Principles of Physics: A Calculus-Based Text
Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 27, Problem 42P
To determine

The angular separation between spectral lines in the visible spectrum.

Expert Solution & Answer
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Answer to Problem 42P

The angular separation between spectral lines of first, second and third order is 5.91°,13.2°and26.5°.

Explanation of Solution

Draw the diagram of spectral lines as per the given data.

Principles of Physics: A Calculus-Based Text, Chapter 27, Problem 42P

Write the equation to find the grating spacing.

    d=1cmngrooves        (I)

Here, d is the grating spacing and ngrooves is the number of grooves per cm.

Write the equation to find the diffraction angle.

    sinθ=mλd        (II)

Here, θ is the diffraction angle λ is the wavelength of light, m is the order and d is the separation between grooves.

Write the equation to find the angular separation between red and blue lines.

    Δθ=θredθblue        (III)

Here, Δθ is the angular separation, θred is the diffraction angle of red wavelength and θblue is the diffraction angle of blue wavelength.

Conclusion:

Substitute 4500grooves/cm for ngrooves in equation (I).

    d=1cm4500grooves/cm=2.22×106m

Substitute 1 for m, 656nm for λ and 2.22×106m for d in equation (II) to find θ for red light.

  sinθred=(1)(656nm(109m1nm))2.22×106m=0.295

Rewrite the expression for θred.

  θred=sin1(0.295)=17.17°

Substitute 1 for m, 434nm for λ and 2.22×106m for d in equation (II) to find θ for blue light.

  sinθblue=(1)(434nm(109m1nm))2.22×106m=0.195

Rewrite the expression for θblue.

  θblue=sin1(0.195)=11.26°

Substitute 17.17° for θred and 11.26° for θblue in equation (III) to find Δθ for first order.

    Δθ=17.17°11.26°=5.91°

Substitute 2 for m, 656nm for λ and 2.22×106m for d in equation (II) to find θ for red light.

  sinθred=(2)(656nm(109m1nm))2.22×106m=0.59

Rewrite the expression for θred.

  θred=sin1(0.59)=36.16°

Substitute 2 for m, 434nm for λ and 2.22×106m for d in equation (II) to find θ for blue light.

  sinθblue=(2)(434nm(109m1nm))2.22×106m=0.39

Rewrite the expression for θblue.

  θblue=sin1(0.39)=22.95°

Substitute 36.16° for θred and 22.95° for θblue in equation (III) to find Δθ for second order.

    Δθ=36.16°22.95°=13.2°

Substitute 3 for m, 656nm for λ and 2.22×106m for d in equation (II) to find θ for red light.

  sinθred=(3)(656nm(109m1nm))2.22×106m=0.885

Rewrite the expression for θred.

  θred=sin1(0.885)=62.25°

Substitute 3 for m, 434nm for λ and 2.22×106m for d in equation (II) to find θ for blue light.

  sinθblue=(3)(434nm(109m1nm))2.22×106m=0.585

Rewrite the expression for θblue.

  θblue=sin1(0.585)=35.8°

Substitute 62.25° for θred and 35.8° for θblue in equation (III) to find Δθ for third order.

    Δθ=62.25°35.8°=26.5°

Thus, the angular separation between spectral lines of first, second and third order is 5.91°,13.2°and26.5°.

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Chapter 27 Solutions

Principles of Physics: A Calculus-Based Text

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