University Physics III · unit UPIII-04
Geometric Optics
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.
Extra challenging questions
A selection from this pathway. Open a topic below for its complete practice and challenge set.
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.
Reflection and plane-mirror images: infer an input with uncertainty
Specular reflection reverses the component of a ray's direction along the local surface normal and leaves the tangential components unchanged, so the reflected ray lies in the plane of incidence with θᵣ = θᵢ measured from that normal. Use the supplied topic model Y_out = deg(atan((h*s/(s+d))/s)), where Y_out is ANGLE OF INCIDENCE (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; deg converts radians to degrees; unmarked trigonometric functions use radians. d is Eye distance d from mirror, measured in m. Keep s (Object distance s from mirror) = 0.8 m; h (Eye height h above object) = 0.3 m fixed. For this exercise use only 0.459375 m ≤ d ≤ 1.64063 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 = 8.6159161 °, with a bounded reading uncertainty ±0.24125 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d 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.
ANGLE OF INCIDENCE against d
Predictions from the supplied Reflection and plane-mirror images 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): ANGLE OF INCIDENCE / °
Horizontal axis (x): d: Eye distance d from mirror / m
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Eye distance d from mirror / m | ANGLE OF INCIDENCE / ° |
|---|---|
| 0.459375 | 13.398904527683145 |
| 0.7546875 | 10.921830749081908 |
| 1.0499999999999998 | 9.211026540816668 |
| 1.3453125 | 7.96060600009535 |
| 1.640625 | 7.0076071289927455 |
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(d) = 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(0.459375) = 13.3989 ° and Y_out(1.64063) = 7.00761 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.459375 + (8.61592 − 13.3989)(1.64063 − 0.459375)/(7.00761 − 13.3989) = 1.34337 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 d = 1.17994 m; substituting back gives Y_out = 8.6159161 °.
- The interpolation error in the input is |d_line − d| = 0.163437 m. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 8.3746661 ° and 8.8571661 °. Sort the roots even if the response decreases: 1.12518 m ≤ d ≤ 1.23783 m.
- The lower and upper input deviations are 0.0547624 m and 0.0578931 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 d ≈ 1.17994 m; compatible interval [1.12518, 1.23783] m. Endpoint interpolation gives 1.34337 m.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Refraction and Snell's law: infer an input with uncertainty
Refraction is the change of direction a wave undergoes at a boundary between media of different phase speed; for light, Snell's law n₁ sin θ₁ = n₂ sin θ₂ fixes the new direction from the refractive indices n = c/v. Use the supplied topic model Y_out = n1*sind(th)/n2, where Y_out is SIN θ₂ = n₁ sin θ₁/n₂ (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees. th is Angle of incidence θ₁, measured in °. Keep n1 (Index of medium 1, n₁) = 1; n2 (Index of medium 2, n₂) = 1.5 fixed. For this exercise use only 8.5 ° ≤ th ≤ 71.5 ° 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 = 0.48701329, with a bounded reading uncertainty ±0.02178. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown th by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0063 °, 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.
SIN θ₂ = n₁ sin θ₁/n₂ against th
Predictions from the supplied Refraction and Snell's law 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): SIN θ₂ = n₁ sin θ₁/n₂
Horizontal axis (x): th: Angle of incidence θ₁ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| th: Angle of incidence θ₁ / ° | SIN θ₂ = n₁ sin θ₁/n₂ |
|---|---|
| 8.5 | 0.09853960741974042 |
| 24.25 | 0.2738125684089848 |
| 40 | 0.4285250731243595 |
| 55.75 | 0.5510598327514591 |
| 71.5 | 0.6322157701374662 |
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(th) = 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(8.5) = 0.0985396 and Y_out(71.5) = 0.632216. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives th_line = 8.5 + (0.487013 − 0.0985396)(71.5 − 8.5)/(0.632216 − 0.0985396) = 54.359 °. 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 th = 46.93 °; substituting back gives Y_out = 0.48701329.
- The interpolation error in the input is |th_line − th| = 7.42898 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.46523329 and 0.50879329. Sort the roots even if the response decreases: 44.2548 ° ≤ th ≤ 49.7462 °.
- The lower and upper input deviations are 2.67524 ° and 2.81623 °. 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 th ≈ 46.93 °; compatible interval [44.2548, 49.7462] °. Endpoint interpolation gives 54.359 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.