/*! 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 68 The second-order maximum produce... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 28: Problem 68

The second-order maximum produced by a diffraction grating with 560 lines per centimeter is at an angle of \(3.1^{\circ}\). (a) What is the wavelength of the light that illuminates the grating? (b) If a grating with a larger number of lines per centimeter is used with this light, is the angle of the second-order maximum greater than or less than \(3.1^{\circ}\) ? Explain.

Short Answer

Expert verified
(a) \( \lambda \approx 482 \) nm. (b) The angle is greater than \( 3.1^{\circ} \).

Step by step solution

01

Convert lines per cm to lines per meter

The diffraction grating has 560 lines per centimeter. To convert this to lines per meter, multiply by 100 (since there are 100 centimeters in a meter): \( 560 \times 100 = 56,000 \) lines per meter. This gives the grating spacing, \( d \), as the reciprocal: \( d = \frac{1}{56,000} \) m.
02

Use the diffraction grating equation

The diffraction grating equation is \( d \sin \theta = m \lambda \), where \( d \) is the spacing between lines, \( \theta \) is the angle of the diffraction order, \( m \) is the order of the maximum, and \( \lambda \) is the wavelength. Given: \( d = \frac{1}{56,000} \) m, \( \theta = 3.1^{\circ} \), \( m = 2 \). We want to solve for \( \lambda \).
03

Calculate \( \lambda \)

Substitute the values into the diffraction equation: \( \frac{1}{56,000} \times \sin(3.1^{\circ}) = 2 \lambda \). Calculate \( \sin(3.1^{\circ}) \approx 0.054 \). Thus, \( \frac{1}{56,000} \times 0.054 = 2 \lambda \). Solving for \( \lambda \), we find \( \lambda \approx \frac{0.054}{2 \times 56,000} \approx 4.82 \times 10^{-7} \) m or 482 nm.
04

Analyze the effect of a larger number of lines per cm

If the number of lines per centimeter increases, \( d \) (the spacing between lines) decreases. Since \( d \sin \theta = m \lambda \), a smaller \( d \) for the same wavelength \( \lambda \) requires \( \sin \theta \) (and hence \( \theta \)) to increase to maintain the equality. Thus, the angle \( \theta \) for the second-order maximum must be greater than \( 3.1^{\circ} \).

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.

Wavelength
In the context of diffraction gratings, the wavelength is a critical concept. The wavelength (\( \lambda \)) is the distance between successive peaks of a wave.
In the exercise we are solving, we are dealing with light waves. Light, like all waves, is characterized by its wavelength, which determines its color.
In this case, the light passing through the diffraction grating is calculated to have a wavelength of 482 nm, which is in the visible spectrum.Wavelength is key because it affects how light behaves when it encounters obstacles or openings.
  • When light hits a diffraction grating, the spread of the light will change depending on its wavelength.
  • Longer wavelengths result in a wider spread and shorter wavelengths result in a narrower spread when passing through a diffraction grating.
Understanding the wavelength also helps us determine other variables, such as the angle of diffraction and the grating spacing in diffraction experiments.
Angle of Diffraction
The angle of diffraction (\( \theta \)) in a diffraction grating experiment is the angle at which light is scattered by the grating.
It is measured from the incidence direction.
In our exercise, the second-order maximum appears at an angle of 3.1 degrees.The angle of diffraction is important because it shows how the light is dispersed based on its wavelength and the grating's properties.
  • According to the diffraction grating equation (\( d \sin \theta = m \lambda \)), the angle of diffraction depends on the wavelength, the order of diffraction, and the grating spacing.
  • A larger wavelength or a higher order will increase the angle, resulting in more spread of light.
Additionally, the angle of diffraction changes if the grating spacing changes. For example, using a more finely spaced grating with the same light will result in a larger angle of diffraction.
Grating Spacing
Grating spacing (\( d \)) refers to the distance between adjacent lines on a diffraction grating.
It's the inverse of the number of lines per meter.
In the initial example, 560 lines per centimeter translates to 56,000 lines per meter, which means our grating spacing is \( d = \frac{1}{56,000} \) meters.This spacing is crucial for determining how the grating will affect light passing through it.
  • The closer the lines are together (smaller \( d \)), the more the light spreads out.
  • The relationship between grating spacing and the angle of diffraction is direct. With smaller spacing, each wavelength bends more, resulting in larger angles of diffraction.
As shown in the solution, reducing the grating spacing increases the angle of diffraction necessary to achieve higher orders of maximum, given the same wavelength.
Thus, the choice of grating spacing is essential for specific applications, determining how effectively certain wavelengths will be diffracted.

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

Images produced by structures within the eye (like lens fibers or cell fragments) are referred to as entoptic images. These images can sometimes take the form of "halos" around a bright light seen against a dark background. The halo in such a case is actually the bright outer rings of a circular diffraction pattern, like Figure \(28-21\), with the central bright spot not visible because it overlaps the direct image of the light. Find the diameter of the eye structure that causes a circular diffraction pattern with the first dark ring at an angle of \(3.7^{\circ}\) when viewed with monochromatic light of wavelength \(630 \mathrm{nm} .\) (Typical eye structures of this type have diameters on the order of \(10 \mu \mathrm{m}\). Also, the index of refraction of the vitreous humor is \(1.336 .\) )

A spy camera is said to be able to read the numbers on a car's license plate. If the numbers on the plate are \(5.0 \mathrm{cm}\) apart, and the spy satellite is at an altitude of \(160 \mathrm{km},\) what must be the diameter of the camera's aperture? (Assume light with a wavelength of \(550 \mathrm{nm} .\) )

A thin layer of magnesium fluoride \((n=1.38)\) is used to (a) What thickness should the coat a flint-glass lens \((n=1.61)\). magnesium fluoride film have if the reflection of \(565-\mathrm{nm}\) light is to be suppressed? Assume that the light is incident at right angles to the film. (b) If it is desired to suppress the reflection of light with a higher frequency, should the coating of magnesium fluoride be made thinner or thicker? Explain.

A microphone is located on the line connecting two speakers that are \(0.845 \mathrm{m}\) apart and oscillating \(180^{\circ}\) out of phase. The microphone is \(2.25 \mathrm{m}\) from the midpoint of the two speakers. What are the lowest two frequencies that produce an interference maximum at the microphone's location?

Find the minimum aperture diameter of a camera that can resolve detail on the ground the size of a person \((2.0 \mathrm{m})\) from an SR-71 Blackbird airplane flying at an altitude of \(27 \mathrm{km}\). (Assume light with a wavelength of \(450 \mathrm{nm}\).)
See all solutions

Recommended explanations on Physics Textbooks

Linear Momentum

Read Explanation

Fundamentals of Physics

Read Explanation

Electricity and Magnetism

Read Explanation

Solid State Physics

Read Explanation

Turning Points in Physics

Read Explanation

Torque and Rotational Motion

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.