University Physics III · topic UPIII-04.08
Thin-lens imaging
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.
Thin-lens imaging: predict a finite change
Thin-lens imaging is the paraxial model in which a lens of negligible thickness maps an object at distance s to an image at distance s′ through 1/s + 1/s′ = 1/f, with magnification m = −s′/s. Use the supplied topic model Y_out = f*s/(s-f), where Y_out is IMAGE DISTANCE s′ (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Object distance s, measured in mm. Keep f (Focal length f) = 75 mm fixed. For this exercise use only 88.7375 mm ≤ s ≤ 360.688 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. Calculate Y_out at s = 148.567 mm and 186.639 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δs across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
IMAGE DISTANCE s′ against s
Predictions from the supplied Thin-lens imaging 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′ / mm
Horizontal axis (x): s: Object distance s / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Object distance s / mm | IMAGE DISTANCE s′ / mm |
|---|---|
| 88.73750000000001 | 484.46314831665114 |
| 156.72500000000002 | 143.82838788620373 |
| 224.7125 | 112.5720130249645 |
| 292.7 | 100.83830960036748 |
| 360.6875 | 94.68934587617589 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of s separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At s = 148.567 mm, substitute into Y_out = f*s/(s-f): Y₀ = 151.461 mm.
- At s = 186.639 mm, the same fixed-parameter model gives Y₁ = 125.385 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = -26.076 mm; the input change is Δs = 38.073 mm.
- Average sensitivity is ΔY_out/Δs = -0.684896 mm per mm. 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 = 151.461 mm, Y₁ = 125.385 mm; ΔY_out = -26.076 mm; average sensitivity = -0.684896 mm per mm.
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.
Thin-lens imaging: infer an input with uncertainty
Thin-lens imaging is the paraxial model in which a lens of negligible thickness maps an object at distance s to an image at distance s′ through 1/s + 1/s′ = 1/f, with magnification m = −s′/s. Use the supplied topic model Y_out = f*s/(s-f), where Y_out is IMAGE DISTANCE s′ (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Object distance s, measured in mm. Keep f (Focal length f) = 75 mm fixed. For this exercise use only 88.7375 mm ≤ s ≤ 360.688 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 = 106.31489 mm, with a bounded reading uncertainty ±1.7438 mm. 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 0.027195 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.
IMAGE DISTANCE s′ against s
Predictions from the supplied Thin-lens imaging 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′ / mm
Horizontal axis (x): s: Object distance s / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Object distance s / mm | IMAGE DISTANCE s′ / mm |
|---|---|
| 88.73750000000001 | 484.46314831665114 |
| 156.72500000000002 | 143.82838788620373 |
| 224.7125 | 112.5720130249645 |
| 292.7 | 100.83830960036748 |
| 360.6875 | 94.68934587617589 |
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(88.7375) = 484.463 mm and Y_out(360.688) = 94.6893 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 88.7375 + (106.315 − 484.463)(360.688 − 88.7375)/(94.6893 − 484.463) = 352.576 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 s = 254.627 mm; substituting back gives Y_out = 106.31489 mm.
- The interpolation error in the input is |s_line − s| = 97.9492 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 104.57109 mm and 108.05869 mm. Sort the roots even if the response decreases: 245.152 mm ≤ s ≤ 265.22 mm.
- The lower and upper input deviations are 9.47508 mm and 10.5926 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 s ≈ 254.627 mm; compatible interval [245.152, 265.22] mm. Endpoint interpolation gives 352.576 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.