University Physics II · topic UPII-13.03
Total internal reflection
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: predict a finite change
Total internal reflection is what happens when light already inside a denser medium meets a boundary with a rarer one beyond the critical angle: no refracted ray exists, and the boundary sends the entire beam back. Use the supplied topic model Y_out = max(0, sign(n1*sind(theta) - 1))/(2*pi*sqrt(max((n1*sind(theta))^2 - 1, 0.0036))), where Y_out is EVANESCENT SKIN (0 = NONE) (λ₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; sind/cosd/tand use degrees. theta is Incidence angle θ₁ in the dense medium, measured in °. Keep n1 (Index n₁ of the dense medium) = 1.5 fixed. For this exercise use only 45 ° ≤ theta ≤ 76.5 ° 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 theta = 51.93 ° and 56.34 °. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δtheta across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
EVANESCENT SKIN (0 = NONE) against theta
Predictions from the supplied Total internal reflection 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): EVANESCENT SKIN (0 = NONE) / λ₀
Horizontal axis (x): theta: Incidence angle θ₁ in the dense medium / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Incidence angle θ₁ in the dense medium / ° | EVANESCENT SKIN (0 = NONE) / λ₀ |
|---|---|
| 45 | 0.4501581580785534 |
| 52.875 | 0.24260451620258364 |
| 60.75 | 0.18850938217253085 |
| 68.625 | 0.16319397209619127 |
| 76.5 | 0.149894092782386 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of theta separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At theta = 51.93 °, substitute into Y_out = max(0, sign(n1*sind(theta) - 1))/(2*pi*sqrt(max((n1*sind(theta))^2 - 1, 0.0036))): Y₀ = 0.253396 λ₀.
- At theta = 56.34 °, the same fixed-parameter model gives Y₁ = 0.212911 λ₀.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0404847 λ₀; the input change is Δtheta = 4.41 °.
- Average sensitivity is ΔY_out/Δtheta = -0.0091802 λ₀ per °. 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 = 0.253396 λ₀, Y₁ = 0.212911 λ₀; ΔY_out = -0.0404847 λ₀; average sensitivity = -0.0091802 λ₀ 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: infer an input with uncertainty
Total internal reflection is what happens when light already inside a denser medium meets a boundary with a rarer one beyond the critical angle: no refracted ray exists, and the boundary sends the entire beam back. Use the supplied topic model Y_out = max(0, sign(n1*sind(theta) - 1))/(2*pi*sqrt(max((n1*sind(theta))^2 - 1, 0.0036))), where Y_out is EVANESCENT SKIN (0 = NONE) (λ₀). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; max(a,b) selects the larger value; sind/cosd/tand use degrees. theta is Incidence angle θ₁ in the dense medium, measured in °. Keep n1 (Index n₁ of the dense medium) = 1.5 fixed. For this exercise use only 45 ° ≤ theta ≤ 76.5 ° 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 = 0.17530323 λ₀, with a bounded reading uncertainty ±0.0038114 λ₀. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown theta 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.
EVANESCENT SKIN (0 = NONE) against theta
Predictions from the supplied Total internal reflection 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): EVANESCENT SKIN (0 = NONE) / λ₀
Horizontal axis (x): theta: Incidence angle θ₁ in the dense medium / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Incidence angle θ₁ in the dense medium / ° | EVANESCENT SKIN (0 = NONE) / λ₀ |
|---|---|
| 45 | 0.4501581580785534 |
| 52.875 | 0.24260451620258364 |
| 60.75 | 0.18850938217253085 |
| 68.625 | 0.16319397209619127 |
| 76.5 | 0.149894092782386 |
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(theta) = 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(45) = 0.450158 λ₀ and Y_out(76.5) = 0.149894 λ₀. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives theta_line = 45 + (0.175303 − 0.450158)(76.5 − 45)/(0.149894 − 0.450158) = 73.8344 °. 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 theta = 64.215 °; substituting back gives Y_out = 0.17530323 λ₀.
- The interpolation error in the input is |theta_line − theta| = 9.61939 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.17149183 λ₀ and 0.17911463 λ₀. Sort the roots even if the response decreases: 63.1037 ° ≤ theta ≤ 65.4404 °.
- The lower and upper input deviations are 1.11134 ° and 1.22538 °. 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 theta ≈ 64.215 °; compatible interval [63.1037, 65.4404] °. Endpoint interpolation gives 73.8344 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.