/*! 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 15 A transparent film \((n=1.43)\) ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 15

A transparent film \((n=1.43)\) is deposited on a glass plate \((n=1.52)\) to form a nonreflecting coating. The film has a thickness that is \(1.07 \times 10^{-7} \mathrm{m} .\) What is the longest possible wavelength (in vacuum) of light for which this film has been designed?

Short Answer

Expert verified
The longest possible wavelength in vacuum is approximately 875 nm.

Step by step solution

01

Understanding the Problem

The aim is to find the longest wavelength in vacuum at which there's destructive interference, making the film non-reflective. For destructive interference, the extra path length difference should ideally be half a wavelength in the medium, removing reflectance.
02

Using the Condition for Destructive Interference

Destructive interference occurs when the optical path difference (twice the thickness, since light reflects from the bottom) equals half a wavelength. This condition can be expressed as: \( 2d = (m + \frac{1}{2}) \frac{\lambda}{n} \), where \( \lambda \) is the wavelength in the film, \( d \) is the thickness of the film, and \( n \) is the refractive index of the film.
03

Calculate Wavelength in the Film

Rearrange the equation to find \( \lambda \): \( \lambda = \frac{2d \, n}{m + \frac{1}{2}} \). Here, \( d = 1.07 \times 10^{-7} \) m and \( n = 1.43 \). We choose \( m = 0 \) to find the longest \( \lambda \).
04

Substitute Values and Calculate

Substitute the given values into the formula: \( \lambda = \frac{2 \times 1.07 \times 10^{-7} \, \times 1.43}{0 + 0.5} = \frac{3.06 \times 10^{-7}}{0.5} = 6.12 \times 10^{-7} \mathrm{m} \).
05

Convert to Wavelength in Vacuum

The calculated wavelength is in the medium (the film). To find the wavelength in vacuum, use the refractive index of the film: \( \lambda_{\text{vacuum}} = \lambda_{\text{film}} \times n = 6.12 \times 10^{-7} \times 1.43 \approx 8.75 \times 10^{-7} \mathrm{m} \).

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 91影视!

Key Concepts

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

Destructive Interference
Destructive interference is a phenomenon where two waves superpose to form a wave with a reduced amplitude. This occurs when the waves are out of phase by half a wavelength (\( \lambda/2 \) causing them to cancel each other out. In the context of optics, such as with nonreflective coatings, this interference reduces reflected light to a minimum, ideally making surfaces appear dark and free of reflection.

In the example of a transparent film on glass, the path difference must be carefully controlled to achieve destructive interference. Specifically:
  • The optical path difference (OPD) must be equal to an odd multiple of half the wavelength of light in the medium.
  • This typically translates to a thickness adjustment in the film.
Through this method, we can effectively "tune" the film to a specific wavelength, neutralizing reflection for that particular wavelength and making the coating nonreflective.
Refractive Index
The refractive index, denoted as \( n \), is a crucial factor in understanding light behavior as it passes from one medium to another. It measures how much the speed of light is reduced inside a material compared to vacuum.

For example, in our problem, we had a film with \( n = 1.43 \) and underlying glass with \( n = 1.52 \). This difference affects how light is refracted and consequently reflected:
  • A higher refractive index means light travels slower in the medium.
  • This also implies that the wavelength of light becomes shorter as it enters the medium from air or vacuum.
Understanding refractive index is essential for calculating the correct conditions for destructive interference since it influences the optical path difference to balance reflection.
Optical Path Difference
The optical path difference (OPD) is the difference in path lengths of two light beams as they traverse different routes. In the case of thin film coatings, the OPD is responsible for determining the condition under which interference occurs.

For a nonreflective coating, like the one discussed, we need to account for the light path as it enters and exits the film. This involves:
  • Calculating the depth or thickness that light travels within the film.
  • Considering the phase shift that may occur upon reflection.
The general formula for OPD is given by:\[OPD = 2d \]where \( d \) is the thickness of the film. As seen in the original solution, fine-tuning the optical path difference is key to achieving the destructive interference which cancels out reflection in nonreflective coatings.

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

The central bright fringe in a single-slit diffraction pattern has a width that equals the distance between the screen and the slit. Find the ratio \(\lambda / W\) of the wavelength \(\lambda\) of the light to the width \(W\) of the slit.

A circular drop of oil lies on a smooth, horizontal surface. The drop is thickest in the center and tapers to zero thickness at the edge. When illuminated from above by blue light \((\lambda=455 \mathrm{nm}), 56\) concentric bright rings are visible, including a bright fringe at the edge of the drop. In addition, there is a bright spot in the center of the drop. When the drop is illuminated from above by red light \((\lambda=637 \mathrm{nm}),\) a bright spot again appears at the center, along with a different number of bright rings. Ignoring the bright spot, how many bright rings appear in red light? Assume that the index of refraction of the oil is the same for both wavelengths. The ability to exhibit interference effects is a fundamental characteristic of any kind of wave. Our understanding of these effects depends on the principle of linear superposition, which we first encountered in Chapter 17\. Only by means of this principle can we understand the constructive and destructive interference of light waves that lie at the heart of every topic in this chapter. Problem 67 serves as a review of the essence of this principle. Problem 68 deals with thin-film interference and reviews the factors that must be considered in such cases.

Two in-phase sources of waves are separated by a distance of \(4.00 \mathrm{m}\). These sources produce identical waves that have a wavelength of \(5.00 \mathrm{m}\). On the line between them, there are two places at which the same type of interference occurs. (a) Is it constructive or destructive interference, and (b) where are the places located?

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

A dark fringe in the diffraction pattern of a single slit is located at an angle of \(\theta_{\mathrm{A}}=34^{\circ} .\) With the same light, the same dark fringe formed with another single slit is at an angle of \(\theta_{\mathrm{B}}=56^{\circ} .\) Find the ratio \(W_{A} / W_{\mathrm{B}}\) of the widths of the two slits.
See all solutions

Recommended explanations on Physics Textbooks

Quantum Physics

Read Explanation

Engineering Physics

Read Explanation

Medical Physics

Read Explanation

Oscillations

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Solid State Physics

Read Explanation
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.