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35.51. 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.)

Short Answer

Expert verified
The coefficient of linear expansion is approximately \( 7.0 \times 10^{-5} \text{°C}^{-1} \).

Step by step solution

01

Understand the Interference Condition

For destructive interference to occur in a thin film when light is reflected, the phase difference should be such that the path difference is half the wavelength, i.e., the optical path length difference is \( \frac{\lambda}{2} \). Given the refractive index of the film \( n = 1.750 \), we have:\[ 2 n t = (m + 0.5) \lambda \]where \( t \) is the thickness of the film, \( m \) is the order of interference, and \( \lambda \) is the wavelength of light.
02

Apply Initial Conditions

For the initial condition at 20°C, the destructive interference condition is given by:\[ 2 \cdot 1.750 \cdot t_0 = (m + 0.5) \times 582.4 \, \text{nm} \]where \( t_0 \) is the initial thickness.
03

Apply Final Conditions

After heating the film to 170°C, the destructive interference condition becomes:\[ 2 \cdot 1.750 \cdot t_f = (m + 0.5) \times 588.5 \, \text{nm} \]where \( t_f \) is the final thickness, considering the same order of interference \( m \).
04

Relate Initial and Final Thicknesses

Since the wavelength changes but the order \( m \) remains constant, we relate the initial and final conditions:\[ 2 \cdot 1.750 \cdot t_0 = (m + 0.5) \times 582.4 \]\[ 2 \cdot 1.750 \cdot t_f = (m + 0.5) \times 588.5 \]Thus,\[ t_f / t_0 = \frac{588.5}{582.4} \approx 1.0105 \]
05

Calculate the Coefficient of Linear Expansion

The coefficient of linear expansion \( \alpha \) is defined as:\[ \alpha = \frac{\Delta t}{t_0 \Delta T} = \frac{t_f - t_0}{t_0 (170 - 20)} \]Substitute \( t_f = 1.0105 \times t_0 \) into the equation:\[ \alpha = \frac{1.0105 \times t_0 - t_0}{t_0 \times 150} = \frac{0.0105}{150} \approx 7.0 \times 10^{-5} \, \text{°C}^{-1} \]
06

Verify Units and Accuracy

Check the calculations for any arithmetic errors and ensure that the final answer is presented with the correct units as the coefficient of linear expansion: \( \text{°C}^{-1} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Refractive Index
When we discuss thin films in optics, the refractive index plays a crucial role. It is a measure of how much the speed of light is reduced inside a medium compared to a vacuum. This property affects how light waves bend and interfere when passing through or reflecting within a material.
In the context of thin film interference, the refractive index determines the phase change of light as it reflects and refracts. For example, if a film has a refractive index of 1.750, it means light travels slower in this medium than in air or a vacuum. This slowing down of light results in changing the wavelength of light inside the medium which is foundational for interference patterns.
The refractive index helps us to understand how the optical path length in the film is affected, which is crucial for calculating interference conditions like destructive interference.
Coefficient of Linear Expansion
The coefficient of linear expansion is essential when considering how materials respond to temperature changes. This coefficient tells us how much a material's length changes for each degree of temperature change.
Mathematically, it is represented as \[ \alpha = \frac{\Delta L}{L_0 \Delta T} \] where - \( \Delta L \) is the change in length - \( L_0 \) is the original length - \( \Delta T \) is the change in temperature.
In this exercise, we calculate the coefficient of linear expansion of a film by considering how its thickness changes when heated from 20°C to 170°C. As materials expand, the thickness of the film changes, affecting the interference conditions of light reflected from the film.
Knowing the coefficient of linear expansion helps in precisely predicting the behavior of thin films under varying temperatures, thus influencing their optical characteristics.
Destructive Interference
Destructive interference occurs when two light waves combine to produce a resultant wave with a reduced amplitude. For thin films, this phenomenon happens due to a combination of phase shifts and optical path differences.
In the exercise, the condition for destructive interference in the film is given by \[ 2nt = (m + 0.5)\lambda \] where:- \( n \) is the refractive index of the film - \( t \) is the thickness - \( m \) is the order of interference (an integer) - \( \lambda \) is the wavelength of light involved.
Both reflected waves from the film surface and the glass surface meet out of phase, cancelling each other out to reduce reflected light intensity observed. This is affected by the film's thickness, refractive index, and wavelength, emphasizing the precise conditions needed for such interference to occur.
Optical Path Length
Optical path length represents the product of the physical path length that light travels through and the refractive index of the medium. It describes how far light travels in each medium, considering its speed reduction in denser materials.
To express this formally, \[ \text{Optical Path Length} = n \times L \] where - \( n \) is the refractive index- \( L \) is the actual distance traveled.
In interference, the difference in optical path lengths determines the phase relationship between two recombining beams. If the optical path lengths differ by multiples of half the wavelength, destructive interference takes place, reducing the light visibility. Accurate calculation of these path lengths is thus critical to understanding and predicting light interference patterns in films.
Wavelength Change with Temperature
When materials like thin films are subjected to temperature changes, their physical properties, such as thickness, change due to thermal expansion. This leads to a change in the effective wavelength of light traveling through the film.
Since a material’s linear dimensions increase with temperature, the interference condition order remains constant, but the effective wavelength must adapt:\[ \text{New Wavelength} = \frac{\text{Old Wavelength} \times t_f}{t_0} \] where:- \( t_f \) is the final thickness - \( t_0 \) is the initial thickness.
This exercise highlights that as the temperature rises, the optical thickness shifts, modifying which wavelengths undergo destructive interference. This subtle interplay between thermal expansion and optical properties is fundamental for designs involving precise optical performance over varied temperatures.

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Most popular questions from this chapter

35.36. What is the thinnest soap film (excluding the case of zero thickness) that appears black when illuminated with light with wavelength 480 \(\mathrm{nm}\) ? The index of refraction of the film is 1.33 , and there is air on both sides of the film.

35.54. White light reflects at normal incidence from the top and bottom surfaces of a glass plate \((n=1.52) .\) There is air above and below the plate. Constructive interference is observed for light whose wavelength in air is 477.0 \(\mathrm{nm}\) . What is the thickness of the plate if the next longer wavelength for which there is constructive interference is 540.6 \(\mathrm{nm}\) ?

35.45. One round face of a \(3.25-\mathrm{m}\) , solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 \(\mathrm{mm}\) apart, are made in the center of the black face. When laser light of wavelength 632.8 \(\mathrm{nm}\) shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 \(\mathrm{mm}\) wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

35.29. Two rectangular pieces of plane glass are laid one upon the other on a table. A thin strip of paper is placed between them at one edge so that a very thin wedge of air is formed The plates are illuminated at normal incidence by 546 -nm light from a mercury-vapor lamp. Interference fringes are formed, with 15.0 fringes per centimeter. Find the angle of the wedge.

35.52. Red light with wavelength 700 \(\mathrm{nm}\) is passed through a two- slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

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