University Physics III · topic UPIII-04.05
Dispersion and chromatic effects
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.
Dispersion and chromatic effects: predict a finite change
Dispersion is the dependence of refractive index on wavelength, n = n(λ): its local measure is the slope dn/dλ, and its size across the visible is the Abbe number V_d = (n_d − 1)/(n_F − n_C), large for a weakly dispersive glass. Use the supplied topic model Y_out = sqrt((hypot((1-(lam0/656.3)^2)+k,(b*lam0/656.3))/hypot((1-(lam0/656.3)^2),(b*lam0/656.3))+1+k*(1-(lam0/656.3)^2)/((1-(lam0/656.3)^2)^2+(b*lam0/656.3)^2))/2), where Y_out is n AT C LINE 656 nm (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; hypot(a,b) means √(a²+b²). lam0 is Resonance wavelength λ₀, measured in nm. Keep b (Damping γ/ω₀ (= 1/Q)) = 0.3; k (Oscillator strength k) = 1.2 fixed. For this exercise use only 153.125 nm ≤ lam0 ≤ 546.875 nm 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 lam0 = 239.75 nm and 294.875 nm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlam0 across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
n AT C LINE 656 nm against lam0
Predictions from the supplied Dispersion and chromatic effects 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): n AT C LINE 656 nm
Horizontal axis (x): lam0: Resonance wavelength λ₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam0: Resonance wavelength λ₀ / nm | n AT C LINE 656 nm |
|---|---|
| 153.125 | 1.504371026330421 |
| 251.5625 | 1.5444102836594582 |
| 350 | 1.6153337110444372 |
| 448.4375 | 1.7344205322272033 |
| 546.875 | 1.8997034647982798 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lam0 separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lam0 = 239.75 nm, substitute into Y_out = sqrt((hypot((1-(lam0/656.3)^2)+k,(b*lam0/656.3))/hypot((1-(lam0/656.3)^2),(b*lam0/656.3))+1+k*(1-(lam0/656.3)^2)/((1-(lam0/656.3)^2)^2+(b*lam0/656.3)^2))/2): Y₀ = 1.53823.
- At lam0 = 294.875 nm, the same fixed-parameter model gives Y₁ = 1.57099.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.0327597; the input change is Δlam0 = 55.125 nm.
- Average sensitivity is ΔY_out/Δlam0 = 0.000594281 1 per nm. Its positive sign means that this response increases 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 = 1.53823, Y₁ = 1.57099; ΔY_out = 0.0327597; average sensitivity = 0.000594281 1 per nm.
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.
Dispersion and chromatic effects: infer an input with uncertainty
Dispersion is the dependence of refractive index on wavelength, n = n(λ): its local measure is the slope dn/dλ, and its size across the visible is the Abbe number V_d = (n_d − 1)/(n_F − n_C), large for a weakly dispersive glass. Use the supplied topic model Y_out = sqrt((hypot((1-(lam0/656.3)^2)+k,(b*lam0/656.3))/hypot((1-(lam0/656.3)^2),(b*lam0/656.3))+1+k*(1-(lam0/656.3)^2)/((1-(lam0/656.3)^2)^2+(b*lam0/656.3)^2))/2), where Y_out is n AT C LINE 656 nm (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; hypot(a,b) means √(a²+b²). lam0 is Resonance wavelength λ₀, measured in nm. Keep b (Damping γ/ω₀ (= 1/Q)) = 0.3; k (Oscillator strength k) = 1.2 fixed. For this exercise use only 153.125 nm ≤ lam0 ≤ 546.875 nm 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 = 1.6605368, with a bounded reading uncertainty ±0.023425. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam0 by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.039375 nm, 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.
n AT C LINE 656 nm against lam0
Predictions from the supplied Dispersion and chromatic effects 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): n AT C LINE 656 nm
Horizontal axis (x): lam0: Resonance wavelength λ₀ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam0: Resonance wavelength λ₀ / nm | n AT C LINE 656 nm |
|---|---|
| 153.125 | 1.504371026330421 |
| 251.5625 | 1.5444102836594582 |
| 350 | 1.6153337110444372 |
| 448.4375 | 1.7344205322272033 |
| 546.875 | 1.8997034647982798 |
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(lam0) = 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(153.125) = 1.50437 and Y_out(546.875) = 1.8997. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lam0_line = 153.125 + (1.66054 − 1.50437)(546.875 − 153.125)/(1.8997 − 1.50437) = 308.666 nm. 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 lam0 = 393.313 nm; substituting back gives Y_out = 1.6605368.
- The interpolation error in the input is |lam0_line − lam0| = 84.6468 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6371118 and 1.6839618. Sort the roots even if the response decreases: 372.094 nm ≤ lam0 ≤ 412.446 nm.
- The lower and upper input deviations are 21.2184 nm and 19.1335 nm. 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 lam0 ≈ 393.313 nm; compatible interval [372.094, 412.446] nm. Endpoint interpolation gives 308.666 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.