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Magazine Chapter 9: Problem 40
Fringes are observed when a parallel beam of light of wavelength \(500 \mathrm{nm}\) is incident perpendicularly onto a wedge-shaped film with an index of refraction of \(1.5 .\) What is the angle of the wedge if the fringe separation is \(\frac{1}{3} \mathrm{cm} ?\)
Short Answer
Expert verified
The angle of the wedge is approximately \(2.87 \times 10^{-3}\) degrees.
Step by step solution
01
Understand the Concept
The problem describes a phenomenon observed in thin film interference, where light waves reflecting off the top and bottom surfaces of a thin film interfere with each other. Constructive interference occurs at specific angles and positions, creating bright fringes. These angles depend on the thickness of the film and the wavelength of the light.
02
Identify the Given Information
We know the wavelength of light \(\lambda = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{meters}\), the index of refraction \(n=1.5\), and the fringe separation \(\Delta x = \frac{1}{3} \, \text{cm} = 3.33 \times 10^{-3} \, \text{meters}\).
03
Know the Relation Between Variables
For fringes in a wedge-shaped film, the separation between two adjacent bright fringes is given by:\[ \Delta x = \frac{\lambda}{2n \tan(\theta)} \], where \(\lambda\) is the wavelength, \(n\) is the refractive index, and \(\theta\) is the angle of the wedge.
04
Solve for the Angle of the Wedge
We rearrange the equation to find the angle \(\theta\):\[ \tan(\theta) = \frac{\lambda}{2n \Delta x} \].Substituting the known values:\[ \tan(\theta) = \frac{500 \times 10^{-9}}{2 \times 1.5 \times 3.33 \times 10^{-3}} \].
05
Calculate the Angle
First, calculate the numerator and denominator separately and then find the tangent:Numerator = \(500 \times 10^{-9} = 5.0 \times 10^{-7}\),Denominator = \(2 \times 1.5 \times 3.33 \times 10^{-3} = 9.99 \times 10^{-3}\).Now divide:\[ \tan(\theta) = \frac{5.0 \times 10^{-7}}{9.99 \times 10^{-3}} \approx 5.005 \times 10^{-5} \].
06
Determine the Angle in Degrees
Using a calculator, calculate the angle:\(\theta = \tan^{-1}(5.005 \times 10^{-5}) \approx 2.87 \times 10^{-3} \, \text{degrees}\).This is a very small angle, as expected for such thin films.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Wavelength of Light
The wavelength of light is a fundamental characteristic that dictates how light behaves in various scenarios, including thin film interference. The wavelength is the distance between successive points of a wave, which translates into how long each complete cycle of the wave is. When dealing with light in a medium, it鈥檚 crucial to consider the wavelength in that medium because it can change due to refraction.
In the given problem, the light has a wavelength of 500 nanometers in air or vacuum. However, when light enters a medium such as a wedge-shaped thin film, its wavelength can alter. This change in wavelength inside the medium is critical for calculating interference patterns as it determines where bright and dark fringes appear. Recall that light travels slower in media other than air, which effectively reduces its wavelength according to the medium's index of refraction.
In the given problem, the light has a wavelength of 500 nanometers in air or vacuum. However, when light enters a medium such as a wedge-shaped thin film, its wavelength can alter. This change in wavelength inside the medium is critical for calculating interference patterns as it determines where bright and dark fringes appear. Recall that light travels slower in media other than air, which effectively reduces its wavelength according to the medium's index of refraction.
Index of Refraction
The index of refraction, denoted as 'n,' is a measure of how much a wave such as light bends when it enters a new medium. This bending of light is crucial for understanding and predicting interference patterns in films.
For the wedge-shaped film described in the problem, the index of refraction is 1.5. This value indicates that light travels at 2/3 its speed in a vacuum within this medium. Consequently, the angle at which light hits the film and its behavior inside the film are significantly affected by its refractive index.
For the wedge-shaped film described in the problem, the index of refraction is 1.5. This value indicates that light travels at 2/3 its speed in a vacuum within this medium. Consequently, the angle at which light hits the film and its behavior inside the film are significantly affected by its refractive index.
- The higher the index, the slower light travels within the medium.
- A greater index of refraction also means a more pronounced bending of light, influencing interference patterns.
Fringe Separation
Fringe separation is the distance between consecutive bright (or dark) interference fringes formed on a thin film. It is an important parameter indicating the spacing of interference patterns due to constructive or destructive interference of light waves.
In the wedge-shaped film setup, fringe separation is determined by the formula: \[\Delta x = \frac{\lambda}{2n \tan(\theta)}\]where \(\Delta x\) is fringe separation, \(\lambda\) is the wavelength of light, \(n\) is the index of refraction, and \(\theta\) is the angle of the wedge. This relation shows that fringe separation is inversely proportional to the tangent of the wedge's angle.
In the wedge-shaped film setup, fringe separation is determined by the formula: \[\Delta x = \frac{\lambda}{2n \tan(\theta)}\]where \(\Delta x\) is fringe separation, \(\lambda\) is the wavelength of light, \(n\) is the index of refraction, and \(\theta\) is the angle of the wedge. This relation shows that fringe separation is inversely proportional to the tangent of the wedge's angle.
- The larger the angle, the closer the fringes appear.
- The greater the refractive index, the smaller the separation for the same angle and wavelength.
- A smaller wavelength results in closer fringe separation.
Angle of Wedge
The angle of the wedge is the angle formed between the top and bottom surfaces of the wedge-shaped film. This angle is crucial because it directly affects the interference pattern observed. Even tiny angles can lead to observable changes in the fringe patterns.
To determine the angle of the wedge, we use the equation:\[\tan(\theta) = \frac{\lambda}{2n \Delta x}\]This relation stems from the requirement for constructive interference at specific points along the film. In our problem, we calculate \(\tan(\theta)\) using the known values for wavelength (\(\lambda = 500 \times 10^{-9} \) m), index of refraction (\(n = 1.5\)), and fringe separation (\(\Delta x = 3.33 \times 10^{-3} \) m).
The calculated \(\tan(\theta)\) allows us to find that \(\theta\) is approximately \(2.87 \times 10^{-3}\) degrees. This minuscule angle exemplifies how precisely thin film interference plays out with extremely small angular dimensions, further demonstrating the critical relationship between angle, wavelength, and refractive index in producing observable fringe patterns.
To determine the angle of the wedge, we use the equation:\[\tan(\theta) = \frac{\lambda}{2n \Delta x}\]This relation stems from the requirement for constructive interference at specific points along the film. In our problem, we calculate \(\tan(\theta)\) using the known values for wavelength (\(\lambda = 500 \times 10^{-9} \) m), index of refraction (\(n = 1.5\)), and fringe separation (\(\Delta x = 3.33 \times 10^{-3} \) m).
The calculated \(\tan(\theta)\) allows us to find that \(\theta\) is approximately \(2.87 \times 10^{-3}\) degrees. This minuscule angle exemplifies how precisely thin film interference plays out with extremely small angular dimensions, further demonstrating the critical relationship between angle, wavelength, and refractive index in producing observable fringe patterns.
