Chapter 35: Problem 40
A thin film of acetone \((n=1.25)\) coats a thick glass plate \((n=1.50)\). White light is incident normal to the film. In the reflections, fully destructive interference occurs at \(600 \mathrm{~nm}\) and fully constructive interference at \(700 \mathrm{~nm}\). Calculate the thickness of the acetone film.
Short Answer
Step by step solution
Identifying Interference Conditions
Setting Up Equations for Path Difference
Expressing m' in Terms of m
Solving for t
Calculate Thickness
Final Calculation
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Interference Conditions
For these phenomena to occur, the thickness of the film, the angle of incidence, and the wavelength of light all play crucial roles. The conditions necessary for interference are guided by the principles of optics, which use specific equations to describe how light waves interfere within a medium.
With destructive interference, the equation is given by \(2nt = (m + \frac{1}{2}) \lambda\), where \(n\) is the refractive index, \(t\) is the film thickness, \(\lambda\) is the wavelength of light, and \(m\) is an integer representing the number of half-wavelength shifts. Constructive interference occurs when the equation \(2nt = m \lambda\) is satisfied, where all terms are as previously defined.
Therefore, interference conditions are crucial to determine how light interacts with a thin film.
Refractive Index
For example, if acetone has a refractive index of 1.25 and glass has a refractive index of 1.50, it implies that light travels slower in acetone than in air or vacuum, and even slower in glass itself. This property affects how light changes direction or refracts as it enters or exits the media.
- Refractive index influences how often light waves interfere constructively or destructively.
- It is essential for calculating the optical path difference that leads to interference patterns.
Destructive Interference
To understand this, recall that destructive interference occurs under the condition \(2nt = (m + \frac{1}{2}) \lambda\). Here, the half-wavelength shift ensures that peaks and troughs of light waves align in such a way that they negate each other's effects.
Key factors in achieving destructive interference include:
- The thickness of the film \(t\).
- The wavelength of light \(\lambda\) that the condition applies to.
- The refractive index \(n\) of the film.
Constructive Interference
This type of interference follows the condition \(2nt = m \lambda\), allowing the waves to reinforce each other by aligning their peaks and troughs. Unlike destructive interference, where waves cancel out, constructive interference leads to the amplification of light.
To achieve constructive interference, consider these elements:
- The film's refractive index \(n\).
- The film's thickness \(t\).
- The wavelength \(\lambda\) at which constructive interference is observed.