University Physics II · topic UPII-13.05
Thin lenses
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 lenses: predict a finite change
A thin lens is modelled as a single refracting plane with one focal length f: rays arriving parallel to the axis leave converging to a point f beyond it when f > 0, or diverging as if from a point |f| in front of it when f < 0. Use the supplied topic model Y_out = 100/f, where Y_out is LENS POWER P (D). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Focal length f, measured in cm. Keep d (Object distance s_o) = 60 cm; h (Object height h_o) = 8 cm fixed. For this exercise use only -17.325 cm ≤ f ≤ -2.31 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 f = -14.0217 cm and -11.9196 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δf across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
LENS POWER P against f
Predictions from the supplied Thin lenses 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): LENS POWER P / D
Horizontal axis (x): f: Focal length f / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Focal length f / cm | LENS POWER P / D |
|---|---|
| -17.325 | -5.772005772005772 |
| -13.57125 | -7.368518006815879 |
| -9.817499999999999 | -10.185892538833716 |
| -6.063749999999999 | -16.491445062873638 |
| -2.3099999999999987 | -43.290043290043315 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of f separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At f = -14.0217 cm, substitute into Y_out = 100/f: Y₀ = -7.1318 D.
- At f = -11.9196 cm, the same fixed-parameter model gives Y₁ = -8.38954 D.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.25774 D; the input change is Δf = 2.1021 cm.
- Average sensitivity is ΔY_out/Δf = -0.598326 D 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 = -7.1318 D, Y₁ = -8.38954 D; ΔY_out = -1.25774 D; average sensitivity = -0.598326 D 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.
Thin lenses: infer an input with uncertainty
A thin lens is modelled as a single refracting plane with one focal length f: rays arriving parallel to the axis leave converging to a point f beyond it when f > 0, or diverging as if from a point |f| in front of it when f < 0. Use the supplied topic model Y_out = 100/f, where Y_out is LENS POWER P (D). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. f is Focal length f, measured in cm. Keep d (Object distance s_o) = 60 cm; h (Object height h_o) = 8 cm fixed. For this exercise use only -17.325 cm ≤ f ≤ -2.31 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 = -12.246123 D, with a bounded reading uncertainty ±0.97112 D. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown f by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0015015 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.
LENS POWER P against f
Predictions from the supplied Thin lenses 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): LENS POWER P / D
Horizontal axis (x): f: Focal length f / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| f: Focal length f / cm | LENS POWER P / D |
|---|---|
| -17.325 | -5.772005772005772 |
| -13.57125 | -7.368518006815879 |
| -9.817499999999999 | -10.185892538833716 |
| -6.063749999999999 | -16.491445062873638 |
| -2.3099999999999987 | -43.290043290043315 |
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(f) = 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(-17.325) = -5.77201 D and Y_out(-2.31) = -43.29 D. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives f_line = -17.325 + (-12.2461 − -5.77201)(-2.31 − -17.325)/(-43.29 − -5.77201) = -14.734 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 f = -8.16585 cm; substituting back gives Y_out = -12.246123 D.
- The interpolation error in the input is |f_line − f| = 6.56816 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -13.217243 D and -11.275003 D. Sort the roots even if the response decreases: -8.86918 cm ≤ f ≤ -7.56587 cm.
- The lower and upper input deviations are 0.703328 cm and 0.599975 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 f ≈ -8.16585 cm; compatible interval [-8.86918, -7.56587] cm. Endpoint interpolation gives -14.734 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.