Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 25, Problem 45P

(a)

To determine

The number of orders of lines that can be seen with the grating.

(a)

Expert Solution
Check Mark

Answer to Problem 45P

The number of orders of lines that can be seen with the grating is 3_.

Explanation of Solution

Given that the width of the grating is 1.600cm, the number of slits is 12000, and the wavelengths of light are λa=440.000nm and λb=440.936nm.

Write the expression for the slit width of the grating.

  d=width of gratingtotal number of slits                                                                                     (I)

Here, d is the slit width.

Write the expression for the diffraction maxima using grating.

  dsinθ=mλ                                                                                                    (II)

Here, θ is the angle of diffraction, λ is the wavelength of light, and m is the order of diffraction.

Deduce the number of orders that can be seen, from equation (II).

  mdsinθλ                                                                                                      (III)

The maximum number of orders is obtained for θ=90° (since sinθ=90°).

Conclusion:

Substitute 1.600cm for width of grating, and 12000 for the total number of slits in equation (I) to find d.

  d=1.600cm12000=1.600cm×1m100cm12000=1.333×106m

Substitute 1.333×106m for d, 90° for θ, and 440.000nm for λ in equation (III) to find m.

  m(1.333×106m)sin90°440.000nm=3.024

Since m must be an integer, it can be taken as m=3.

Therefore, the number of orders of lines that can be seen with the grating is 3_.

(b)

To determine

The angular separation θbθa between the lines in each order.

(b)

Expert Solution
Check Mark

Answer to Problem 45P

The angular separation θbθa between the lines in each order are θb1θa1=0.04°_, θb2θa2=0.11°_, and θb3θa3=0.91°_.

Explanation of Solution

Given that the width of the grating is 1.600cm, the number of slits is 12000, and the wavelengths of light are λa=440.000nm and λb=440.936nm.

Solve equation (II) for θ.

  θ=sin1(mλd)                                                                                          (IV)

The angles corresponding to each wavelengths for each orders can be computed and their difference gives the angular separation.

Conclusion:

The value of 1/d can be computed from equation (I) as,

  1d=120001.600cm=120001.600cm×1m100cm=7.500×105/m

Substitute 1 for m, 440.000nm for λ, and 7.500×105/m for 1/d in equation (IV) to find θ corresponding to order 1 of λa, (θa1).

  θa1=sin1[(1)(440.000nm)(7.500×105/m)]=sin1[(1)(440.000nm×1m1×109m)(7.500×105/m)]=19.27°

Substitute 2 for m, 440.000nm for λ, and 7.500×105/m for 1/d in equation (IV) to find θ corresponding to order 2 of λa, (θa2)

  θa2=sin1[(2)(440.000nm)(7.500×105/m)]=sin1[(2)(440.000nm×1m1×109m)(7.500×105/m)]=41.30°

Substitute 3 for m, 440.000nm for λ, and 7.500×105/m for 1/d in equation (IV) to find θ corresponding to order 3 of λa, (θa3)

  θa3=sin1[(3)(440.000nm)(7.500×105/m)]=sin1[(3)(440.000nm×1m1×109m)(7.500×105/m)]=81.89°

Substitute 1 for m, 440.936nm for λ, and 7.500×105/m for 1/d in equation (IV) to find θ corresponding to order 1 of λb, (θb1).

  θb1=sin1[(1)(440.936nm)(7.500×105/m)]=sin1[(1)(440.936nm×1m1×109m)(7.500×105/m)]=19.31°

Substitute 2 for m, 440.936nm for λ, and 7.500×105/m for 1/d in equation (IV) to find θ corresponding to order 2 of λb, (θb2)

  θb2=sin1[(2)(440.936nm)(7.500×105/m)]=sin1[(2)(440.936nm×1m1×109m)(7.500×105/m)]=41.41°

Substitute 3 for m, 440.936nm for λ, and 7.500×105/m for 1/d in equation (IV) to find θ corresponding to order 3 of λb, (θb3)

  θb3=sin1[(3)(440.936nm)(7.500×105/m)]=sin1[(3)(440.936nm×1m1×109m)(7.500×105/m)]=82.80°

Compute the angular separation for each orders.

  θb1θa1=19.31°19.27°=0.04°

  θb2θa2=41.41°41.30°=0.11°

  θb3θb3=82.80°81.89°=0.91°

Therefore, the angular separation θbθa between the lines in each order are θb1θa1=0.04°_, θb2θa2=0.11°_, and θb3θa3=0.91°_.

(c)

To determine

The order of the spectrum which best resolves the two lines.

(c)

Expert Solution
Check Mark

Answer to Problem 45P

The third order spectrum best resolves the two lines.

Explanation of Solution

The angular separation for different orders of the spectrum are obtained as θb1θa1=0.04°, θb2θa2=0.11°, and θb3θa3=0.91°.

The higher resolution corresponds to larger angular separation in the diffraction process. Among the obtained values of the angular separations, the third order spectrum has the largest angular separation. Thus, the two lines are best resolved in the third-order spectrum.

Conclusion:

Therefore, the third order spectrum best resolves the two lines.

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
A uniform ladder of length L and weight w is leaning against a vertical wall. The coefficient of static friction between the ladder and the floor is the same as that between the ladder and the wall. If this coefficient of static friction is μs : 0.535, determine the smallest angle the ladder can make with the floor without slipping. ° = A 14.0 m uniform ladder weighing 480 N rests against a frictionless wall. The ladder makes a 55.0°-angle with the horizontal. (a) Find the horizontal and vertical forces (in N) the ground exerts on the base of the ladder when an 850-N firefighter has climbed 4.10 m along the ladder from the bottom. horizontal force magnitude 342. N direction towards the wall ✓ vertical force 1330 N up magnitude direction (b) If the ladder is just on the verge of slipping when the firefighter is 9.10 m from the bottom, what is the coefficient of static friction between ladder and ground? 0.26 × You appear to be using 4.10 m from part (a) for the position of the…
Your neighbor designs automobiles for a living. You are fascinated with her work. She is designing a new automobile and needs to determine how strong the front suspension should be. She knows of your fascination with her work and your expertise in physics, so she asks you to determine how large the normal force on the front wheels of her design automobile could become under a hard stop, ma when the wheels are locked and the automobile is skidding on the road. She gives you the following information. The mass of the automobile is m₂ = 1.10 × 103 kg and it can carry five passengers of average mass m = 80.0 kg. The front and rear wheels are separated by d = 4.45 m. The center of mass of the car carrying five passengers is dCM = 2.25 m behind the front wheels and hcm = 0.630 m above the roadway. A typical coefficient of kinetic friction between tires and roadway is μk 0.840. (Caution: The braking automobile is not in an inertial reference frame. Enter the magnitude of the force in N.)…
John is pushing his daughter Rachel in a wheelbarrow when it is stopped by a brick 8.00 cm high (see the figure below). The handles make an angle of 0 = 17.5° with the ground. Due to the weight of Rachel and the wheelbarrow, a downward force of 403 N is exerted at the center of the wheel, which has a radius of 16.0 cm. Assume the brick remains fixed and does not slide along the ground. Also assume the force applied by John is directed exactly toward the center of the wheel. (Choose the positive x-axis to be pointing to the right.) (a) What force (in N) must John apply along the handles to just start the wheel over the brick? (No Response) N (b) What is the force (magnitude in kN and direction in degrees clockwise from the -x-axis) that the brick exerts on the wheel just as the wheel begins to lift over the brick? magnitude (No Response) KN direction (No Response) ° clockwise from the -x-axis

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. 36PCh. 25 - Prob. 28PCh. 25 - Prob. 32PCh. 25 - Prob. 29PCh. 25 - Prob. 30PCh. 25 - Prob. 31PCh. 25 - Prob. 34PCh. 25 - Prob. 33PCh. 25 - Prob. 35PCh. 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. 60PCh. 25 - Prob. 61PCh. 25 - Prob. 59PCh. 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. 91PCh. 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. 93PCh. 25 - Prob. 92PCh. 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
Spectra Interference: Crash Course Physics #40; Author: CrashCourse;https://www.youtube.com/watch?v=-ob7foUzXaY;License: Standard YouTube License, CC-BY