/*! 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 45 35.45. One round face of a \(3.2... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 35: Problem 45

35.45. One round face of a \(3.25-\mathrm{m}\) , solid, cylindrical plastic pipe is covered with a thin black coating that completely blocks light. The opposite face is covered with a fluorescent coating that glows when it is struck by light. Two straight, thin, parallel scratches, 0.225 \(\mathrm{mm}\) apart, are made in the center of the black face. When laser light of wavelength 632.8 \(\mathrm{nm}\) shines through the slits perpendicular to the black face, you find that the central bright fringe on the opposite face is 5.82 \(\mathrm{mm}\) wide, measured between the dark fringes that border it on either side. What is the index of refraction of the plastic?

Short Answer

Expert verified
Index of refraction \( n \) is approximately 1.49.

Step by step solution

01

Understand the Problem

We have a cylindrical plastic pipe with two parallel scratches on one face acting as a double slit. A laser light passes through the slits, creating an interference pattern on the opposite face. We need to find the index of refraction (\( n \) of the plastic.
02

Determine the Formula

Since the light forms an interference pattern, we can use the double-slit diffraction formula: \[ d \sin(\theta) = m \lambda \ \]. Here, \( d \) is the distance between the slits, \( \lambda \) is the wavelength of the light, and \( m \) is the order of the fringe.
03

Calculate Slit Separation in Meters

The distance between scratches (slit separation) \( d \) is given as 0.225 mm, which we convert to meters: \( d = 0.225 \times 10^{-3} \) m.
04

Use the Geometry of the Fringe Pattern

The central fringe width is 5.82 mm, or 5.82 \( \times 10^{-3} \) m. This width corresponds to the angular width caused by the light refracting through the plastic, given by \( 2 \times \theta \). Find \( heta \) using \( \tan(\theta) \approx \sin(\theta) = \frac{x}{L} \). Here, \( x \) is the fringe width, and \( L \) is the length of the pipe.
05

Calculate \( \theta \)

The full length \( L \) of the cylinder is 3.25 m. Thus, \( \theta = \arctan(\frac{5.82 \times 10^{-3}}{3.25}) \).
06

Apply the Diffraction Formula

Since we have a central maximum, use \( m = 1 \). Plug the values into the equation:\[ d \sin(\theta) = \lambda_{ ext{effective}} = \frac{\lambda}{n} \].
07

Solve for Index of Refraction \( n \)

Re-arrange the equation to solve for \( n \): \[ n = \frac{\lambda}{d \sin(\theta)} \]. Using \( \lambda = 632.8 \times 10^{-9} \) m, substitute the values and calculate \( n \).

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.

Double-Slit Diffraction
Double-slit diffraction is an important concept in wave mechanics and optics. It involves the phenomenon where a light wave passes through two narrow slits and produces a pattern of bright and dark fringes on a screen. This happens due to the wave nature of light. Each slit acts as a source of light waves, spreading out from both slits.
These waves overlap or "interfere" when they reach a screen, resulting in a pattern of alternating bright and dark bands. This is because light waves reinforce each other (constructive interference) or cancel each other out (destructive interference), depending on their phases.
  • The distance between the slits, denoted by \(d\), influences the spacing of these fringes.
  • The light鈥檚 wavelength, \(\lambda\), also plays a crucial role in determining the interference pattern.
Understanding the geometry of how these bands form allows us to calculate other values, such as the angle \(\theta\), essential for determining the index of refraction in materials, as seen in the given problem.
Interference Pattern
An interference pattern results from the superposition of two or more waves, which constructively or destructively interfere with one another. In the context of double-slit diffraction, this pattern appears as a series of alternating light and dark bands on a screen. The dark bands occur when the waves subtract from each other (destructive interference), and the bright bands occur when they add together (constructive interference).
  • The formula \(d \sin(\theta) = m \lambda\) is used to describe the conditions under which these fringes appear, where \(m\) is the order of the fringe.
  • The central bright fringe (central maximum) is particularly prominent and occurs at \(m=0\).
The width of the central fringe is affected by the wavelength of the light and can be calculated using the geometry of the setup. This width allows us to relate light's behavior with materials it travels through, especially when determining the index of refraction.
Laser Light Wavelength
The wavelength of laser light is a key parameter in interference and diffraction phenomena. In this exercise, a laser with a wavelength of 632.8 nm is used. Characteristics of Laser Light:
  • Monochromatic: Laser light has a single wavelength, which is highly useful in producing clear and distinct interference patterns.
  • Coherent: Laser waves are in-phase, meaning their crests and troughs match up over long distances, essential for stable interference patterns.
  • Directional: Laser light spreads minimally, maintaining its intensity over greater distances, ideal for precision applications.
The wavelength of light influences how it interacts with matter. When light passes through a medium like the plastic pipe in our problem, knowing its wavelength helps us determine how much the light slows down inside the medium, quantified by the index of refraction \(n\). This exercise shows how the wavelength and the slit separation are used to find the interference pattern and, ultimately, calculate \(n\).

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

35.57. Two thin parallel slits are made in an opaque sheet of film. When a monochromatic beam of light is shone through them at normal incidence, the first bright fringes in the transmitted light occur in air at \(\pm 18.0^{\circ}\) with the original direction of the light beam on a distant screen when the apparatus is in air. When the apparatus is immersed in a liquid, the same bright fringes now occur at \(\pm 12.6^{\circ} .\) Find the index of refraction of the liquid.

35.52. Red light with wavelength 700 \(\mathrm{nm}\) is passed through a two- slit apparatus. At the same time, monochromatic visible light with another wavelength passes through the same apparatus. As a result, most of the pattern that appears on the screen is a mixture of two colors; however, the center of the third bright fringe \((m=3)\) of the red light appears pure red, with none of the other color. What are the possible wavelengths of the second type of visible light? Do you need to know the slit spacing to answer this question? Why or why not?

35.31. A uniform film of \(\mathrm{TiO}_{2}, 1036 \mathrm{nm}\) thick and having index of refraction \(2.62,\) is spread uniformly over the surface of crown glass of refractive index \(1.52 .\) Light of wavelength 520.0 \(\mathrm{nm}\) falls at normal incidence onto the film from air. You want to increase the thickness of this film so that the reflected light cancels. (a) What is the minimum thickness of \(\mathrm{TiO}_{2}\) that you must add so the reflected light cancels as desired? (b) After you make the adjustment in part (a), what is the path difference between the light reflected off the top of the film and the light that cancels it after traveling through the film? Express your answer in (i) nanometers and (ii) wave-lengths of the light in the TiO_ 2 film.

35.55. A source \(S\) of monochrounatic light and a detector \(D\) are both located in air a distance \(h\) above a horizontal plane sheet of glass, and are separated by a horizontal distance \(x\) . Waves reaching \(D\) directly from \(S\) interfere with waves that reflect off the glass. The distance \(x\) is small compared to \(h\) so that the reflection is at close to normal incidence. (a) Show that the condition for constructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=\left(m+\frac{1}{2}\right) \lambda,\) and the condition for destructive interference is \(\sqrt{x^{2}+4 h^{2}}-x=m \lambda\) (Hint: Take into account the phase change on reflection.) (b) Let \(h=24 \mathrm{cm}\) and \(x=14 \mathrm{cm} .\) What is the longest wavelength for which there will be constructive interference?

35.25. Points \(A\) and \(B\) are 56.0 m apart along an east-west line. At each of these points, a radio transmitter is emitting a \(12.5-\mathrm{MHz}\) signal horizontally. These transmitters are in phase with other and emit their beams uniformly in a horizontal plane. A receiver is taken 0.500 \(\mathrm{km}\) north of the \(A B\) line and initially placed at point \(C\) . directly opposite the midpoint of \(A B\) . The receiver can be moved only along an east-west direction but, due to its limited sensitivity, it must always remain within a range so that the intensity of the signal it receives from the transmitter is no less than \(\frac{1}{4}\) of its maximum value. How far from point \(C\) (along an east-west line) can the receiver be moved and always be able to pick up the signal?
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Magnetism and Electromagnetic Induction

Read Explanation

Astrophysics

Read Explanation

Nuclear Physics

Read Explanation

Circular Motion and Gravitation

Read Explanation

Mechanics and Materials

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.