University Physics III · topic UPIII-03.02
Phase difference, path difference, and coherence
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.
Phase difference, path difference, and coherence: predict a finite change
The phase difference is the single number δ saying how far one arriving wave's cycle sits behind another's at the same point, built from the path difference and from whatever offset the sources already carried. Use the supplied topic model Y_out = exp(-sig*sig/2), where Y_out is COHERENCE FACTOR (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. sig is Phase jitter σ, measured in rad. Keep dr (Path difference Δr / λ) = 0.35 λ; p0 (Source phase offset Δφ₀) = 0.5 π rad fixed. For this exercise use only 0.31875 rad ≤ sig ≤ 2.68125 rad 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 sig = 0.8385 rad and 1.16925 rad. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsig across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
COHERENCE FACTOR against sig
Predictions from the supplied Phase difference, path difference, and coherence 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): COHERENCE FACTOR
Horizontal axis (x): sig: Phase jitter σ / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Phase jitter σ / rad | COHERENCE FACTOR |
|---|---|
| 0.31875 | 0.9504680027199568 |
| 0.9093749999999999 | 0.6613438052942324 |
| 1.5 | 0.32465246735834974 |
| 2.0906249999999997 | 0.11243764863785047 |
| 2.68125 | 0.027473021800949925 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of sig separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sig = 0.8385 rad, substitute into Y_out = exp(-sig*sig/2): Y₀ = 0.703603.
- At sig = 1.16925 rad, the same fixed-parameter model gives Y₁ = 0.50481.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.198793; the input change is Δsig = 0.33075 rad.
- Average sensitivity is ΔY_out/Δsig = -0.601036 1 per rad. 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 = 0.703603, Y₁ = 0.50481; ΔY_out = -0.198793; average sensitivity = -0.601036 1 per rad.
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.
Phase difference, path difference, and coherence: infer an input with uncertainty
The phase difference is the single number δ saying how far one arriving wave's cycle sits behind another's at the same point, built from the path difference and from whatever offset the sources already carried. Use the supplied topic model Y_out = exp(-sig*sig/2), where Y_out is COHERENCE FACTOR (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; exp(x) means eˣ. sig is Phase jitter σ, measured in rad. Keep dr (Path difference Δr / λ) = 0.35 λ; p0 (Source phase offset Δφ₀) = 0.5 π rad fixed. For this exercise use only 0.31875 rad ≤ sig ≤ 2.68125 rad 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 = 0.21254958, with a bounded reading uncertainty ±0.027761. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sig by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00023625 rad, 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.
COHERENCE FACTOR against sig
Predictions from the supplied Phase difference, path difference, and coherence 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): COHERENCE FACTOR
Horizontal axis (x): sig: Phase jitter σ / rad
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sig: Phase jitter σ / rad | COHERENCE FACTOR |
|---|---|
| 0.31875 | 0.9504680027199568 |
| 0.9093749999999999 | 0.6613438052942324 |
| 1.5 | 0.32465246735834974 |
| 2.0906249999999997 | 0.11243764863785047 |
| 2.68125 | 0.027473021800949925 |
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(sig) = 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(0.31875) = 0.950468 and Y_out(2.68125) = 0.027473. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sig_line = 0.31875 + (0.21255 − 0.950468)(2.68125 − 0.31875)/(0.027473 − 0.950468) = 2.20753 rad. 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 sig = 1.75987 rad; substituting back gives Y_out = 0.21254958.
- The interpolation error in the input is |sig_line − sig| = 0.447653 rad. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.18478858 and 0.24031058. Sort the roots even if the response decreases: 1.68868 rad ≤ sig ≤ 1.83768 rad.
- The lower and upper input deviations are 0.0711932 rad and 0.0778099 rad. 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 sig ≈ 1.75987 rad; compatible interval [1.68868, 1.83768] rad. Endpoint interpolation gives 2.20753 rad.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.