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Chapter 40: Problem 31

A mixture of yellow light of wavelength \(580 \mathrm{~nm}\) and blue light of wavelength \(450 \mathrm{~nm}\) is incident normally on an air film \(290 \mathrm{~nm}\) thick. What is the color of the reflected light?

Short Answer

Expert verified
The reflected light is blue.

Step by step solution

01

Understanding Interference Conditions

To determine the color of the reflected light, we need to consider the concept of thin film interference. Constructive interference occurs when the path difference between the reflected rays is an integer multiple of the wavelength, while destructive interference occurs when the path difference is a half-integer multiple of the wavelength.
02

Calculate Path Difference and Phase Change

For a thin film of air with thickness \(t = 290 \, \text{nm}\), the path difference for normal incidence is \(2t\). For destructive interference, the condition is \[2t = \left(m + \frac{1}{2}\right) \lambda_{m}\] where \(m\) is an integer and \(\lambda_{m}\) is the wavelength in the film.
03

Determine Wavelengths in the Film

Since air is the medium, the refractive index \(n = 1\). Thus, the wavelengths in the film remain the same, \(\lambda_{yellow} = 580 \, \text{nm}\) and \(\lambda_{blue} = 450 \, \text{nm}\).
04

Evaluate Yellow Light Interference

Substitute \(\lambda_{yellow} = 580 \, \text{nm}\) into the destructive interference equation: \[2(290) = \left(m + \frac{1}{2}\right)580\] leads to \[580m + 290 = 580\] which simplifies to \(m = 0.5\). This value for \(m\) leads to destructive interference, meaning yellow light is not reflected.
05

Evaluate Blue Light Interference

Substitute \(\lambda_{blue} = 450 \, \text{nm}\) into the destructive interference equation: \[2(290) = \left(m + \frac{1}{2}\right)450\] leads to \[580 = 225m + 225\] approximating to \(m = 1.56\), which is not an integer. This shows constructive interference, allowing blue light to be reflected.
06

Conclusion

Since the yellow light undergoes destructive interference and is not reflected, and blue light undergoes constructive interference, the reflected light will primarily be blue. Thus, the reflected light's color is dominated by the blue wavelength.

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Key Concepts

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

Wavelength
Wavelength refers to the distance between two consecutive points of a wave that are in phase, such as two adjacent peaks or troughs. This characteristic is crucial in determining the color of light, as different colors correspond to different wavelengths. For example, in the visible spectrum, blue light has shorter wavelengths, typically around 450 nm, while yellow light has longer wavelengths, around 580 nm.
Understanding the concept of wavelength is essential when studying light interference, especially in phenomena like thin film interference, where variations in wavelength directly affect the type of interference occurring. The problem at hand involves analyzing how these wavelengths interact when light reflects from a thin air film. - Yellow light: 580 nm - Blue light: 450 nm These wavelengths play a key role in determining whether the light undergoes constructive or destructive interference.
Constructive Interference
Constructive interference occurs when two or more waves overlap and combine to create a wave with a larger amplitude. This happens when the path difference between the waves is an integer multiple of the wavelength. For the waves reflecting off a thin film, the condition for constructive interference can be described mathematically as:\[2t = m\lambda\] where \(t\) is the thickness of the film, \(\lambda\) is the wavelength, and \(m\) is an integer (0, 1, 2, ...).
In the current problem, we consider the interference of blue light reflecting off a 290 nm air film. As shown in the solution, the calculation leads to constructive interference (since the path difference can approximate an integer multiple). - The absence of phase shift in air and the path difference satisfying the condition results in the blue wavelength being prominently reflected.- Constructive interference in this context allows the observer to see a dominant blue color in the reflected light.
Destructive Interference
Destructive interference arises when waves overlap to create a wave with a reduced, or even zero, amplitude. This occurs when the path difference between the waves is a half-integer multiple of the wavelength. The formula for this condition is:\[2t = \left(m + \frac{1}{2}\right)\lambda\] where \(m\) is an integer and \(\lambda\) is the wavelength. When this condition holds, the waves cancel each other out. In the scenario with the air film, yellow light meets the equation for destructive interference.
Calculations using the wavelength of yellow light (580 nm) with the given thickness lead to a path difference that coincides with a half-integer multiple. As a result, the yellow light is not reflected. - This leads to the absence of yellow light in the reflection.- Understanding destructive interference helps clarify why certain colors are diminished or absent when viewing a thin film under particular conditions.

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Most popular questions from this chapter

In the double-slit setup using monochromatic illumination, what is the value of the path-length difference for the first bright bands above and below the central band? Explain your answer.

Monochromatic light from a point source illuminates two narrow, horizontal, parallel slits. The centers of the two slits are \(a=0.80\) \(\mathrm{mm}\) apart, as shown in Fig. \(40-1\). An interference pattern forms on the screen, \(50 \mathrm{~cm}\) away. In the pattern, the bright and dark fringes are evenly spaced. The distance \(\mathrm{y}_{1}\) shown is \(0.304 \mathrm{~mm}\). Compute the wavelength \(\lambda\) of the light. Notice first that Fig. \(40-1\) is not to scale. The rays from the slits would actually be nearly parallel. We can therefore use the result of \(\underline{\text { Problem } 40.2}\) with \(\left(r_{1}-r_{2}\right)=m \lambda\) at the maxima (bright spots), where \(m=0, \pm 1, \pm 2, \ldots\) Then $$ \sin \theta=\frac{\left(r_{1}-r_{2}\right)}{a} \quad \text { becomes } \quad m \lambda=a \sin \theta_{n} $$ Or, alternatively, we could use the grating equation, since a double slit is simply a grating with two lines. Both approaches result in \(m \lambda=a \sin \theta_{m}\) We know that the distance from the central maximum to the first maximum on either side is \(0.304 \mathrm{~mm}\). Therefore, from Fig. \(40-1\), $$ \sin \theta_{1}=\frac{0.0304 \mathrm{~cm}}{50 \mathrm{~cm}}=0.000608 $$ Then, for \(m=1\), \(m \lambda=a \sin \theta_{m} \quad\) becomes \(\quad(1) \lambda=\left(0.80 \times 10^{-3} \mathrm{~m}\right)\left(6.08 \times 10^{-4}\right)\) from which \(\lambda=486 \mathrm{~nm}\), or to two significant figures, \(0.49 \times 10^{3}\) \(\mathrm{nm} .\)

How far apart are the diffracting planes in a NaCl crystal for which X-rays of wavelength \(1.54 \AA\) make a glancing angle of \(15^{\circ} 54^{\prime}\) in the first order?

A single slit of width \(D=0.10 \mathrm{~mm}\) is illuminated by parallel light of wavelength \(600 \mathrm{~nm}\), and diffraction bands are observed on a screen \(40 \mathrm{~cm}\) from the slit. How far is the third dark band from the central bright band? (Refer to \(\underline{\text { Fig. } 40-7 . \text { .) }}\) For a single slit, the locations of dark bands are given by the equation \(m^{\prime} \lambda=D \sin \theta_{m}^{\prime}\). Then $$ \sin \theta_{3}=\frac{3 \lambda}{D}=\frac{3\left(6.00 \times 10^{-7} \mathrm{~m}\right)}{0.10 \times 10^{-3} \mathrm{~m}}=0.018 \quad \text { or } \quad \theta_{3}=1.0^{\circ} $$ From the figure, \(\tan \theta_{3}=\mathrm{y} / 40 \mathrm{~cm}\), and \(\mathrm{so}\) $$ y=(40 \mathrm{~cm})\left(\tan \theta_{3}\right)=(40 \mathrm{~cm})(0.018)=0.72 \mathrm{~cm} $$

Green light of wavelength \(540 \mathrm{~nm}\) is diffracted by a grating ruled with 2000 lines/cm. (a) Compute the angular deviation of the third-order image. (b) Is a 10th-order image possible? (a) \(\quad \sin \theta_{3}=\frac{3 \lambda}{a}=\frac{3\left(5.40 \times 10^{-5} \mathrm{~cm}\right)}{5.00 \times 10^{-4} \mathrm{~cm}}=0.324 \quad\) or \(\quad \theta=18.9^{\circ}\) (b) $$ \sin \theta_{10}=\frac{10 \lambda}{a}=\frac{10\left(5.40 \times 10^{-5} \mathrm{~cm}\right)}{5.00 \times 10^{-4} \mathrm{~cm}}=1.08 \quad \text { (impossible) } $$ Since the value of \(\theta_{10}\) cannot exceed 1 , a 10 th-order image is impossible.
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