/*! 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 51 Violet light (wavelength \(=410 ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 27: Problem 51

Violet light (wavelength \(=410 \mathrm{nm}\) ) and red light (wavelength \(=\) \(660 \mathrm{nm}\) ) lie at opposite ends of the visible spectrum. (a) For each wavelength, find the angle \(\theta\) that locates the first-order maximum produced by a grating with 3300 lines/cm. This grating converts a mixture of all colors between violet and red into a rainbow-like dispersion between the two angles. Repeat the calculation above for (b) the second-order maximum and (c) the thirdorder maximum. (d) From your results, decide whether there is an overlap between any of the "rainbows" and, if so, specify which orders overlap.

Short Answer

Expert verified
No overlap occurs between any orders.

Step by step solution

01

Convert Units

First, convert the wavelength of light from nanometers (nm) to meters (m), because we generally use meters in physics problems. For violet light: \[\lambda_{\text{violet}} = 410 \text{ nm} = 410 \times 10^{-9} \text{ m}\]For red light: \[\lambda_{\text{red}} = 660 \text{ nm} = 660 \times 10^{-9} \text{ m}\]
02

Calculate Grating Equation

The grating equation is given by:\[d \cdot \sin \theta = m \cdot \lambda\]where:- \(d\) is the distance between adjacent lines on the grating,- \(\theta\) is the diffraction angle,- \(m\) is the order of the maximum,- \(\lambda\) is the wavelength of the light.Convert 3300 lines/cm to meters to find \(d\):\[d = \frac{1}{3300 \times 100} = \frac{1}{330000} \text{ m}\]
03

Calculate First-Order Maximum (\(m=1\))

For the first-order maximum, set \(m = 1\):Violet light:\[\frac{1}{330000} \cdot \sin \theta = 1 \cdot 410 \times 10^{-9}\]\[\sin \theta_{\text{violet}} = 410 \times 10^{-9} \cdot 330000\]\[\\theta_{\text{violet}} = \sin^{-1}(0.1353) \approx 7.8^\circ\]Red light:\[\frac{1}{330000} \cdot \sin \theta = 1 \cdot 660 \times 10^{-9}\]\[\sin \theta_{\text{red}} = 660 \times 10^{-9} \cdot 330000\]\[\theta_{\text{red}} = \sin^{-1}(0.2178) \approx 12.6^\circ\]
04

Calculate Second-Order Maximum (\(m=2\))

For the second-order maximum, set \(m = 2\):Violet light:\[\frac{1}{330000} \cdot \sin \theta = 2 \cdot 410 \times 10^{-9}\]\[\sin \theta_{\text{violet}} = 0.2706\]\[\theta_{\text{violet}} = \sin^{-1}(0.2706) \approx 15.7^\circ\]Red light:\[\frac{1}{330000} \cdot \sin \theta = 2 \cdot 660 \times 10^{-9}\]\[\sin \theta_{\text{red}} = 0.4356\]\[\theta_{\text{red}} = \sin^{-1}(0.4356) \approx 25.9^\circ\]
05

Calculate Third-Order Maximum (\(m=3\))

For the third-order maximum, set \(m = 3\):Violet light:\[\frac{1}{330000} \cdot \sin \theta = 3 \cdot 410 \times 10^{-9}\]\[\sin \theta_{\text{violet}} = 0.4059\]\[\theta_{\text{violet}} = \sin^{-1}(0.4059) \approx 24.0^\circ\]Red light:\[\frac{1}{330000} \cdot \sin \theta = 3 \cdot 660 \times 10^{-9}\]\[\sin \theta_{\text{red}} = 0.6534\]\[\theta_{\text{red}} = \sin^{-1}(0.6534) \approx 40.8^\circ\]
06

Check for Overlapping Orders

Compare the angles calculated for each order. Two orders will overlap if their angle predictions coincide or are very close: - First-order angles do not overlap (7.8掳 to 12.6掳). - Second-order angles do not overlap (15.7掳 to 25.9掳). - Third-order angles do not overlap (24.0掳 to 40.8掳). There are no overlaps between these orders.

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.

Violet Light
Violet light, one of the colors visible to the human eye, occupies the shorter end of the visible light spectrum. It has a wavelength of approximately 410 nm. A fascinating aspect of violet light is its ability to refract more compared to other colors in the spectrum. This occurs due to its shorter wavelength and higher frequency. In the context of diffraction gratings, violet light will diffract at different angles when compared to longer wavelength lights, such as red.
  • Shorter wavelength: Tends to diffract at larger angles.
  • Higher frequency: Results in more rapid oscillations of the light wave.
When a diffraction grating is used, the difference in angles at which violet and red diffract contributes to the phenomenon of spectral separation, similar to a rainbow.
Red Light
Red light is at the opposite end of the visible spectrum from violet, with a wavelength around 660 nm. Its longer wavelength means that it diffracts at smaller angles compared to violet light when passed through a diffraction grating. This longer wavelength also correlates with a lower frequency than violet light.
  • Longer wavelength: Leads to a smaller diffraction angle.
  • Lower frequency: Waves oscillate more slowly compared to violet light.
Understanding the behavior of red light in diffraction is crucial for observing how different colors separate naturally, producing a spectrum based on their wavelengths.
First-order Maximum
The first-order maximum refers to the initial level of diffraction for a specific light color as it passes through a grating. It occurs when the path difference between adjacent light waves is equal to one full wavelength. For violet light, this results in an angle of approximately 7.8掳 for the first-order maximum, while red light is observed at about 12.6掳.To determine the first-order maximum:
  • Identify the order of the maximum, here set as \(m = 1\).
  • Use the grating formula: \(d \cdot \sin \theta = m \cdot \lambda\).
This concept is crucial for understanding how each wavelength focuses at different angles, creating a clear and separated image of the color spectrum.
Second-order Maximum
The second-order maximum is another significant diffraction point and relates to situations where the path difference equals two wavelengths. For violet light, this second order occurs at approximately 15.7掳, whereas for red light, it reaches about 25.9掳.Understanding how to calculate the second-order maximum involves similar steps:
  • Set the order of maximum \(m = 2\).
  • Apply the formula: \(d \cdot \sin \theta = m \cdot \lambda\) with \(m=2\).
The second-order maximum gives insight into how multiple wavelengths interact with the grating, offering additional opportunities for separating and analyzing different colors within a spectrum.
Third-order Maximum
The third-order maximum occurs when the path difference through the grating is equal to three wavelengths. For violet light, this third-order is seen at around 24.0掳, and for red light, it extends to approximately 40.8掳.To determine the third-order maximum:
  • Set the order \(m = 3\).
  • Utilize the formula: \(d \cdot \sin \theta = m \cdot \lambda\), replacing \(m\) with 3.
The third-order maximum provides a further separation in colors, helping to clarify and enhance the observation of the spectrum produced by a diffraction grating. By studying these various orders, one gains an understanding of how light behaves in different circumstances, leading to a more comprehensive grasp of diffraction phenomena.

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 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?

It is claimed that some professional baseball players can see which way the ball is spinning as it travels toward home plate. One way to judge this claim is to estimate the distance at which a batter can first hope to resolve two points on opposite sides of a baseball, which has a diameter of \(0.0738 \mathrm{m} .\) (a) Estimate this distance, assuming that the pupil of the eye has a diameter of \(2.0 \mathrm{mm}\) and the wavelength of the light is \(550 \mathrm{nm}\) in vacuum. (b) Considering that the distance between the pitcher's mound and home plate is \(18.4 \mathrm{m},\) can you rule out the claim based on your answer to part (a)?

A soap film \((n=1.33)\) is \(375 \mathrm{nm}\) thick and coats a flat piece of glass \((n=1.52) .\) Thus, air is on one side of the film and glass is on the other side, as the figure illustrates. Sunlight, whose wavelengths (in vacuum) extend from 380 to \(750 \mathrm{nm},\) travels through air and strikes the film nearly perpendicularly. Concepts: (i) What, if any, phase change occurs when light, traveling in air, reflects from the air-film interface? (ii) What, if any, phase change occurs when light, traveling in the film, reflects from the film-glass interface? (iii) Is the wavelength of the light in the film greater than, smaller than, or equal to the wavelength in a vacuum? Calculations: For what wavelengths in the range of 380 to 750 nm does constructive interference cause the film to look bright in reflected light?

For a wavelength of \(420 \mathrm{nm},\) a diffraction grating produces a bright fringe at an angle of \(26^{\circ} .\) For an unknown wavelength, the same grating produces a bright fringe at an angle of \(41^{\circ} .\) In both cases the bright fringes are of the same order \(m .\) What is the unknown wavelength?

In a setup like that in Figure \(27.7,\) a wavelength of \(625 \mathrm{nm}\) is used in a Young's double-slit experiment. The separation between the slits is \(d=\) \(1.4 \times 10^{-5} \mathrm{m} .\) The total width of the screen is \(0.20 \mathrm{m} .\) In one version of the setup, the separation between the double slit and the screen is \(L_{\mathrm{A}}=0.35 \mathrm{m}\) whereas in another version it is \(L_{\mathrm{B}}=0.50 \mathrm{m} .\) On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting.
See all solutions

Recommended explanations on Physics Textbooks

Electrostatics

Read Explanation

Electricity

Read Explanation

Fluids

Read Explanation

Translational Dynamics

Read Explanation

Astrophysics

Read Explanation

Particle Model of Matter

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.