/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 13 A nonreflective coating of magne... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A nonreflective coating of magnesium fluoride \((n=1.38)\) covers the glass \((n=1.52)\) of a camera lens. Assuming that the coating prevents reflection of yellow-green light (wavelength in vacuum \(=565 \mathrm{nm}\) ), determine the minimum nonzero thickness that the coating can have.

Short Answer

Expert verified
The minimum thickness is approximately 74.16 nm.

Step by step solution

01

Understanding the Problem

We need to find the minimum thickness of a magnesium fluoride coating that eliminates reflection. The problem involves two materials: magnesium fluoride with a refractive index of 1.38 and glass with a refractive index of 1.52. The light we are considering has a wavelength of 565 nm in a vacuum.
02

Identifying the Condition for Minimum Thickness

For destructive interference and to prevent reflection, the thickness of the coating should satisfy the condition for destructive interference. This condition can be given by: \(2nt = (m + 1/2) \lambda'\), where \(n\) is the refractive index of the coating, \(t\) is the thickness, \(m\) is an integer (0 for minimum thickness), and \(\lambda'\) is the wavelength of light within the coating.
03

Calculating the Wavelength within the Coating

The wavelength of light within the coating, \(\lambda'\), is related to the wavelength in vacuum by \(\lambda' = \frac{\lambda_0}{n}\). Here, \(\lambda_0 = 565 \text{ nm}\). Therefore, \(\lambda' = \frac{565}{1.38} \approx 409.42 \text{ nm}\).
04

Applying the Condition for Minimum Thickness

Using \(m = 0\) for minimum nonzero thickness, substitute \(\lambda'\) into the destructive interference equation: \(2 \cdot 1.38 \cdot t = (0 + 1/2) \cdot 409.42 \.\)Thus, \(2.76t = 204.71 \.\)
05

Solving for the Minimum Thickness

Solve the equation \(2.76t = 204.71\) for \(t\): \(t = \frac{204.71}{2.76} \approx 74.16 \text{ nm}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Thin Film Interference
Thin film interference occurs when light waves reflected by the upper and lower boundaries of a thin film, such as a road with water or a bubble, interfere with one another. This interference is due to differences in the path length traveled by the two waves.
Visible colors can emerge due to this interference process, creating the colorful patterns often seen on soap bubbles or oil slicks.
This phenomenon is not only beautiful but also practical.
Engineers use thin film interference in coatings to reduce glare on camera lenses or eyeglasses by canceling out certain wavelengths of light.
  • The key to achieving this cancellation is the constructive and destructive interference between reflected waves.
  • The film’s thickness and the wavelength of light define the kind of interference that occurs.
  • By precisely controlling these variables, reflections at specific wavelengths can be minimized or enhanced.
For a non-reflective coating on a lens, we aim for destructive interference to eliminate reflected light, resulting in a clearer image.
Refractive Index
The refractive index of a material describes how much it reduces the speed of light traveling through it compared to the speed of light in a vacuum.
This is a vital factor in determining how light behaves as it passes through different media.
For instance, glass, water, and air all have different refractive indices, affecting the bending, or refraction, of light passing between them.
  • The refractive index is denoted as "n" and typically has values greater than 1.
  • A higher refractive index means light travels slower in the material and bends more upon entry.
  • In our exercise, magnesium fluoride has a refractive index of 1.38, while the glass of the lens has a refractive index of 1.52.
Understanding refractive indices allow for engineering surfaces with specific properties, such as creating coatings that reduce glare through thin film interference.
Wavelength in Medium
Light's wavelength changes when it enters a medium different from a vacuum due to the material's refractive index.
The wavelength in a medium, denoted as \lambda', is calculated by dividing the wavelength in a vacuum (\lambda_0) by the medium's refractive index "n".
This change in wavelength is crucial for understanding how light interacts within a medium. As in our example, where yellow-green light has a wavelength of 565 nm in vacuum:
  • \[ \lambda' = \frac{\lambda_0}{n} \]
  • The wavelength inside the magnesium fluoride, with a refractive index of 1.38, becomes approximately 409.42 nm.
  • This modification affects the interference patterns within the thin film.
These insights are applied in designing non-reflective coatings, enabling us to fine-tune optical performances in devices like camera lenses.
Destructive Interference
Destructive interference occurs when two waves combine to form a resultant wave of reduced amplitude.
This happens when the waves are out of phase, effectively canceling each other out.
In the context of thin films, this principle is used to eliminate reflective light.
  • For destructive interference within a thin film, the conditions are typically defined by:\[2nt = (m + 1/2)\lambda',\]
  • Where "2nt" represents the path difference due to the thin film, "m" is an integer, and "\lambda'" is the wavelength of light in the film.
  • This condition ensures that reflected light waves are out of phase by half a wavelength, causing them to cancel.
By applying this concept, engineers can craft coatings that minimize specific wavelength reflections, enhancing the optical performance of devices like camera lenses.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A diffraction grating has 2604 lines per centimeter, and it produces a principal maximum at \(\theta=30.0^{\circ} .\) The grating is used with light that contains all wavelengths between 410 and 660 \(\mathrm{nm}\) . What is (are) the wavelength(s) of the incident light that could have produced this maximum?

At most, how many bright fringes can be formed on either side of the central bright fringe when light of wavelength 625 \(\mathrm{nm}\) falls on a double slit whose slit separation is \(3.76 \times 10^{-6} \mathrm{m} ?\)

Light waves with two different wavelengths, 632 nm and 474 nm, pass simultaneously through a single slit whose width is 7.15 \(\times 10^{-5} \mathrm{m}\) and strike a screen 1.20 \(\mathrm{m}\) from the slit. Two diffraction patterns are formed on the screen. What is the distance (in \(\mathrm{cm}\) ) between the common center of the diffraction patterns and the first occurrence of \(\mathrm{a}\) dark fringe from one pattern falling on top of a dark fringe from the other pattern?

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.

A uniform layer of water \((n=1.33)\) lies on a glass plate \((n=1.52) .\) Light shines perpendicularly on the layer. Because of constructive interference, the layer looks maximally bright when the wavelength of the light is 432 \(\mathrm{nm}\) in vacuum and also when it is 648 \(\mathrm{nm}\) in vacuum. \((\mathrm{a})\) Obtain the minimum thickness of the film. (b) Assuming that the film has the minimum thickness and that the visible spectrum extends from 380 to \(750 \mathrm{nm},\) determine the visible wavelength(s) in vacuum for which the film appears completely dark.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.