Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 38, Problem 38.28P

Three discrete spectral lines occur at angles of 10.1°, 13.7°, and 14.8° in the first-order spectrum of a grating spectrometer. (a) If the grating has 3 660 slits/cm, what are the wavelengths of the light? (b) At what angles are these lines found in the second-order spectrum?

(a)

Expert Solution
Check Mark
To determine
The wavelengths of the light.

Answer to Problem 38.28P

The wavelengths of light are 487.52nm , 658.41nm and 710.14nm at spectral angles of 10.1° , 13.7° and 14.8° respectively.

Explanation of Solution

Given info: Angles of spectral lines are 10.1° , 13.7° and 14.8° , Slits on the grating are 3600slits/cm .

The width of slit can be given as,

d=1N

Here,

N is the number of slits per length.

d is the width of the slit.

Substitute 3600slits/cm for N in the above equation,

d=13600slits/cm=(2.78×104cm)(1m100cm)=2.78×106m

The condition for first order diffraction grating can be given as,

λ=dsinθ (1)

Here,

θ is the angle of spectral line.

λ is the wavelength of light.

For the angle of 10.1° :

Substitute 2.78×106m for d , λ1 for λ and 10.1° for θ in the equation (1),

λ1=(2.78×106m)(sin10.1°)=487.52×109m=487.52nm

Thus, the wavelength of the light is 487.52nm at angle of 10.1° .

For the angle of 13.7° :

Substitute 2.78×106m for d , λ2 for λ and 13.7° for θ in the equation (1),

λ2=(2.78×106m)(sin13.7°)=658.41×109m=658.41nm

Thus, the wavelength of the light is 658.41nm at angle of 13.7° .

For the angle of 14.8° :

Substitute 2.78×106m for d , λ3 for λ and 14.8° for θ in the equation (1),

λ3=(2.78×106m)(sin14.8°)=710.14×109m=710.14nm

Thus, the wavelength of the light is 710.14nm at angle of 14.8° .

Conclusion:

Therefore, the wavelengths of light are 487.52nm , 658.41nm and 710.14nm at spectral angles of 10.1° , 13.7° and 14.8° respectively.

(b)

Expert Solution
Check Mark
To determine
The angles for the lines in the second order spectrum.

Answer to Problem 38.28P

The angles for the second order lines are 20.53° , 28.27° and 30.72° for the wavelengths of 487.52nm , 658.41nm and 710.14nm respectively.

Explanation of Solution

Given info: Angles of spectral lines are 10.1° , 13.7° and 14.8° , Slits on the grating are 3600slits/cm .

The condition for first order diffraction grating can be given as,

dsinθ=2λ (1)

θ is the angle of spectral line.

λ is the wavelength of light.

Rearrange the above equation for θ ,

θ=sin1(2λd) (2)

For the wavelength 487.52nm :

Substitute 2.78×106m for d , 487.52×109m for λ and θ1 for θ in the equation (2),

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 38, Problem 38.28P , additional homework tip  1 θ1=sin1(2(487.52×109m)(2.78×106m))=20.53°

Thus, the angle of the spectral line is 20.53° for wavelength of 487.52nm .

For the wavelength 658.41nm :

Substitute 2.78×106m for d , 658.41×109m for λ and θ2 for θ in the equation (2),

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 38, Problem 38.28P , additional homework tip  2 θ2=sin1(2(658.41×109m)(2.78×106m))=28.27°

Thus, the angle of the spectral line is 28.27° for wavelength of 658.41nm .

For the wavelength 710.14nm :

Substitute 2.78×106m for d , 710.14×109m for λ and θ3 for θ in the equation (2),

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 38, Problem 38.28P , additional homework tip  3 θ3=sin1(2(710.14×109m)(2.78×106m))=30.72°

Thus, the angle of the spectral line is 30.72° for wavelength of 710.14nm .

Conclusion:

Therefore, the angles of lines are 20.53° , 28.27° and 30.72° for the wavelengths of 487.52nm , 658.41nm and 710.14nm respectively.

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Chapter 38 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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