Concept explainers
(a)
The number of waves constituting the laser light and the wavelength of each waves precise to eight digits.
(a)
Answer to Problem 64P
The laser light is made up of
Explanation of Solution
The reflecting surface of the mirror is metallic in nature. So the waves formed has nodes at both the ends. The distance between two nodes is
Write the expression for the distance between
Here,
The active medium between the reflecting mirrors amplify lights with the wavelengths in between
Write the expression for the midpoint of the wavelength range.
Here,
Using these values gathered, a trial value of number of nodes can be calculated.
Use expression (I) to find a trial value of
Here,
Now assume some values for
Use expression (I) to find the wavelength of the first wave of laser light.
Here,
Write the expression for wavelength of second wave.
Here,
Write the expression for wavelength of third wave.
Here,
Write the expression for wavelength of fourth wave.
Here,
Conclusion:
Substitute
Substitute
The trial value of
Therefore the possible
Substitute
Substitute
Substitute
Substitute
The active medium between the reflecting mirrors amplify lights with the wavelengths in between
Therefore, the laser light is made up of
(b)
The root mean square speed for neon atom at
(b)
Answer to Problem 64P
The root mean square value of speed for the neon atom at
Explanation of Solution
Write the expression for the rms speed of the atom.
Here,
Write the expression to convert temperature from Celsius scale to Kelvin scale.
Here,
Conclusion:
Substitute
Substitute
Therefore, the root mean square value of speed for the neon atom at
(c)
Show that the Doppler effect for light emitted by the moving Neon atoms makes the bandwidth of the amplifier larger than
(c)
Answer to Problem 64P
It is shown that the Doppler effect for light emitted by the moving Neon atoms makes the bandwidth of the amplifier larger than
Explanation of Solution
Write the expression for the Doppler shift in the
Here,
Write the expression for frequency after Doppler shift.
Here,
Write the expression for frequency before Doppler shift.
Here,
Use expressions (XII) and (XIII) in (X) and solve for
Conclusion:
Substitute
Therefore, the new wavelength is greater than
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Chapter 42 Solutions
Physics for Scientists and Engineers with Modern, Revised Hybrid (with Enhanced WebAssign Printed Access Card for Physics, Multi-Term Courses)
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- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage Learning