trip

The round trip path length in the film is twice the thickness because light travels to the back of the film and returns. The thickness is \( 8.76 \ \, \mu m = 8760 \ nm \). Hence, the round trip path length is \( 2 \times 8760 \ nm = 17520 \ nm \).

Step 3: Calculate the Number of Waves in the Round Trip

The number of wavelengths in the round trip is given by \( n_{waves} = \frac{\text{Path length}}{\lambda_{film}} = \frac{17520}{480} = 36.5 \).

Step 4: Calculate the Phase Difference

The phase difference between the two beams can be determined by considering both path difference and reflection phase shifts. Each reflection at the film boundary introduces a phase shift of \( \pi \) (or half a wavelength), and since there are two reflections, this contributes to an additional \( 2 \pi \) to the phase difference. Hence, the total phase difference is \( 2 \pi \times 36.5 + 2 \pi \), which effectively simplifies due to periodicity to \( \pi \).

Key Concepts

Wavelength in Medium

When light travels from one medium to another, its speed changes, therefore altering its wavelength. This phenomenon is crucial in understanding thin film interference. The change in wavelength is due to the refractive index of the medium.

In a medium, the wavelength is given by the formula:

• \(\lambda_{medium} = \frac{\lambda_{air}}{n}\)

where \(n\) is the refractive index. In this exercise, light with an initial wavelength of 648 nm in air enters a medium with a refractive index of 1.35. This results in a new wavelength inside the medium:

• \(\lambda_{film} = \frac{648}{1.35} \ nm \approx 480 \ nm\)

Understanding this change is essential for calculating how light waves interact within different materials, which is the basis for creating vibrant colors in thin films.

Phase Difference

The phase difference between two light beams is significant in thin film interference, governing how they combine. This difference is determined by the light's path length and possible phase shifts due to reflections.

For light reflected off a surface, a phase shift of \(\pi\) (equivalent to half a wavelength) occurs if it reflects off a medium of higher refractive index. In the given scenario, both reflections from the film's surfaces induce phase shifts of \(\pi\). Therefore, their cumulative contribution to phase difference is \(2\pi\).

Also, the path length contributes to the phase difference. Since the number of complete wavelengths that fit into the round trip path length is 36.5, it adds up to

• \(2\pi \times 36.5\)

This simplifies, due to the mod 2\(\pi\) periodicity of phase, to just \(\pi\). Such phase differences determine whether light waves constructively or destructively interfere.

Refractive Index

The refractive index, represented as \(n\), is a dimensionless number describing how fast light travels through a medium. It plays a pivotal role in determining the wavelength of light within a material. A higher refractive index means light slows down more in that medium, resulting in a shorter wavelength than in air.

For example, the refractive index for the given thin film is 1.35. This index value implies that light travels 1.35 times slower in the film than in a vacuum. It affects how light bends and reflects at the boundaries of different substances, which is why understanding it is crucial for calculations such as wavelength in the medium and phase differences in thin film interference.

Overall, the refractive index helps predict how light will behave as it moves between different materials, affecting everything from color appearance to optical focusing properties.

Path Length Calculation

Thin film interference involves calculating the path length that light travels within a film. In particular, for a round trip path from one surface of a film back to the same surface, light travels twice the film's thickness. This total path length can be used to find the number of wavelengths that light travels, impacting interference effects.

For a film with a thickness of 8.76 µm (which converts to 8760 nm), the round trip path length is

• \(2 \times 8760 \ nm = 17520 \ nm\)

By dividing this path length by the wavelength of light in the film (480 nm in this case), one calculates the number of waves contained in this path:

• \(\frac{17520}{480} = 36.5\)

Knowing this helps in deciding how the light waves interfere with each other—constructively or destructively—resulting in visible patterns of light and color.