University Physics III · topic UPIII-04.04
Total internal reflection and guided light
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.
Total internal reflection and guided light: predict a finite change
Total internal reflection is the regime in which a ray inside the higher-index medium meets the boundary beyond the critical angle sin θ_c = n₂/n₁, so no refracted ray exists and the ray model hands the entire beam to the reflected ray. Use the supplied topic model Y_out = 3335.64*n1*(1/sqrt(1-(sind(alpha)/n1)^2)-1), where Y_out is LAG BEHIND AXIAL RAY (ns km⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees. alpha is Launch angle θ_a in air, measured in °. Keep n1 (Core index n₁) = 1.48; dn (Index step Δn = n₁ − n₂) = 0.02 fixed. For this exercise use only 4.25 ° ≤ alpha ≤ 35.75 ° 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 alpha = 11.18 ° and 15.59 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δalpha across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LAG BEHIND AXIAL RAY against alpha
Predictions from the supplied Total internal reflection and guided light 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): LAG BEHIND AXIAL RAY / ns km⁻¹
Horizontal axis (x): alpha: Launch angle θ_a in air / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| alpha: Launch angle θ_a in air / ° | LAG BEHIND AXIAL RAY / ns km⁻¹ |
|---|---|
| 4.25 | 6.200704350995274 |
| 12.125 | 50.48181056835567 |
| 20 | 137.3493831387047 |
| 27.875 | 266.4573750452527 |
| 35.75 | 436.3909007759868 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of alpha separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At alpha = 11.18 °, substitute into Y_out = 3335.64*n1*(1/sqrt(1-(sind(alpha)/n1)^2)-1): Y₀ = 42.9182 ns km⁻¹.
- At alpha = 15.59 °, the same fixed-parameter model gives Y₁ = 83.4634 ns km⁻¹.
- Signed change: ΔY_out = Y₁ − Y₀ = 40.5452 ns km⁻¹; the input change is Δalpha = 4.41 °.
- Average sensitivity is ΔY_out/Δalpha = 9.19393 ns km⁻¹ per °. 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 = 42.9182 ns km⁻¹, Y₁ = 83.4634 ns km⁻¹; ΔY_out = 40.5452 ns km⁻¹; average sensitivity = 9.19393 ns km⁻¹ per °.
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.
Total internal reflection and guided light: infer an input with uncertainty
Total internal reflection is the regime in which a ray inside the higher-index medium meets the boundary beyond the critical angle sin θ_c = n₂/n₁, so no refracted ray exists and the ray model hands the entire beam to the reflected ray. Use the supplied topic model Y_out = 3335.64*n1*(1/sqrt(1-(sind(alpha)/n1)^2)-1), where Y_out is LAG BEHIND AXIAL RAY (ns km⁻¹). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; sind/cosd/tand use degrees. alpha is Launch angle θ_a in air, measured in °. Keep n1 (Core index n₁) = 1.48; dn (Index step Δn = n₁ − n₂) = 0.02 fixed. For this exercise use only 4.25 ° ≤ alpha ≤ 35.75 ° 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 = 189.00071 ns km⁻¹, with a bounded reading uncertainty ±27.42 ns km⁻¹. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown alpha by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00315 °, 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.
LAG BEHIND AXIAL RAY against alpha
Predictions from the supplied Total internal reflection and guided light 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): LAG BEHIND AXIAL RAY / ns km⁻¹
Horizontal axis (x): alpha: Launch angle θ_a in air / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| alpha: Launch angle θ_a in air / ° | LAG BEHIND AXIAL RAY / ns km⁻¹ |
|---|---|
| 4.25 | 6.200704350995274 |
| 12.125 | 50.48181056835567 |
| 20 | 137.3493831387047 |
| 27.875 | 266.4573750452527 |
| 35.75 | 436.3909007759868 |
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(alpha) = 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(4.25) = 6.2007 ns km⁻¹ and Y_out(35.75) = 436.391 ns km⁻¹. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives alpha_line = 4.25 + (189.001 − 6.2007)(35.75 − 4.25)/(436.391 − 6.2007) = 17.6352 °. 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 alpha = 23.465 °; substituting back gives Y_out = 189.00071 ns km⁻¹.
- The interpolation error in the input is |alpha_line − alpha| = 5.82976 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 161.58071 ns km⁻¹ and 216.42071 ns km⁻¹. Sort the roots even if the response decreases: 21.6939 ° ≤ alpha ≤ 25.113 °.
- The lower and upper input deviations are 1.77106 ° and 1.64801 °. 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 alpha ≈ 23.465 °; compatible interval [21.6939, 25.113] °. Endpoint interpolation gives 17.6352 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.