University Physics III · topic UPIII-04.09
Multi-element systems and optical instruments
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 systems and optical instruments: predict a finite change
A multi-element system is analysed by treating each element's image as the object of the next, so an instrument's overall effect is the product of one-element results, reported for instruments used by eye as an angular magnification. Use the supplied topic model Y_out = -fo/fe, where Y_out is MAGNIFICATION −f_o/f_e (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. fe is Eyepiece focal length f_e, measured in cm. Keep fo (Objective focal length f_o) = 40 cm; th (Field angle θ of the star) = 1.5 ° fixed. For this exercise use only 2.85 cm ≤ fe ≤ 9.15 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 fe = 4.236 cm and 5.118 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δfe across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
MAGNIFICATION −f_o/f_e against fe
Predictions from the supplied Multi-element systems and optical instruments 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): MAGNIFICATION −f_o/f_e
Horizontal axis (x): fe: Eyepiece focal length f_e / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| fe: Eyepiece focal length f_e / cm | MAGNIFICATION −f_o/f_e |
|---|---|
| 2.85 | -14.035087719298245 |
| 4.425 | -9.03954802259887 |
| 6 | -6.666666666666667 |
| 7.574999999999999 | -5.280528052805281 |
| 9.149999999999999 | -4.371584699453552 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of fe separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At fe = 4.236 cm, substitute into Y_out = -fo/fe: Y₀ = -9.44287.
- At fe = 5.118 cm, the same fixed-parameter model gives Y₁ = -7.81555.
- Signed change: ΔY_out = Y₁ − Y₀ = 1.62732; the input change is Δfe = 0.882 cm.
- Average sensitivity is ΔY_out/Δfe = 1.84503 1 per cm. Its positive sign means that this response increases 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 = -9.44287, Y₁ = -7.81555; ΔY_out = 1.62732; average sensitivity = 1.84503 1 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 systems and optical instruments: infer an input with uncertainty
A multi-element system is analysed by treating each element's image as the object of the next, so an instrument's overall effect is the product of one-element results, reported for instruments used by eye as an angular magnification. Use the supplied topic model Y_out = -fo/fe, where Y_out is MAGNIFICATION −f_o/f_e (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. fe is Eyepiece focal length f_e, measured in cm. Keep fo (Objective focal length f_o) = 40 cm; th (Field angle θ of the star) = 1.5 ° fixed. For this exercise use only 2.85 cm ≤ fe ≤ 9.15 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 = -5.9763932, with a bounded reading uncertainty ±0.24072. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown fe by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.00063 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.
MAGNIFICATION −f_o/f_e against fe
Predictions from the supplied Multi-element systems and optical instruments 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): MAGNIFICATION −f_o/f_e
Horizontal axis (x): fe: Eyepiece focal length f_e / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| fe: Eyepiece focal length f_e / cm | MAGNIFICATION −f_o/f_e |
|---|---|
| 2.85 | -14.035087719298245 |
| 4.425 | -9.03954802259887 |
| 6 | -6.666666666666667 |
| 7.574999999999999 | -5.280528052805281 |
| 9.149999999999999 | -4.371584699453552 |
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(fe) = 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(2.85) = -14.0351 and Y_out(9.15) = -4.37158. The target lies between them. On the selected branch the response is strictly increasing.
- Endpoint interpolation gives fe_line = 2.85 + (-5.97639 − -14.0351)(9.15 − 2.85)/(-4.37158 − -14.0351) = 8.10377 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 fe = 6.693 cm; substituting back gives Y_out = -5.9763932.
- The interpolation error in the input is |fe_line − fe| = 1.41077 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = -6.2171132 and -5.7356732. Sort the roots even if the response decreases: 6.43385 cm ≤ fe ≤ 6.9739 cm.
- The lower and upper input deviations are 0.259146 cm and 0.280898 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 fe ≈ 6.693 cm; compatible interval [6.43385, 6.9739] cm. Endpoint interpolation gives 8.10377 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.