University Physics III · topic UPIII-04.01
Fermat's principle and the ray laws
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.
Fermat's principle and the ray laws: predict a finite change
Fermat's principle states that between two points light follows a route whose optical path length ∫ n ds, and so its travel time, is stationary against small deformations of the route: unchanged to first order, not necessarily least. Use the supplied topic model Y_out = 33.356*(n1*sqrt(16+xc^2)+n2*sqrt(16+(6-xc)^2)), where Y_out is TRAVEL TIME T(x) (ps). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. xc is Crossing point x, measured in cm. Keep n1 (Index n₁ above) = 1; n2 (Index n₂ below) = 1.5 fixed. For this exercise use only 0.6375 cm ≤ xc ≤ 3.7875 cm 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 xc = 1.3305 cm and 1.7715 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δxc across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
TRAVEL TIME T(x) against xc
Predictions from the supplied Fermat's principle and the ray laws 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): TRAVEL TIME T(x) / ps
Horizontal axis (x): xc: Crossing point x / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xc: Crossing point x / cm | TRAVEL TIME T(x) / ps |
|---|---|
| 0.6375 | 469.8364902845865 |
| 1.4249999999999998 | 445.6973429551044 |
| 2.2124999999999995 | 428.093827792164 |
| 3 | 416.95000000000005 |
| 3.7874999999999996 | 412.45783945119405 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of xc separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At xc = 1.3305 cm, substitute into Y_out = 33.356*(n1*sqrt(16+xc^2)+n2*sqrt(16+(6-xc)^2)): Y₀ = 448.246 ps.
- At xc = 1.7715 cm, the same fixed-parameter model gives Y₁ = 437.155 ps.
- Signed change: ΔY_out = Y₁ − Y₀ = -11.0914 ps; the input change is Δxc = 0.441 cm.
- Average sensitivity is ΔY_out/Δxc = -25.1506 ps per cm. 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 = 448.246 ps, Y₁ = 437.155 ps; ΔY_out = -11.0914 ps; average sensitivity = -25.1506 ps per cm.
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.
Fermat's principle and the ray laws: infer an input with uncertainty
Fermat's principle states that between two points light follows a route whose optical path length ∫ n ds, and so its travel time, is stationary against small deformations of the route: unchanged to first order, not necessarily least. Use the supplied topic model Y_out = 33.356*(n1*sqrt(16+xc^2)+n2*sqrt(16+(6-xc)^2)), where Y_out is TRAVEL TIME T(x) (ps). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. xc is Crossing point x, measured in cm. Keep n1 (Index n₁ above) = 1; n2 (Index n₂ below) = 1.5 fixed. For this exercise use only 0.6375 cm ≤ xc ≤ 3.7875 cm 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 = 422.39157 ps, with a bounded reading uncertainty ±1.4901 ps. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown xc by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000315 cm, 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.
TRAVEL TIME T(x) against xc
Predictions from the supplied Fermat's principle and the ray laws 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): TRAVEL TIME T(x) / ps
Horizontal axis (x): xc: Crossing point x / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| xc: Crossing point x / cm | TRAVEL TIME T(x) / ps |
|---|---|
| 0.6375 | 469.8364902845865 |
| 1.4249999999999998 | 445.6973429551044 |
| 2.2124999999999995 | 428.093827792164 |
| 3 | 416.95000000000005 |
| 3.7874999999999996 | 412.45783945119405 |
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(xc) = 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(0.6375) = 469.836 ps and Y_out(3.7875) = 412.458 ps. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives xc_line = 0.6375 + (422.392 − 469.836)(3.7875 − 0.6375)/(412.458 − 469.836) = 3.24215 cm. 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 xc = 2.559 cm; substituting back gives Y_out = 422.39157 ps.
- The interpolation error in the input is |xc_line − xc| = 0.683154 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 420.90147 ps and 423.88167 ps. Sort the roots even if the response decreases: 2.46075 cm ≤ xc ≤ 2.66467 cm.
- The lower and upper input deviations are 0.0982527 cm and 0.105666 cm. 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 xc ≈ 2.559 cm; compatible interval [2.46075, 2.66467] cm. Endpoint interpolation gives 3.24215 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.