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After a minor oil spill, a think film of oil \((n=1.40)\) of thickness 450 nm floats on the water surface in a bay. (a) What predominant color is seen by a bird flying overhead? (b) What predominant color is seen by a seal swimming underwater?

Short Answer

Expert verified
(a) The predominant color seen by the bird flying overhead is red. (b) The predominant color seen by the seal swimming underwater is greenish-yellow.

Step by step solution

01

Determine the conditions for constructive interference

To find the predominant color seen by the bird and the seal, we need to use the condition for constructive interference: For thin film interference, the condition for constructive interference is given by: For the bird: \( 2nt = m\lambda \) For the seal: \( 2nt = (m+\frac{1}{2})\lambda \) Where \(m\) is an integer, \(n\) is the refractive index of oil, \(t\) is the thickness of the oil film, and \(\lambda\) is the wavelength of light.
02

Calculate the wavelength of light for constructive interference

We need to find the minimum value of \(m\) that satisfies the above equations and results in visible light (approx. 400 nm to 700 nm). (a) For the bird: \( 2nt = m\lambda \) \( \lambda = \frac{2nt}{m} \) (b) For the seal: \( 2nt = (m+\frac{1}{2})\lambda \) \( \lambda = \frac{2nt}{m+\frac{1}{2}} \)
03

Find the minimum value of m and the corresponding predominant color

(a) For the bird, we'll start with the smallest integer value of \(m=1\). \(\lambda = \frac{2(1.40)(450 \times 10^{-9} m)}{1}\) \(\lambda = 1260 \times 10^{-9} m = 1260 nm\) This is not in the range of visible light, so we'll increment m to 2. \(\lambda = \frac{2(1.40)(450 \times 10^{-9} m)}{2}\) \(\lambda = 630 \times 10^{-9} m = 630 nm\) This is in the range of visible light (400 nm to 700 nm), so this is the predominant wavelength seen by the bird. The color corresponding to a wavelength of 630 nm is red. (b) For the seal, we'll start with the smallest integer value of \(m=1\). \(\lambda = \frac{2(1.40)(450 \times 10^{-9}m)}{1+\frac{1}{2}}\) \(\lambda = 840 \times 10^{-9} m = 840 nm\) This is not in the range of visible light, so we'll increment m to 2. \(\lambda = \frac{2(1.40)(450 \times 10^{-9}m)}{2+\frac{1}{2}}\) \(\lambda = 560 \times 10^{-9} m = 560 nm\) This is in the range of visible light (400 nm to 700 nm), so this is the predominant wavelength seen by the seal. The color corresponding to a wavelength of 560 nm is greenish-yellow.
04

Answer

(a) The predominant color seen by the bird flying overhead is red. (b) The predominant color seen by the seal swimming underwater is greenish-yellow.

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Key Concepts

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

Constructive Interference
When discussing thin film interference, particularly in relation to colors produced by an oil spill, the principle of constructive interference plays a pivotal role. Constructive interference occurs when waves overlap in a manner that causes their amplitudes to add together, resulting in an increased amplitude. In the context of light, this implies that certain wavelengths of light reinforce each other when reflecting off different layers of a film, such as oil, leading to the appearance of specific vibrant colors.

For a thin film of oil floating on water, constructive interference happens for specific wavelengths of light that satisfy the equation \( 2nt = m\lambda \) for reflection from the top layer of the film, and \( 2nt = (m+\frac{1}{2})\lambda \) for reflection from the bottom layer at the oil-water boundary. The path difference created by reflections at differing interfaces leads to constructive interference, culminating in bright hues being visible. To identify the predominant color, we must find the wavelengths that occur within the visible spectrum and apply the equations iteratively, starting with the smallest integer values of \( m \), until we identify a suitable wavelength.
Wavelength of Light
Wavelength of light is a core aspect of understanding the phenomenon of thin film interference. It refers to the distance between two consecutive crests or troughs of a wave. The wavelength determines the color of light, with each color corresponding to a different range of wavelengths; red light has a longer wavelength compared to blue light.

To uncover which colors will dominate in the interference pattern of the oil film, one must calculate the wavelength of light that results in constructive interference. For example, when light reflects off the surface of an oil film, the equation \( \lambda = \frac{2nt}{m} \) (for a bird viewing from above) or \( \lambda = \frac{2nt}{m+\frac{1}{2}} \) (for a seal viewing from below) is used to find the effective wavelengths. The visible spectrum is typically between 400 nm and 700 nm, so only the wavelengths that fall within this range will correspond to the predominant color visible to the observer.
Refractive Index
The refractive index (represented by \( n \)) is a dimensionless number that describes how much light bends, or refracts, when entering a material. It is defined as the ratio of the speed of light in a vacuum to the speed of light in the material. Different materials have different refractive indices, which influences how light behaves when transitioning from one medium to another.

In the case of thin film interference, the refractive index of the oil (\( n=1.40 \)) is vital to determining the subsequent interference pattern. The refractive index not only dictates how much the light bends but also influences the phase change that occurs upon reflection. When light reflects off the boundary of a medium with a higher refractive index, such as going from air to oil, it typically undergoes a half-wavelength phase shift. This phase change is an integral component when calculating the conditions for constructive or destructive interference and hence affects the color observed by the bird flying overhead or the seal swimming underwater.

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Most popular questions from this chapter

An inventor notices that a soap bubble is dark at its thinnest and realizes that destructive interference is taking place for all wavelengths. How could she use this knowledge to make a nonreflective coating for lenses that is effective at all wavelengths? That is, what limits would there be on the index of refraction and thickness of the coating? How might this be impractical?

Two identical pieces of rectangular plate glass are used to measure the thickness of a hair. The glass plates are in direct contact at one edge and a single hair is placed between them hear the opposite edge. When illuminated with a sodium lamp \((\lambda=589 \mathrm{nm}),\) the hair is seen between the 180 th and 181 st dark fringes. What are the lower and upper limits on the hair's diameter?

A film of soapy water ( \(n=1.33\) ) on top of a plastic cutting board has a thickness of \(233 \mathrm{nm} .\) What color is most strongly reflected if it is illuminated perpendicular to its surface?

An oil slick on water is \(120 \mathrm{nm}\) thick and illuminated by white light incident at a \(45^{\circ}\) angle to its surface. What color does the oil appear (what is the most constructively reflected wavelength), given its index of refraction is \(1.40 ?\)

An effect analogous to two-slit interference can occur with sound waves, instead of light. In an open field, two speakers placed \(1.30 \mathrm{m}\) apart are powered by a singlefunction generator producing sine waves at \(1200-\mathrm{Hz}\) frequency. A student walks along a line \(12.5 \mathrm{m}\) away and parallel to the line between the speakers. She hears an alternating pattern of loud and quiet, due to constructive and destructive interference. What is (a) the wavelength of this sound and (b) the distance between the central maximum and the first maximum (loud) position along this line?

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