Physics
Physics
3rd Edition
ISBN: 9781259233616
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 25, Problem 51P

(a)

To determine

Whether the central maximum be narrower or wider if red light is used instead of blue in the single-slit experiment.

(a)

Expert Solution
Check Mark

Answer to Problem 51P

The central maximum will be wider for red light_ than it is for the blue.

Explanation of Solution

Given that the width of the central maximum when blue light is used is 2.0cm.

Write the condition for the diffraction minima in a single-slit diffraction.

    asinθ=mλ                                                                                                              (I)

Here, a is the slit width, θ is the angle of diffraction, m is the order of the diffraction, λ is the wavelength of light used.

Equation (I) is used mainly for dark spots on the screen. The width of the central maximum can be quantified as the distance between the minima on each side with m=1,

    asinθ=λ                                                                                                                (II)

From equation (II) it is clear that larger wavelengths give larger angles for the first minimum.

In this case, the width of the central bright fringe is twice the distance from the center of the screen to the first minimum so that the fringe is wider for greater wavelengths. The wavelength of the red light is greater than the wavelength of the blue light so that the central maximum is wider for the red light than it is for the blue light.

Conclusion:

Therefore, the central maximum will be wider for red light_ than it is for the blue.

(b)

To determine

The width of the central maximum when red light is used.

(b)

Expert Solution
Check Mark

Answer to Problem 51P

The width of the central maximum when red light is used is 3.3cm_.

Explanation of Solution

Given that the wavelength of the blue light is 0.43μm, and the wavelength of the red light is 0.70μm.

Let the width of the central maximum is W, referring to the figure 25.31 it can be seen that,

    W=2x                                                                                                                   (III)

Here, W is the width of the central maximum, x is the distance between first minimum and the central maximum.

Write the expression for the distance x.

    x=Dtanθ                                                                                                             (IV)

Here, D is the distance between slit and the screen.

Use equation (IV) in equation (III),

    W=2Dtanθ                                                                                                             (V)

The angle θ is assumed to be small so that equation (V) becomes (use the approximation tanθsinθ),

    W2Dsinθ

Solve equation (II) for sinθ.

    sinθ=λa                                                                                                                (VI)

Use equation (VI) in equation (V) to find W.

    W=2Dλa                                                                                                             (VII)

Apply equation (VII) for the red light,

    WR=2DλRa                                                                                                         (VIII)

Apply equation (VII) for the blue light,

    WB=2DλBa                                                                                                            (IX)

Divide equation (VIII) by (IX) and solve for WR.

    WRWB=2DλRa2DλBaWR=WBλRλB                                                                                                            (X)

Conclusion:

Substitute 2.0cm for WB, 0.70μm for λR and 0.43μm for λB in equation (X) to find WR.

    WR=(2.0cm)(0.70μm0.43μm)=3.3cm

Therefore, the width of the central maximum when red light is used is 3.3cm_.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
air is pushed steadily though a forced air pipe at a steady speed of 4.0 m/s. the pipe measures 56 cm by 22 cm. how fast will air move though a narrower portion of the pipe that is also rectangular and measures 32 cm by 22 cm
No chatgpt pls will upvote
13.87 ... Interplanetary Navigation. The most efficient way to send a spacecraft from the earth to another planet is by using a Hohmann transfer orbit (Fig. P13.87). If the orbits of the departure and destination planets are circular, the Hohmann transfer orbit is an elliptical orbit whose perihelion and aphelion are tangent to the orbits of the two planets. The rockets are fired briefly at the depar- ture planet to put the spacecraft into the transfer orbit; the spacecraft then coasts until it reaches the destination planet. The rockets are then fired again to put the spacecraft into the same orbit about the sun as the destination planet. (a) For a flight from earth to Mars, in what direction must the rockets be fired at the earth and at Mars: in the direction of motion, or opposite the direction of motion? What about for a flight from Mars to the earth? (b) How long does a one- way trip from the the earth to Mars take, between the firings of the rockets? (c) To reach Mars from the…

Chapter 25 Solutions

Physics

Ch. 25.7 - Prob. 25.8PPCh. 25.8 - Prob. 25.9PPCh. 25 - Prob. 1CQCh. 25 - Prob. 2CQCh. 25 - Prob. 3CQCh. 25 - Prob. 4CQCh. 25 - Prob. 5CQCh. 25 - Prob. 6CQCh. 25 - Prob. 7CQCh. 25 - Prob. 8CQCh. 25 - Prob. 9CQCh. 25 - Prob. 10CQCh. 25 - Prob. 11CQCh. 25 - 12. In Section 25.3 we studied interference due to...Ch. 25 - Prob. 13CQCh. 25 - Prob. 14CQCh. 25 - Prob. 15CQCh. 25 - Prob. 16CQCh. 25 - Prob. 17CQCh. 25 - Prob. 18CQCh. 25 - Prob. 19CQCh. 25 - Prob. 20CQCh. 25 - Prob. 21CQCh. 25 - Prob. 1MCQCh. 25 - Prob. 2MCQCh. 25 - Prob. 3MCQCh. 25 - Prob. 4MCQCh. 25 - Prob. 5MCQCh. 25 - Prob. 6MCQCh. 25 - 7. Coherent light of a single frequency passes...Ch. 25 - Prob. 8MCQCh. 25 - Prob. 9MCQCh. 25 - Prob. 10MCQCh. 25 - Prob. 1PCh. 25 - Prob. 2PCh. 25 - Prob. 3PCh. 25 - Prob. 4PCh. 25 - Prob. 5PCh. 25 - Prob. 6PCh. 25 - Prob. 7PCh. 25 - Prob. 8PCh. 25 - Prob. 9PCh. 25 - Prob. 10PCh. 25 - Prob. 11PCh. 25 - Prob. 12PCh. 25 - Prob. 13PCh. 25 - Prob. 14PCh. 25 - Prob. 15PCh. 25 - 16. A transparent film (n = 1.3) is deposited on a...Ch. 25 - 17. A camera lens (n = 1.50) is coated with a thin...Ch. 25 - 18. A soap film has an index of refraction n =...Ch. 25 - Prob. 19PCh. 25 - Prob. 20PCh. 25 - Prob. 21PCh. 25 - Prob. 22PCh. 25 - Prob. 23PCh. 25 - Prob. 24PCh. 25 - Prob. 25PCh. 25 - Prob. 26PCh. 25 - Prob. 27PCh. 25 - Prob. 28PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 32PCh. 25 - Prob. 33PCh. 25 - Prob. 34PCh. 25 - Prob. 35PCh. 25 - Prob. 36PCh. 25 - Prob. 37PCh. 25 - Prob. 38PCh. 25 - Prob. 39PCh. 25 - Prob. 40PCh. 25 - Prob. 41PCh. 25 - Prob. 42PCh. 25 - Prob. 43PCh. 25 - 44. ✦ White light containing wavelengths from 400...Ch. 25 - Prob. 45PCh. 25 - Prob. 46PCh. 25 - 47. The central bright fringe in a single-slit...Ch. 25 - Prob. 48PCh. 25 - Prob. 49PCh. 25 - Prob. 50PCh. 25 - Prob. 51PCh. 25 - Prob. 52PCh. 25 - Prob. 53PCh. 25 - Prob. 54PCh. 25 - Prob. 55PCh. 25 - Prob. 56PCh. 25 - Prob. 57PCh. 25 - Prob. 58PCh. 25 - Prob. 59PCh. 25 - Prob. 60PCh. 25 - Prob. 61PCh. 25 - Prob. 62PCh. 25 - 63. ✦ If you shine a laser with a small aperture...Ch. 25 - Prob. 64PCh. 25 - Prob. 65PCh. 25 - Prob. 66PCh. 25 - Prob. 67PCh. 25 - Prob. 68PCh. 25 - Prob. 69PCh. 25 - 70. Coherent green light with a wavelength of 520...Ch. 25 - Prob. 71PCh. 25 - Prob. 72PCh. 25 - Prob. 73PCh. 25 - Prob. 74PCh. 25 - Prob. 75PCh. 25 - Prob. 76PCh. 25 - Prob. 77PCh. 25 - Prob. 78PCh. 25 - Prob. 79PCh. 25 - Prob. 80PCh. 25 - Prob. 81PCh. 25 - Prob. 82PCh. 25 - Prob. 83PCh. 25 - Prob. 84PCh. 25 - Prob. 85PCh. 25 - Prob. 86PCh. 25 - Prob. 87PCh. 25 - Prob. 88PCh. 25 - Prob. 89PCh. 25 - Prob. 90PCh. 25 - Prob. 91PCh. 25 - Prob. 92PCh. 25 - Prob. 93PCh. 25 - Prob. 94PCh. 25 - Prob. 95PCh. 25 - Prob. 96PCh. 25 - Prob. 97P
Knowledge Booster
Background pattern image
Physics
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.
Similar questions
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
College Physics
Physics
ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Text book image
University Physics (14th Edition)
Physics
ISBN:9780133969290
Author:Hugh D. Young, Roger A. Freedman
Publisher:PEARSON
Text book image
Introduction To Quantum Mechanics
Physics
ISBN:9781107189638
Author:Griffiths, David J., Schroeter, Darrell F.
Publisher:Cambridge University Press
Text book image
Physics for Scientists and Engineers
Physics
ISBN:9781337553278
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Text book image
Lecture- Tutorials for Introductory Astronomy
Physics
ISBN:9780321820464
Author:Edward E. Prather, Tim P. Slater, Jeff P. Adams, Gina Brissenden
Publisher:Addison-Wesley
Text book image
College Physics: A Strategic Approach (4th Editio...
Physics
ISBN:9780134609034
Author:Randall D. Knight (Professor Emeritus), Brian Jones, Stuart Field
Publisher:PEARSON
Diffraction of light animation best to understand class 12 physics; Author: PTAS: Physics Tomorrow Ambition School;https://www.youtube.com/watch?v=aYkd_xSvaxE;License: Standard YouTube License, CC-BY