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Chapter 27: Problem 12

You are standing in air and are looking at a flat piece of glass \((n=1.52)\) on which there is a layer of transparent plastic \((n=1.61)\) . Light whose wavelength is 589 nm is vacuum is incident nearly perpendicularly on the coated glass and reflects into your eyes. The layer of plastic looks dark. Find the two smallest possible nonzero values for the thickness of the layer.

Short Answer

Expert verified
The smallest thicknesses are approximately 114.0 nm and 342.0 nm.

Step by step solution

01

Understanding Thin Film Interference

When light reflects off a thin film, such as the layer of plastic on the glass, some of the light reflects off the top surface and some reflects off the bottom surface of the film. Constructive or destructive interference occurs depending on the thickness of the film, leading to bright or dark appearances, respectively. To determine the thickness for destructive interference (dark appearance), we set up the condition for destructive interference.
02

Condition for Destructive Interference

For a film with a higher refractive index than the surrounding medium, destructive interference occurs when the optical path difference, including phase change upon reflection, is an odd multiple of half-wavelengths inside the film. The formula is:\[2nt = (m + 0.5) \lambda, \quad m = 0, 1, 2, \ldots\]where \(n\) is the refractive index of the plastic, \(t\) is the thickness, \(m\) is the order of interference, and \(\lambda\) is the wavelength of light in a vacuum.
03

Find the Wavelength in the Plastic

Calculate the wavelength of light inside the plastic layer using its refractive index:\[\lambda' = \frac{\lambda}{n} = \frac{589 \text{ nm}}{1.61} \approx 365.2 \text{ nm}\]This is the wavelength of the light while it travels within the plastic medium.
04

Calculate the Smallest Thickness for m=0

Using the condition for destructive interference for \(m=0\), we plug in our values:\[2n t = \frac{1}{2} \cdot 365.2 \text{ nm}\]\(n = 1.61\), so:\[t = \frac{1}{2} \cdot \frac{365.2 \text{ nm}}{2 \times 1.61} \approx 114.0 \text{ nm}\]
05

Calculate the Next Thickness for m=1

Using \(m=1\), we determine the next possible thickness:\[2n t = \frac{3}{2} \cdot 365.2 \text{ nm}\]Solving for \(t\):\[t = \frac{3}{2} \cdot \frac{365.2 \text{ nm}}{2 \times 1.61} \approx 342.0 \text{ nm}\]
06

Conclusion

The two smallest nonzero thicknesses of the plastic layer giving rise to destructive interference (dark appearance) are approximately 114.0 nm and 342.0 nm.

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Key Concepts

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

Optical Path Difference
In thin film interference, the optical path difference determines whether light will undergo constructive or destructive interference. When a light wave enters a thin film, it splits: some reflects off the top surface, and some proceeds through the film, reflecting off the bottom surface. Because the light travels different distances, these reflected waves can interfere with each other. The optical path difference must take into account both distances travelled by the light and any phase changes that occur. This is a crucial factor in calculating whether the interference will result in darkness or brightness in the reflected light.
Destructive Interference
Destructive interference happens when two or more light waves combine to produce a decrease in amplitude, resulting in darkness. For a thin film where the refractive index of the film is higher than the surrounding medium, destructive interference occurs when the optical path difference is an odd multiple of half-wavelengths. The formula used is:
  • \[2nt = (m + 0.5) \lambda\]
This equation describes the condition under which a light wave reflecting off a surface and another wave reflecting off a lower layer will cancel each other out. The parameter \(m\) represents the interference order, starting from 0 upward. This concept helps explain why the plastic layer appears dark to an observer.
Refractive Index
The refractive index measures how much a material slows down light compared to a vacuum. In this exercise, the plastic layer on glass has a refractive index of 1.61, higher than that of air, which is 1. This refractive index is critical in determining the speed and bending of light within a medium. When solving for interference in a thin film, you need to adjust the wavelength of the light wave to account for its decreased speed. Additionally, any phase changes upon reflection, particularly at interfaces where light moves from a lower to a higher refractive index, must be considered.
Wavelength in Medium
When light enters a medium with a different refractive index, its speed changes, affecting its wavelength. The new wavelength, \(\lambda'\), in the medium is calculated by dividing the original wavelength, \(\lambda\), by the refractive index, \(n\), of the medium:
  • \[\lambda' = \frac{\lambda}{n}\]
For the given problem, the wavelength of light changes as it enters the plastic. This altered wavelength is a key factor in calculating the conditions for destructive interference. By knowing \(\lambda'\); you can correctly set up the interference equations needed to find suitable thicknesses for achieving darkness.

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Most popular questions from this chapter

Two gratings \(A\) and \(B\) have slit separations \(d_{A}\) and \(d_{B},\) respectively. They are used with the same light and the same observation screen. When grating \(A\) is replaced with grating \(B,\) it is observed that the first order maximum of \(A\) is exactly replaced by the second-order maximum of B. (a) Determine the ratio \(d_{\mathrm{B}} / d_{\mathrm{A}}\) of the spacings between the slits of the gratings. (b) Find the next two principal maxima of grating A and the principal maxima of \(\mathrm{B}\) that exactly replace them when the gratings are switched. Identify these maxima by their order numbers.

The same diffraction grating is used with two different wave- lengths of light, \(\lambda_{\mathrm{A}}\) and \(\lambda_{\mathrm{B}}\) . The fourth-order principal maximum of light \(\mathrm{A}\) exactly overlaps the third-order principal maximum of light B. Find the ratio \(\lambda_{\mathrm{A}} / \lambda_{\mathrm{B}}\) .

A slit has a width of \(W_{1}=2.3 \times 10^{-6} \mathrm{m} .\) When light with a wavelength of \(\lambda_{1}=510 \mathrm{nm}\) passes through this slit, the width of the central bright fringe on a flat observation screen has a certain value. With the screen kept in the same place, this slit is replaced with a second slit (width \(W_{2} ),\) and a wavelength of \(\lambda_{2}=740 \mathrm{nm}\) is used. The width of the central bright fringe on the screen is observed to be unchanged. Find \(W_{2} .\)

In a Young's double-slit experiment the separation \(y\) between the second- order bright fringe and the central bright fringe on a flat screen is \(0.0180 \mathrm{~m}\) when the light has a wavelength of \(425 \mathrm{nm} .\) Assume that the angles that locate the fringes on the screen are small enough so that \(\sin \theta \approx \tan \theta .\) Find the separation \(y\) when the light has a wavelength of \(585 \mathrm{nm}\).

In a Young's double-slit experiment, the angle that locates the second dark fringe on either side of the central bright fringe is \(5.4^{\circ} .\) Find the ratio \(d / \lambda\) of the slit separation \(d\) to the wavelength \(\lambda\) of the light.
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