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Chapter 26: Problem 20

\(\bullet\) The lenses of a particular set of binoculars have a coating with index of refraction \(n=1.38,\) and the glass itself has \(n=1.52 .\) If the lenses reflect a wavelength of 525 \(\mathrm{nm}\) the most strongly, what is the minimum thickness of the coating?

Short Answer

Expert verified
The minimum thickness of the coating is approximately 190.22 nm.

Step by step solution

01

Understanding the problem

The problem involves a lens with a coating that causes constructive interference for a specific wavelength, 525 nm. Our objective is to determine the minimum thickness of this coating that will produce this constructive interference, given the indices of refraction for the coating and the glass.
02

Condition for Maximum Reflection

For constructive interference due to thin-film interference, the optical path difference between two waves must be equal to an integer multiple of the wavelength divided by the index of refraction, i.e., \(2nt = m\lambda\), where \(n\) is the index of refraction of the coating. For maximum reflection, when \(m=1\), we have \(2nt = \lambda\).
03

Calculating Minimum Thickness

Plug the known values into the equation from Step 2. With \(n=1.38\) and \(\lambda=525\ nm\), solve for \(t\) (thickness):
04

Solving for Thickness

Rearrange the equation \(2nt = \lambda\) to solve for \(t\). Substitute the given values: \[2(1.38)t = 525 \t = \frac{525}{2 \times 1.38} \t = \frac{525}{2.76} \t \approx 190.22 \m\]
05

Conclusion

Thus, the minimum thickness of the coating needed for the lenses to reflect the 525 nm wavelength most strongly is approximately 190.22 nm.

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Key Concepts

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

Constructive Interference
Constructive interference is a phenomenon where two or more waves superimpose to produce a wave of greater amplitude. This occurs when the crests and troughs of the waves align perfectly. Think of it as when two friends are paddling in a canoe鈥攊f they paddle in sync, they can move faster! This perfect alignment leads to an increase in the intensity or brightness of light, called constructive interference.
  • For thin films, constructive interference can cause certain wavelengths to become more pronounced, explaining why we see vivid colors on soap bubbles or oil slicks.
  • The condition for constructive interference in thin-film interference can be mathematically represented as: \[2nt = m\lambda\]where \(n\) is the index of refraction, \(t\) is the thickness of the film, \(m\) is an integer (1, 2, 3, ...), and \(\lambda\) is the wavelength of light.
  • In the context of the problem, we seek the minimum thickness that makes \(m=1\) and focuses on the visible wavelength of 525 nm.
By using the equation for \(m=1\), we ensure that the reflection from the coated surface is in phase with the one from the glass, thereby amplifying the wavelength of 525 nm.
Index of Refraction
The index of refraction, commonly represented by \(n\), is a measure of how much a material can bend light. In simple terms, it's how slow light travels in a material compared to in a vacuum, where it travels fastest. An easy way to picture this is to imagine a car slowing down when it hits mud.
  • A higher index of refraction means light travels slower. For example, the glass in the binoculars has an index of 1.52, meaning light travels 1.52 times slower than in a vacuum.
  • The coating on the lens has a slightly lower index of 1.38, which impacts how much the light wave is slowed down and thus its interference pattern.
  • Knowing the indices is essential because they help determine the phase difference between the light waves, crucial for calculating the required thickness for constructive interference.
This index is crucial for designing optical devices like lenses and coatings, because it influences how colors are enhanced or suppressed.
Optical Path Difference
Optical path difference (OPD) involves comparing the journey two light waves take when passing through different media. Essentially, it's the difference in their travel paths, adjusted for the speed change due to varying indices of refraction.
  • The concept is pivotal in understanding phenomena like thin-film interference, where the reflection of light is intensified or diminished based on this difference.
  • In the provided solution, the relationship \(2nt = m\lambda\) specifies the OPD in terms of the thickness \(t\), the index \(n\), and the wavelength \(\lambda\).
  • The factor of 2 accounts for the light traveling through the film, reflecting off the surface, and coming back through the film again鈥攂asically a round trip. This round trip is why the OPD is crucial in crafting the interference conditions.
This mathematical relationship ensures the light waves constructively interfere by being perfectly in phase, leading to enhanced reflection, particularly at the specified wavelength of 525 nm.

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Most popular questions from this chapter

\(\bullet$$\bullet\) Resolution of the eye, I. Even if the lenses of our eyes functioned perfectly, our vision would still be limited due to diffraction of light at the pupil. Using Rayleigh's criterion, what is the smallest object a person can see clearly at his near point of 25.0 \(\mathrm{cm}\) with a pupil 2.00 \(\mathrm{mm}\) in diameter and light of wavelength 550 \(\mathrm{nm}\) ? (To get a reasonable estimate without having to go through complicated calculations, we'll ignore the effect of the fluid in the eye.) Based upon your answer, does it seem that diffraction plays a significant role in limiting our visual acuity?

\(\bullet$$\bullet\) Resolution of telescopes. Due to blurring caused by atmospheric distortion, the best resolution that can be obtained by a normal, earth-based, visible-light telescope is about 0.3 arcsecond (there are 60 arcminutes in a degree and 60 arcsec- onds in an arcminute).(a) Using Rayleigh's criterion, calculate the diameter of an earth-based telescope that gives this resolu- tion with \(550-\) nm light. (b) Increasing the telescope diameter beyond the value found in part (a) will increase the light- gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10 -m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope \((5-m\) diameter) on Palomar Mountain in California? In what ways are they not superior? Explain.

\(\bullet$$\bullet\) Thickness of human hair. Although we have discussed single-slit diffraction only for a slit, a similar result holds when light bends around a straight, thin object, such as a strand of hair. In that case, \(a\) is the width of the strand. From actual laboratory measurements on a human hair, it was found that when a beam of light of wavelength 632.8 \(\mathrm{nm}\) was shone on a single strand of hair, and the diffracted light was viewed on a screen 1.25 \(\mathrm{m}\) away, the first dark fringes on either side of the central bright spot were 5.22 \(\mathrm{cm}\) apart. How thick was this strand of hair?

\(\bullet\) Monochromatic light from a distant source is incident on a slit 0.750 mm wide. On a screen 2.00 m away, the distance from the central maximum of the diffraction pattern to the first minimum is measured to be 1.35 \(\mathrm{mm} .\) Calculate the wave- length of the light.

\(\bullet\) A beam of laser light of wavelength 632.8 nm falls on a thin slit 0.00375 \(\mathrm{mm}\) wide. After the light passes through the slit, at what angles relative to the original direction of the beam is it completely cancelled when viewed far from the slit?
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