University Physics II · unit UPII-13
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.
The Ray Model & Reflection: infer an input with uncertainty
The ray model represents light as straight lines drawn perpendicular to its wavefronts, and the law of reflection fixes what one of them does at a smooth surface: it leaves in the plane of the incident ray and the normal, on the opposite side of the normal, at the same angle. Use the supplied topic model Y_out = theta-alpha, where Y_out is ANGLE OF INCIDENCE (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. theta is Incident ray angle θ₀, measured in °. Keep alpha (Mirror tilt α) = 0 ° fixed. For this exercise use only 25.3125 ° ≤ theta ≤ 64.6875 ° 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 = 49.33125 °, with a bounded reading uncertainty ±2.3034 °. 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 0.0039375 °, 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 theta
Predictions from the supplied The Ray Model & 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): ANGLE OF INCIDENCE / °
Horizontal axis (x): theta: Incident ray angle θ₀ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Incident ray angle θ₀ / ° | ANGLE OF INCIDENCE / ° |
|---|---|
| 25.3125 | 25.3125 |
| 35.15625 | 35.15625 |
| 45 | 45 |
| 54.84375 | 54.84375 |
| 64.6875 | 64.6875 |
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(25.3125) = 25.3125 ° and Y_out(64.6875) = 64.6875 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 25.3125 + (49.3312 − 25.3125)(64.6875 − 25.3125)/(64.6875 − 25.3125) = 49.3312 °. 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 = 49.3312 °; substituting back gives Y_out = 49.33125 °.
- The interpolation error in the input is |theta_line − theta| = 0 °. This is within the requested numerical tolerance.
- Invert both limits Y_out = 47.02785 ° and 51.63465 °. Sort the roots even if the response decreases: 47.0279 ° ≤ theta ≤ 51.6346 °.
- The lower and upper input deviations are 2.3034 ° and 2.3034 °. They agree on this branch; this does not establish linearity for changes in other inputs or outside the interval. This interval includes reading uncertainty only, not model error or uncertainty in fixed inputs.
Answer theta ≈ 49.3312 °; compatible interval [47.0279, 51.6346] °. Endpoint interpolation gives 49.3312 °.
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 makes on crossing a boundary where its speed changes, and Snell's law fixes the new direction: n₁ sin θ₁ = n₂ sin θ₂, with both angles measured from the normal to the surface. Use the supplied topic model Y_out = theta-deg(asin(sind(theta)/n2)), where Y_out is DEVIATION θ₁ − θ₂ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sind/cosd/tand use degrees; deg converts radians to degrees; unmarked trigonometric functions use radians. theta is Angle of incidence θ₁, measured in °. Keep n2 (Index of medium 2, n₂) = 1.33 fixed. For this exercise use only 8.5 ° ≤ theta ≤ 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 = 13.613526 °, with a bounded reading uncertainty ±1.7241 °. 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 0.0063 °, 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.
DEVIATION θ₁ − θ₂ against theta
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): DEVIATION θ₁ − θ₂ / °
Horizontal axis (x): theta: Angle of incidence θ₁ / °
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| theta: Angle of incidence θ₁ / ° | DEVIATION θ₁ − θ₂ / ° |
|---|---|
| 8.5 | 2.1192585489469105 |
| 24.25 | 6.2623994773037275 |
| 40 | 11.098915460291888 |
| 55.75 | 17.32453339145195 |
| 71.5 | 26.01840141398673 |
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(8.5) = 2.11926 ° and Y_out(71.5) = 26.0184 °. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives theta_line = 8.5 + (13.6135 − 2.11926)(71.5 − 8.5)/(26.0184 − 2.11926) = 38.7998 °. 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 = 46.93 °; substituting back gives Y_out = 13.613526 °.
- The interpolation error in the input is |theta_line − theta| = 8.13022 °. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 11.889426 ° and 15.337626 °. Sort the roots even if the response decreases: 42.2685 ° ≤ theta ≤ 51.2216 °.
- The lower and upper input deviations are 4.66152 ° and 4.29163 °. 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 ≈ 46.93 °; compatible interval [42.2685, 51.2216] °. Endpoint interpolation gives 38.7998 °.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
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.
Spherical Mirrors: infer an input with uncertainty
A spherical mirror is a polished cap of a sphere of radius R, and within the paraxial region near its axis it sends every ray that arrived parallel to that axis to a single focal point at a distance f = R/2 from the vertex: really through it, in front of the mirror, if the cap is concave, and only by backward extension, behind the mirror, if it is convex. Use the supplied topic model Y_out = s*r/(2*s-r), where Y_out is IMAGE DISTANCE s′ (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Object distance s, measured in cm. Keep r (Radius of curvature R) = 40 cm fixed. For this exercise use only 45.1 cm ≤ s ≤ 82.9 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 = 28.305993 cm, with a bounded reading uncertainty ±0.292 cm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown s by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00378 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.
IMAGE DISTANCE s′ against s
Predictions from the supplied Spherical Mirrors 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): IMAGE DISTANCE s′ / cm
Horizontal axis (x): s: Object distance s / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Object distance s / cm | IMAGE DISTANCE s′ / cm |
|---|---|
| 45.1 | 35.93625498007968 |
| 54.550000000000004 | 31.577424023154844 |
| 64 | 29.09090909090909 |
| 73.45 | 27.483629560336762 |
| 82.9 | 26.359300476947535 |
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(s) = 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(45.1) = 35.9363 cm and Y_out(82.9) = 26.3593 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 45.1 + (28.306 − 35.9363)(82.9 − 45.1)/(26.3593 − 35.9363) = 75.2165 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 s = 68.158 cm; substituting back gives Y_out = 28.305993 cm.
- The interpolation error in the input is |s_line − s| = 7.05845 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 28.013993 cm and 28.597993 cm. Sort the roots even if the response decreases: 66.5225 cm ≤ s ≤ 69.9127 cm.
- The lower and upper input deviations are 1.63551 cm and 1.7547 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 s ≈ 68.158 cm; compatible interval [66.5225, 69.9127] cm. Endpoint interpolation gives 75.2165 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.