University Physics II · unit UPII-14
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.
Wave model of light and coherence: infer an input with uncertainty
Two light fields are coherent when the phase difference between them stays constant for at least as long as the detector averages, which is the condition for their cross term, and so for any visible interference pattern, to survive. Use the supplied topic model Y_out = 2*sqrt(r)/(1+r), where Y_out is PEAK VISIBILITY V(0) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; sqrt(x) means √x. r is Intensity ratio I₂/I₁, dimensionless. Keep lam (Wavelength λ₀) = 0.6 μm; dlam (Spectral width Δλ) = 60 nm fixed. For this exercise use only 0.150938 ≤ r ≤ 0.899062 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 = 0.96969274, with a bounded reading uncertainty ±0.0043341. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown r by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000748125, 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.
PEAK VISIBILITY V(0) against r
Predictions from the supplied Wave model of light and coherence 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): PEAK VISIBILITY V(0)
Horizontal axis (x): r: Intensity ratio I₂/I₁
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| r: Intensity ratio I₂/I₁ | PEAK VISIBILITY V(0) |
|---|---|
| 0.1509375 | 0.6751135603766518 |
| 0.33796875 | 0.8690050462115415 |
| 0.5249999999999999 | 0.9502542128648812 |
| 0.7120312499999999 | 0.9857523822498959 |
| 0.8990624999999999 | 0.9985864745637851 |
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(r) = 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(0.150938) = 0.675114 and Y_out(0.899062) = 0.998586. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives r_line = 0.150938 + (0.969693 − 0.675114)(0.899062 − 0.150938)/(0.998586 − 0.675114) = 0.832237. 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 r = 0.607294; substituting back gives Y_out = 0.96969274.
- The interpolation error in the input is |r_line − r| = 0.224944. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.96535864 and 0.97402684. Sort the roots even if the response decreases: 0.586135 ≤ r ≤ 0.630746.
- The lower and upper input deviations are 0.0211586 and 0.0234522. 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 r ≈ 0.607294; compatible interval [0.586135, 0.630746]. Endpoint interpolation gives 0.832237.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Young's Double-Slit Interference: infer an input with uncertainty
Young's double-slit interference is the fringe pattern formed when light from two coherent narrow slits recombines on a screen, with the brightness at every point fixed by the path difference d sin θ between the two routes. Use the supplied topic model Y_out = deg(asin(0.000001*lam/sep)), where Y_out is ANGLE OF FIRST MAX θ₁ (°). 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. sep is Slit separation d, measured in mm. Keep lam (Wavelength λ) = 600 nm; scr (Slit-to-screen distance L) = 2 m fixed. For this exercise use only 0.2425 mm ≤ sep ≤ 0.5575 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 = 0.079092324 °, with a bounded reading uncertainty ±0.0026143 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sep by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 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.
ANGLE OF FIRST MAX θ₁ against sep
Predictions from the supplied Young's Double-Slit Interference 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 FIRST MAX θ₁ / °
Horizontal axis (x): sep: Slit separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sep: Slit separation d / mm | ANGLE OF FIRST MAX θ₁ / ° |
|---|---|
| 0.2425 | 0.14176289807532558 |
| 0.32125 | 0.10701163484662408 |
| 0.3999999999999999 | 0.0859437014985321 |
| 0.4787499999999999 | 0.07180673985823813 |
| 0.5574999999999999 | 0.06166363111084516 |
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(sep) = 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(0.2425) = 0.141763 ° and Y_out(0.5575) = 0.0616636 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sep_line = 0.2425 + (0.0790923 − 0.141763)(0.5575 − 0.2425)/(0.0616636 − 0.141763) = 0.48896 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 sep = 0.43465 mm; substituting back gives Y_out = 0.079092324 °.
- The interpolation error in the input is |sep_line − sep| = 0.0543096 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.076478024 ° and 0.081706624 °. Sort the roots even if the response decreases: 0.420743 mm ≤ sep ≤ 0.449508 mm.
- The lower and upper input deviations are 0.0139071 mm and 0.0148579 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 sep ≈ 0.43465 mm; compatible interval [0.420743, 0.449508] mm. Endpoint interpolation gives 0.48896 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Multiple Slits & Diffraction Gratings: infer an input with uncertainty
A diffraction grating is a large number of identical, equally spaced slits, and it sends sharp principal maxima in the directions where the path difference between neighbouring slits is a whole number of wavelengths. Use the supplied topic model Y_out = q, where Y_out is ENVELOPE ZERO AT m = (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. q is Spacing / slit width d/a, dimensionless. Keep n (Slits illuminated N) = 5; r (Spacing / wavelength d/λ) = 3.5 fixed. For this exercise use only 2.2125 ≤ q ≤ 3.7875 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 = 3.17325, with a bounded reading uncertainty ±0.092137. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown q by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0001575, 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.
ENVELOPE ZERO AT m = against q
Predictions from the supplied Multiple Slits & Diffraction Gratings 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): ENVELOPE ZERO AT m =
Horizontal axis (x): q: Spacing / slit width d/a
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| q: Spacing / slit width d/a | ENVELOPE ZERO AT m = |
|---|---|
| 2.2125 | 2.2125 |
| 2.6062499999999997 | 2.6062499999999997 |
| 3 | 3 |
| 3.39375 | 3.39375 |
| 3.7874999999999996 | 3.7874999999999996 |
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(q) = 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(2.2125) = 2.2125 and Y_out(3.7875) = 3.7875. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives q_line = 2.2125 + (3.17325 − 2.2125)(3.7875 − 2.2125)/(3.7875 − 2.2125) = 3.17325. 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 q = 3.17325; substituting back gives Y_out = 3.17325.
- The interpolation error in the input is |q_line − q| = 4.44089e-16. This is within the requested numerical tolerance.
- Invert both limits Y_out = 3.081113 and 3.265387. Sort the roots even if the response decreases: 3.08111 ≤ q ≤ 3.26539.
- The lower and upper input deviations are 0.092137 and 0.092137. 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 q ≈ 3.17325; compatible interval [3.08111, 3.26539]. Endpoint interpolation gives 3.17325.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.
Single-Slit Diffraction: infer an input with uncertainty
Single-slit diffraction is the pattern formed when light passing through one aperture of width a interferes with itself, because wavelets leaving different parts of the opening travel different distances to the same point. Use the supplied topic model Y_out = deg(asin(lam*0.000001/a)), where Y_out is ANGLE θ₁ = arcsin(λ/a) (°). 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. a is Slit width a, measured in mm. Keep lam (Wavelength λ) = 630 nm fixed. For this exercise use only 0.07125 mm ≤ a ≤ 0.22875 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 = 0.21572644 °, with a bounded reading uncertainty ±0.0086892 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00001575 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.
ANGLE θ₁ = arcsin(λ/a) against a
Predictions from the supplied Single-Slit Diffraction 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 θ₁ = arcsin(λ/a) / °
Horizontal axis (x): a: Slit width a / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Slit width a / mm | ANGLE θ₁ = arcsin(λ/a) / ° |
|---|---|
| 0.07125000000000001 | 0.506621915257732 |
| 0.110625 | 0.3262963725125847 |
| 0.15000000000000002 | 0.24064298144884724 |
| 0.18937500000000002 | 0.1906080933289963 |
| 0.22875 | 0.15779841191491334 |
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(a) = 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(0.07125) = 0.506622 ° and Y_out(0.22875) = 0.157798 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 0.07125 + (0.215726 − 0.506622)(0.22875 − 0.07125)/(0.157798 − 0.506622) = 0.202594 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 a = 0.167325 mm; substituting back gives Y_out = 0.21572644 °.
- The interpolation error in the input is |a_line − a| = 0.0352695 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.20703724 ° and 0.22441564 °. Sort the roots even if the response decreases: 0.160846 mm ≤ a ≤ 0.174347 mm.
- The lower and upper input deviations are 0.00647866 mm and 0.00702247 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 a ≈ 0.167325 mm; compatible interval [0.160846, 0.174347] mm. Endpoint interpolation gives 0.202594 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.