University Physics I · topic UPI-EXT-WS.05
Superposition, reflection, and interference
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.
Superposition, Reflection & Interference: predict a finite change
Superposition: where two waves overlap, the medium's displacement is the sum of the displacements each wave would produce alone, and both waves continue afterwards as if the other had never been there. Use the supplied topic model Y_out = sqrt(1+q^2+2*q*cos(2*pi*dr/lam)), where Y_out is AMPLITUDE A (A₁). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. lam is Wavelength λ, measured in m. Keep dr (Path difference Δr) = 0.25 m; q (Amplitude ratio A₂/A₁) = 1 fixed. For this exercise use only 0.659375 m ≤ lam ≤ 1.84062 m 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 lam = 0.91925 m and 1.08462 m. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δlam across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
AMPLITUDE A against lam
Predictions from the supplied Superposition, Reflection & Interference 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): AMPLITUDE A / A₁
Horizontal axis (x): lam: Wavelength λ / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Wavelength λ / m | AMPLITUDE A / A₁ |
|---|---|
| 0.659375 | 0.741230116139565 |
| 0.9546875 | 1.3605249486594666 |
| 1.25 | 1.618033988749895 |
| 1.5453125 | 1.7471993171531155 |
| 1.840625 | 1.8206713798930576 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of lam separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At lam = 0.91925 m, substitute into Y_out = sqrt(1+q^2+2*q*cos(2*pi*dr/lam)): Y₀ = 1.31336 A₁.
- At lam = 1.08462 m, the same fixed-parameter model gives Y₁ = 1.49817 A₁.
- Signed change: ΔY_out = Y₁ − Y₀ = 0.184809 A₁; the input change is Δlam = 0.165375 m.
- Average sensitivity is ΔY_out/Δlam = 1.11751 A₁ per m. 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 = 1.31336 A₁, Y₁ = 1.49817 A₁; ΔY_out = 0.184809 A₁; average sensitivity = 1.11751 A₁ per m.
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.
Superposition, Reflection & Interference: infer an input with uncertainty
Superposition: where two waves overlap, the medium's displacement is the sum of the displacements each wave would produce alone, and both waves continue afterwards as if the other had never been there. Use the supplied topic model Y_out = sqrt(1+q^2+2*q*cos(2*pi*dr/lam)), where Y_out is AMPLITUDE A (A₁). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; unmarked trigonometric functions use radians. lam is Wavelength λ, measured in m. Keep dr (Path difference Δr) = 0.25 m; q (Amplitude ratio A₂/A₁) = 1 fixed. For this exercise use only 0.659375 m ≤ lam ≤ 1.84062 m 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 = 1.6847135 A₁, with a bounded reading uncertainty ±0.020394 A₁. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown lam by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000118125 m, 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.
AMPLITUDE A against lam
Predictions from the supplied Superposition, Reflection & Interference 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): AMPLITUDE A / A₁
Horizontal axis (x): lam: Wavelength λ / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| lam: Wavelength λ / m | AMPLITUDE A / A₁ |
|---|---|
| 0.659375 | 0.741230116139565 |
| 0.9546875 | 1.3605249486594666 |
| 1.25 | 1.618033988749895 |
| 1.5453125 | 1.7471993171531155 |
| 1.840625 | 1.8206713798930576 |
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(lam) = 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(0.659375) = 0.74123 A₁ and Y_out(1.84062) = 1.82067 A₁. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives lam_line = 0.659375 + (1.68471 − 0.74123)(1.84062 − 0.659375)/(1.82067 − 0.74123) = 1.69184 m. 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 lam = 1.37994 m; substituting back gives Y_out = 1.6847135 A₁.
- The interpolation error in the input is |lam_line − lam| = 0.311906 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.6643195 A₁ and 1.7051075 A₁. Sort the roots even if the response decreases: 1.33615 m ≤ lam ≤ 1.42814 m.
- The lower and upper input deviations are 0.0437839 m and 0.0482043 m. 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 lam ≈ 1.37994 m; compatible interval [1.33615, 1.42814] m. Endpoint interpolation gives 1.69184 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.