Physics
Physics
5th Edition
ISBN: 9781260486919
Author: GIAMBATTISTA
Publisher: MCG
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Chapter 25, Problem 86P
To determine

The sketch of the grating spectra with its labels.

Expert Solution & Answer
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Answer to Problem 86P

The diffraction spectra is given in figure 1.

Explanation of Solution

The wavelength of the red light is 690nm, wavelength of the blue light is 460nm, the grating-screen distance is 2.0m and the slit density is 10000.0slits/cm.

Write the expression for slit separation

    d=1dn                                                                                                    (I)

Here, dn is the slit density and d is the slit separation.

Substitute 10000.0slits/cm for dn in (I) to find d

  d=110000.0slits/cm=110000.0slits/cm1021cm=1.00000×106m

Write the expression for diffraction maxima

  dsinθ=mλ

Here, λ is the wavelength, m is the order of the fringe and θ is the angular spacing for fringes.

Rearrange for θ

    θ=sin1(mλd)                                                                                       (II)

Write the expression for θ

  tanθ=ΔxD

Here, Δx is the distance from the central fringe and D is the grating-screen distance.

Rearrange for Δx

  Δx=Dtanθ                                                                                          (III)

Substitute (II) in (III)

  Δx=Dtan(sin1(mλd))                                                                       (IV)

Substitute 0 for m, 2.0m for D, 690nm for λ and 1.00000×106m for d in (IV) to find Δx for central red fringe.

  Δx=690nmtan(sin10×690nm1.00000×106m)=0

Substitute ±1 for m, 2.0m for D, 690nm for λ and 1.00000×106m for d in (IV) to find Δx for first order red light.

  Δx=690nmtan(sin1(±1)690nm1.00000×106m)=690nmtan(sin1(±1)690×109m1.00000×106m)=±1.9nm

Substitute 0 for m, 2.0m for D, 460nm for λ and 1.00000×106m for d in (IV) to find Δx for central blue fringe.

  Δx=460nmtan(sin10×690nm1.00000×106m)=0

Substitute ±1 for m, 2.0m for D, 460nm for λ and 1.00000×106m for d in (IV) to find Δx for first order blue light.

  Δx=460nmtan(sin1(±1)690nm1.00000×106m)=460nmtan(sin1(±1)690×109m1.00000×106m)=±1.0m

Substitute ±2 for m, 2.0m for D, 460nm for λ and 1.00000×106m for d in (IV) to find Δx for second order blue light.

  Δx=460nmtan(sin1(±2)690nm1.00000×106m)=460nmtan(sin1(±2)690×109m1.00000×106m)=±4.7m

The blue light and red light overlap for the central fringe and hence it appears purple.

The grating spectra drawn using the above information is given below:

Physics, Chapter 25, Problem 86P

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