University Physics II · topic UPII-13.06
Multi-element imaging and sign conventions
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.
Multi-element imaging and sign conventions: predict a finite change
Multi-element imaging solves an optical system one element at a time, in the order the light meets them, taking the image formed by each element as the object for the next. Use the supplied topic model Y_out = 10*p/(p-10), where Y_out is FIRST IMAGE dᵢ₁ (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Object distance dₒ₁, measured in cm. Keep sep (Lens separation L) = 9 cm; f2 (Focal length of lens 2, f₂) = 6 cm fixed. For this exercise use only 15.9563 cm ≤ p ≤ 23.0437 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. Calculate Y_out at p = 17.5155 cm and 18.5078 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δp across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIRST IMAGE dᵢ₁ against p
Predictions from the supplied Multi-element imaging and sign conventions 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): FIRST IMAGE dᵢ₁ / cm
Horizontal axis (x): p: Object distance dₒ₁ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Object distance dₒ₁ / cm | FIRST IMAGE dᵢ₁ / cm |
|---|---|
| 15.95625 | 26.789087093389295 |
| 17.728125 | 22.939749292357465 |
| 19.5 | 20.526315789473685 |
| 21.271875 | 18.871638480731907 |
| 23.04375 | 17.66650694777192 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of p separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At p = 17.5155 cm, substitute into Y_out = 10*p/(p-10): Y₀ = 23.3058 cm.
- At p = 18.5078 cm, the same fixed-parameter model gives Y₁ = 21.754 cm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.55185 cm; the input change is Δp = 0.99225 cm.
- Average sensitivity is ΔY_out/Δp = -1.56397 cm per cm. 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 = 23.3058 cm, Y₁ = 21.754 cm; ΔY_out = -1.55185 cm; average sensitivity = -1.56397 cm per cm.
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.
Multi-element imaging and sign conventions: infer an input with uncertainty
Multi-element imaging solves an optical system one element at a time, in the order the light meets them, taking the image formed by each element as the object for the next. Use the supplied topic model Y_out = 10*p/(p-10), where Y_out is FIRST IMAGE dᵢ₁ (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. p is Object distance dₒ₁, measured in cm. Keep sep (Lens separation L) = 9 cm; f2 (Focal length of lens 2, f₂) = 6 cm fixed. For this exercise use only 15.9563 cm ≤ p ≤ 23.0437 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 = 19.727981 cm, with a bounded reading uncertainty ±0.30922 cm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown p by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00070875 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.
FIRST IMAGE dᵢ₁ against p
Predictions from the supplied Multi-element imaging and sign conventions 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): FIRST IMAGE dᵢ₁ / cm
Horizontal axis (x): p: Object distance dₒ₁ / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| p: Object distance dₒ₁ / cm | FIRST IMAGE dᵢ₁ / cm |
|---|---|
| 15.95625 | 26.789087093389295 |
| 17.728125 | 22.939749292357465 |
| 19.5 | 20.526315789473685 |
| 21.271875 | 18.871638480731907 |
| 23.04375 | 17.66650694777192 |
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(p) = 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(15.9563) = 26.7891 cm and Y_out(23.0437) = 17.6665 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives p_line = 15.9563 + (19.728 − 26.7891)(23.0437 − 15.9563)/(17.6665 − 26.7891) = 21.4422 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 p = 20.2796 cm; substituting back gives Y_out = 19.727981 cm.
- The interpolation error in the input is |p_line − p| = 1.16253 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 19.418761 cm and 20.037201 cm. Sort the roots even if the response decreases: 19.9629 cm ≤ p ≤ 20.6171 cm.
- The lower and upper input deviations are 0.316688 cm and 0.337482 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 p ≈ 20.2796 cm; compatible interval [19.9629, 20.6171] cm. Endpoint interpolation gives 21.4422 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.