University Physics III · unit UPIII-05
Physical 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.
Coherence, phase & superposition of light: infer an input with uncertainty
Two light waves are coherent at a detector when their phase difference stays fixed for longer than the detector averages; their fields then add to make fringes, and the fringe visibility measures how completely they do so. Use the supplied topic model Y_out = 2*sqrt(ratio)/(1+ratio)*abs(sin(pi*max(path,0.01)*band/360)/(pi*max(path,0.01)*band/360)), where Y_out is VISIBILITY V (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x; abs(x) means |x|; max(a,b) selects the larger value; unmarked trigonometric functions use radians. ratio is Intensity ratio I₂/I₁, dimensionless. Keep path (Path difference Δ) = 4 μm; band (Source linewidth Δλ) = 20 nm fixed. For this exercise use only 0.10625 ≤ ratio ≤ 0.89375 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 = 0.88892705, with a bounded reading uncertainty ±0.0045522. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown ratio by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00007875, 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.
VISIBILITY V against ratio
Predictions from the supplied Coherence, phase & superposition of 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): VISIBILITY V
Horizontal axis (x): ratio: Intensity ratio I₂/I₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| ratio: Intensity ratio I₂/I₁ | VISIBILITY V |
|---|---|
| 0.10625 | 0.5425894174913162 |
| 0.303125 | 0.7780095623023046 |
| 0.5 | 0.8680682592366252 |
| 0.6968749999999999 | 0.9059155713299115 |
| 0.8937499999999999 | 0.9192751387850172 |
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(ratio) = 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(0.10625) = 0.542589 and Y_out(0.89375) = 0.919275. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives ratio_line = 0.10625 + (0.888927 − 0.542589)(0.89375 − 0.10625)/(0.919275 − 0.542589) = 0.830304. 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 ratio = 0.586625; substituting back gives Y_out = 0.88892705.
- The interpolation error in the input is |ratio_line − ratio| = 0.243679. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.88437485 and 0.89347925. Sort the roots even if the response decreases: 0.564685 ≤ ratio ≤ 0.610992.
- The lower and upper input deviations are 0.0219396 and 0.0243665. 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 ratio ≈ 0.586625; compatible interval [0.564685, 0.610992]. Endpoint interpolation gives 0.830304.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Two-source interference & double-slit intensity: infer an input with uncertainty
Two-source interference is the far-field pattern of two equal, in-phase point sources: a screen intensity I = 4I₀ cos²(φ/2) whose phase difference φ = 2πd sinθ/λ is set by the slit separation, the angle and the wavelength. Use the supplied topic model Y_out = 0.002*lam/d, where Y_out is FRINGE SPACING λL/d (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Slit separation d, measured in mm. Keep lam (Wavelength λ) = 633 nm; y (Probe position y on the screen) = 1.25 mm fixed. For this exercise use only 0.131875 mm ≤ d ≤ 0.368125 mm 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 = 4.5871643 mm, with a bounded reading uncertainty ±0.17222 mm. 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 0.000023625 mm, 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.
FRINGE SPACING λL/d against d
Predictions from the supplied Two-source interference & double-slit intensity 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): FRINGE SPACING λL/d / mm
Horizontal axis (x): d: Slit separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Slit separation d / mm | FRINGE SPACING λL/d / mm |
|---|---|
| 0.13187500000000002 | 9.599999999999998 |
| 0.19093750000000004 | 6.630441898527003 |
| 0.25 | 5.064 |
| 0.3090625 | 4.0962588473205255 |
| 0.36812500000000004 | 3.4390492359932083 |
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(0.131875) = 9.6 mm and Y_out(0.368125) = 3.43905 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.131875 + (4.58716 − 9.6)(0.368125 − 0.131875)/(3.43905 − 9.6) = 0.324099 mm. 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 = 0.275987 mm; substituting back gives Y_out = 4.5871643 mm.
- The interpolation error in the input is |d_line − d| = 0.0481115 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.4149443 mm and 4.7593843 mm. Sort the roots even if the response decreases: 0.266001 mm ≤ d ≤ 0.286753 mm.
- The lower and upper input deviations are 0.0099867 mm and 0.0107658 mm. 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 ≈ 0.275987 mm; compatible interval [0.266001, 0.286753] mm. Endpoint interpolation gives 0.324099 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Thin Films, Wedges & Newton's Rings: infer an input with uncertainty
Thin-film interference is the superposition of the light reflected from a film's two faces, whose phase difference is the optical path 2nt cos θ_t counted in wavelengths, plus a half-cycle for each reflection off a denser medium. Use the supplied topic model Y_out = 100*(((1-nf)/(1+nf))^2+((1-((1-nf)/(1+nf))^2)*(nf-ns)/(nf+ns))^2+2*((1-nf)/(1+nf))*(1-((1-nf)/(1+nf))^2)*(nf-ns)/(nf+ns)*cos(4*pi*nf*t/550)), where Y_out is R AT 550 nm (%). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; unmarked trigonometric functions use radians. t is Film thickness t, measured in nm. Keep nf (Film index n_f) = 1.38; ns (Substrate index n_s) = 1.52 fixed. For this exercise use only 310.325 nm ≤ t ≤ 363.875 nm 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 = 2.5001094 %, with a bounded reading uncertainty ±0.14412 %. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown t by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.005355 nm, 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.
R AT 550 nm against t
Predictions from the supplied Thin Films, Wedges & Newton's Rings 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): R AT 550 nm / %
Horizontal axis (x): t: Film thickness t / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| t: Film thickness t / nm | R AT 550 nm / % |
|---|---|
| 310.325 | 1.3645030338248338 |
| 323.7125 | 1.7046285577859859 |
| 337.1 | 2.231878668679646 |
| 350.4875 | 2.8536961833453516 |
| 363.875 | 3.460922893167949 |
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(t) = 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(310.325) = 1.3645 % and Y_out(363.875) = 3.46092 %. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives t_line = 310.325 + (2.50011 − 1.3645)(363.875 − 310.325)/(3.46092 − 1.3645) = 339.332 nm. 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 t = 342.991 nm; substituting back gives Y_out = 2.5001094 %.
- The interpolation error in the input is |t_line − t| = 3.65809 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.3559894 % and 2.6442294 %. Sort the roots even if the response decreases: 339.864 nm ≤ t ≤ 346.061 nm.
- The lower and upper input deviations are 3.12639 nm and 3.07076 nm. 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 t ≈ 342.991 nm; compatible interval [339.864, 346.061] nm. Endpoint interpolation gives 339.332 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
The Michelson Interferometer & Optical Metrology: infer an input with uncertainty
A Michelson interferometer divides a beam's amplitude at a beam splitter, returns the two halves from separate mirrors and reads the path difference 2d between the arms as a fringe pattern that shifts by one fringe for every λ/2 of mirror travel. Use the supplied topic model Y_out = lam^2/(1000*dl), where Y_out is COHERENCE LENGTH λ²/Δλ (μm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dl is Spectral width Δλ, measured in nm. Keep lam (Centre wavelength λ) = 633 nm; d (Mirror displacement d) = 0.95 μm fixed. For this exercise use only 32.7687 nm ≤ dl ≤ 268.231 nm 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 = 2.2714683 μm, with a bounded reading uncertainty ±0.11665 μm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dl by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0235462 nm, 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.
COHERENCE LENGTH λ²/Δλ against dl
Predictions from the supplied The Michelson Interferometer & Optical Metrology 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): COHERENCE LENGTH λ²/Δλ / μm
Horizontal axis (x): dl: Spectral width Δλ / nm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dl: Spectral width Δλ / nm | COHERENCE LENGTH λ²/Δλ / μm |
|---|---|
| 32.76875 | 12.227777989700552 |
| 91.63437499999999 | 4.372693107799339 |
| 150.5 | 2.6623853820598007 |
| 209.36562499999997 | 1.9138242010836308 |
| 268.23125 | 1.4938192324719808 |
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(dl) = 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(32.7687) = 12.2278 μm and Y_out(268.231) = 1.49382 μm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dl_line = 32.7687 + (2.27147 − 12.2278)(268.231 − 32.7687)/(1.49382 − 12.2278) = 251.173 nm. 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 dl = 176.401 nm; substituting back gives Y_out = 2.2714683 μm.
- The interpolation error in the input is |dl_line − dl| = 74.7717 nm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 2.1548183 μm and 2.3881183 μm. Sort the roots even if the response decreases: 167.784 nm ≤ dl ≤ 185.95 nm.
- The lower and upper input deviations are 8.61648 nm and 9.54937 nm. 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 dl ≈ 176.401 nm; compatible interval [167.784, 185.95] nm. Endpoint interpolation gives 251.173 nm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.