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Magazine Chapter 28: Problem 41
A thin coating \((t=340.0 \mathrm{nm}, n=1.480)\) is placed on a glass lens. Which visible \((400 \mathrm{nm}<\lambda<700 \mathrm{nm})\) wavelength(s) will be absent in the reflected beam if (a) the glass has an index of refraction \(n=1.350,\) and \((b)\) the glass has an index of refraction \(n=1.675 ?\)
Short Answer
Expert verified
The absent wavelength in both part (a) and (b) is 671.4 nm.
Step by step solution
01
Understand the Question
We need to find which wavelengths within the visible spectrum (400 nm to 700 nm) are absent in the reflected beam due to destructive interference. The situation involves a thin film with a refractive index placed on a glass lens, and we also have two different cases of different refractive indices for the glass itself.
02
Identify the Relevant Formula
For thin film interference, wavelengths get canceled due to destructive interference when we have:\[ 2nt = (m+\frac{1}{2})\lambda_n \]where \( n \) is the refractive index of the film, \( t \) is the thickness of the film, \( \lambda_n \) is the effective wavelength within the film, and \( m \) is an integer. In the film, \( \lambda_n = \frac{\lambda}{n} \), where \( \lambda \) is the wavelength in air.
03
Determine the Effective Wavelength
First we determine the effective wavelength inside the film for each light wave:\[ \lambda_n = \frac{\lambda}{n} = \frac{\lambda}{1.480} \]
04
Implement Glass Index Conditions - Part (a)
For part (a), we have the condition where the refractive index of glass is smaller than that of the film \((n=1.350)\). The formula for destructive interference remains:\[ 2nt = (m+\frac{1}{2})\lambda_n \]\[ 2 \cdot 1.480 \cdot 340.0 \text{ nm} = (m+\frac{1}{2}) \frac{\lambda}{1.480}\]Solve for \(\lambda\):\[ \lambda = \frac{2 \cdot 1.480^2 \cdot 340.0}{m + \frac{1}{2}} \]
05
Calculate Absent Wavelength for Part (a)
Substitute different \(m\) values and solve for \(\lambda\) within \(400 \text{ nm} \leq \lambda \leq 700 \text{ nm}\). Using \(m=1\):\[ \lambda = \frac{2 \cdot 1.480^2 \cdot 340.0}{1 + \frac{1}{2}} = 671.4 \text{ nm} \]This is the absent wavelength for part (a) as it falls within the given visible range.
06
Implement Glass Index Conditions - Part (b)
For part (b), we have the condition where glass has a larger refractive index (\(n=1.675\)) with respect to the film. Now the film undergoes a phase shift, and the formula remains:\[ 2nt = (m+\frac{1}{2})\lambda_n \]\[ 2 \cdot 1.480 \cdot 340.0 = (m+\frac{1}{2}) \frac{\lambda}{1.480}\]Solve for \(\lambda\):\[ \lambda = \frac{2 \cdot 1.480^2 \cdot 340.0}{m + \frac{1}{2}} \]
07
Calculate Absent Wavelength for Part (b)
Substitute different \(m\) values and solve for \(\lambda\) within \(400 \text{ nm} \leq \lambda \leq 700 \text{ nm}\). Using \(m=1\):\[ \lambda = \frac{2 \cdot 1.480^2 \cdot 340.0}{1 + \frac{1}{2}} = 671.4 \text{ nm} \]This is the absent wavelength for part (b) as it also falls within the specified range.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Destructive Interference
Thin film interference occurs due to the interaction of light waves reflecting off the top surface and the bottom surface of a film. These interactions can lead to either constructive or destructive interference. In the case of destructive interference, specific wavelengths of light are eliminated or 'absent' from the reflected light spectrum.
For destructive interference to occur in a thin film, the reflected waves need to be out of phase by half a wavelength. This means that when one wave has a crest, the other has a trough. Mathematically, the condition for destructive interference in a thin film of refractive index \( n \) and thickness \( t \) is defined by the formula:
\[ 2nt = (m + \frac{1}{2})\lambda_n \]
where:
This condition ensures that certain wavelengths destructively interfere and do not appear in the reflected light.
For destructive interference to occur in a thin film, the reflected waves need to be out of phase by half a wavelength. This means that when one wave has a crest, the other has a trough. Mathematically, the condition for destructive interference in a thin film of refractive index \( n \) and thickness \( t \) is defined by the formula:
\[ 2nt = (m + \frac{1}{2})\lambda_n \]
where:
- \( n \) is the refractive index of the film,
- \( t \) is the thickness of the film,
- \( \lambda_n \) is the wavelength of light within the film, and
- \( m \) is an integer representing the order of the interference.
This condition ensures that certain wavelengths destructively interfere and do not appear in the reflected light.
Refractive Index
The refractive index is a measure of how much light bends, or refracts, when it enters a material. Each medium has its unique refractive index, which affects the speed and wavelength of light passing through it.
In our problem scenario, we deal with a thin film of refractive index \( n = 1.480 \). The refractive index alters the wavelength of light inside the material compared to its wavelength in air. This change is calculated as:
\[ \lambda_n = \frac{\lambda}{n} \]
where \( \lambda \) is the wavelength in air.
The refractive index also impacts whether there is a phase shift upon reflection. When light reflects from a medium with a higher refractive index than the medium it comes from, it undergoes a phase shift of half a wavelength (180 degrees). Conversely, if the light reflects from a medium with a lower refractive index, no phase shift occurs. This phenomenon plays a crucial role in determining whether constructive or destructive interference will take place.
In our problem scenario, we deal with a thin film of refractive index \( n = 1.480 \). The refractive index alters the wavelength of light inside the material compared to its wavelength in air. This change is calculated as:
\[ \lambda_n = \frac{\lambda}{n} \]
where \( \lambda \) is the wavelength in air.
The refractive index also impacts whether there is a phase shift upon reflection. When light reflects from a medium with a higher refractive index than the medium it comes from, it undergoes a phase shift of half a wavelength (180 degrees). Conversely, if the light reflects from a medium with a lower refractive index, no phase shift occurs. This phenomenon plays a crucial role in determining whether constructive or destructive interference will take place.
Wavelength Calculation
Calculating the absent wavelength due to destructive interference involves first determining the effective wavelength inside the thin film. For this, the refractive index of the film is crucial:
\[ \lambda_n = \frac{\lambda}{1.480} \]
Once we know the effective wavelength within the film, we use the formula for destructive interference:
\[ 2nt = (m + \frac{1}{2})\lambda_n \]
From here, solving for the wavelength \( \lambda \) in air gives us:
\[ \lambda = \frac{2n^2t}{m + \frac{1}{2}} \]
This equation allows us to calculate which wavelengths will not be reflected due to destructive interference.
For visible light, which ranges from 400 nm to 700 nm, only specific wavelengths determined by the index \( m \) will be absent. By plugging different integer values of \( m \) into our wavelength calculation formula, we can determine these absent wavelengths. In the problem at hand, repeated calculations confirm the absent wavelength as 671.4 nm within the visible range.
\[ \lambda_n = \frac{\lambda}{1.480} \]
Once we know the effective wavelength within the film, we use the formula for destructive interference:
\[ 2nt = (m + \frac{1}{2})\lambda_n \]
From here, solving for the wavelength \( \lambda \) in air gives us:
\[ \lambda = \frac{2n^2t}{m + \frac{1}{2}} \]
This equation allows us to calculate which wavelengths will not be reflected due to destructive interference.
For visible light, which ranges from 400 nm to 700 nm, only specific wavelengths determined by the index \( m \) will be absent. By plugging different integer values of \( m \) into our wavelength calculation formula, we can determine these absent wavelengths. In the problem at hand, repeated calculations confirm the absent wavelength as 671.4 nm within the visible range.
