University Physics V · topic UPV-13.09
Line profiles, saturation, and Doppler-free laser spectroscopy
Connect the ideas, make your working visible, and test the result. Use the hints when you need a starting point; open the full solution after your attempt.
Further practice
Work through a fresh case before tackling the extension.
Line profiles and Doppler-free spectroscopy: predict a finite change
A line profile is the distribution of absorption strength over frequency, and Doppler-free spectroscopy is any scheme that reports the homogeneous width of that profile by making the whole sample respond as a single velocity class. Use the supplied topic model Y_out = fg/(gam*(1+sqrt(1+sat))/2), where Y_out is DOPPLER / DIP WIDTH (×). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. gam is Natural width Γ/2π, measured in MHz. Keep fg (Doppler FWHM of the pedestal) = 1300 MHz; sat (Saturation parameter S = I/I_sat) = 2 fixed. For this exercise use only 4.275 MHz ≤ gam ≤ 13.725 MHz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. Calculate Y_out at gam = 6.354 MHz and 7.677 MHz. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δgam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
DOPPLER / DIP WIDTH against gam
Predictions from the supplied Line profiles and Doppler-free spectroscopy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): DOPPLER / DIP WIDTH / ×
Horizontal axis (x): gam: Natural width Γ/2π / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gam: Natural width Γ/2π / MHz | DOPPLER / DIP WIDTH / × |
|---|---|
| 4.275 | 222.61194148293342 |
| 6.6375 | 143.37718265002493 |
| 9 | 105.74067220439339 |
| 11.3625 | 83.75498788466803 |
| 13.725 | 69.33814570779894 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of gam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At gam = 6.354 MHz, substitute into Y_out = fg/(gam*(1+sqrt(1+sat))/2): Y₀ = 149.774 ×.
- At gam = 7.677 MHz, the same fixed-parameter model gives Y₁ = 123.963 ×.
- Signed change: ΔY_out = Y₁ − Y₀ = -25.811 ×; the input change is Δgam = 1.323 MHz.
- Average sensitivity is ΔY_out/Δgam = -19.5095 × per MHz. Its negative sign means that this response decreases over the specified interval.
- This is a secant slope between two settings. It equals a point derivative only for a locally linear relation or in a justified small-interval limit.
Answer Y₀ = 149.774 ×, Y₁ = 123.963 ×; ΔY_out = -25.811 ×; average sensitivity = -19.5095 × per MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Extra challenging questions
Work through the linked stages and explain what each result means.
Line profiles and Doppler-free spectroscopy: infer an input with uncertainty
A line profile is the distribution of absorption strength over frequency, and Doppler-free spectroscopy is any scheme that reports the homogeneous width of that profile by making the whole sample respond as a single velocity class. Use the supplied topic model Y_out = fg/(gam*(1+sqrt(1+sat))/2), where Y_out is DOPPLER / DIP WIDTH (×). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. gam is Natural width Γ/2π, measured in MHz. Keep fg (Doppler FWHM of the pedestal) = 1300 MHz; sat (Saturation parameter S = I/I_sat) = 2 fixed. For this exercise use only 4.275 MHz ≤ gam ≤ 13.725 MHz on this continuous branch. A grid of test settings is an input restriction for this exercise, not by itself a claim of physical quantisation. This calculation retains the topic model's stated assumptions; it does not extend them to a different physical setup. A reading gives Y_out = 94.792176 ×, with a bounded reading uncertainty ±3.8181 ×. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown gam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000945 MHz, using a numerical bracket if required. Find the full input interval consistent with the bounded reading uncertainty and compare it with a symmetric input-error assumption. Do not assume that the supplied calibration points prove monotonicity outside this branch.
DOPPLER / DIP WIDTH against gam
Predictions from the supplied Line profiles and Doppler-free spectroscopy model, with all other inputs fixed as stated in the question. 5 calibration points support a trend comparison; they are not measured data or a straight-line law.
Vertical axis (y): DOPPLER / DIP WIDTH / ×
Horizontal axis (x): gam: Natural width Γ/2π / MHz
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| gam: Natural width Γ/2π / MHz | DOPPLER / DIP WIDTH / × |
|---|---|
| 4.275 | 222.61194148293342 |
| 6.6375 | 143.37718265002493 |
| 9 | 105.74067220439339 |
| 11.3625 | 83.75498788466803 |
| 13.725 | 69.33814570779894 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
The model's endpoint predictions bracket the reading. Solve Y_out(gam) = Y_measured and then solve again at Y_measured − u and Y_measured + u; a nonlinear inverse can give unequal input deviations.
Reveal the worked solution
- Endpoint predictions are Y_out(4.275) = 222.612 × and Y_out(13.725) = 69.3381 ×. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives gam_line = 4.275 + (94.7922 − 222.612)(13.725 − 4.275)/(69.3381 − 222.612) = 12.1556 MHz. This estimates the inverse of the endpoint chord, rather than assuming that chord is the physical model.
- Use the actual equation, not the chord, and bracket its root. Evaluating the midpoint and retaining the sign-changing half-interval gives gam = 10.0395 MHz; substituting back gives Y_out = 94.792176 ×.
- The interpolation error in the input is |gam_line − gam| = 2.11615 MHz. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 90.974076 × and 98.610276 ×. Sort the roots even if the response decreases: 9.65078 MHz ≤ gam ≤ 10.4608 MHz.
- The lower and upper input deviations are 0.38872 MHz and 0.421349 MHz. They differ because the inverse response is nonlinear; a symmetric error bar would only be a local approximation. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer gam ≈ 10.0395 MHz; compatible interval [9.65078, 10.4608] MHz. Endpoint interpolation gives 12.1556 MHz.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.