University Physics II · topic UPII-13.07
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.
Optical instruments and angular magnification: predict a finite change
Angular magnification is the angle an object subtends at the eye when viewed through an instrument, divided by the angle the same object subtends when it is held at the unaided eye's near point. Use the supplied topic model Y_out = fo/fe, where Y_out is MAGNIFICATION f_o / f_e (×). 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 mm. Keep fo (Objective focal length f_o) = 180 mm; theta (Incoming field angle θ) = 2.5 ° fixed. For this exercise use only 24.25 mm ≤ fe ≤ 55.75 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 fe = 31.18 mm and 35.59 mm. 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 Optical instruments and angular magnification 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 / mm
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 / mm | MAGNIFICATION f_o / f_e / × |
|---|---|
| 24.25 | 7.422680412371134 |
| 32.125 | 5.603112840466926 |
| 40 | 4.5 |
| 47.875 | 3.759791122715405 |
| 55.75 | 3.2286995515695067 |
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 = 31.18 mm, substitute into Y_out = fo/fe: Y₀ = 5.77293 ×.
- At fe = 35.59 mm, the same fixed-parameter model gives Y₁ = 5.0576 ×.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.715331 ×; the input change is Δfe = 4.41 mm.
- Average sensitivity is ΔY_out/Δfe = -0.162207 × 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 = 5.77293 ×, Y₁ = 5.0576 ×; ΔY_out = -0.715331 ×; average sensitivity = -0.162207 × 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.
Optical instruments and angular magnification: infer an input with uncertainty
Angular magnification is the angle an object subtends at the eye when viewed through an instrument, divided by the angle the same object subtends when it is held at the unaided eye's near point. Use the supplied topic model Y_out = fo/fe, where Y_out is MAGNIFICATION f_o / f_e (×). 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 mm. Keep fo (Objective focal length f_o) = 180 mm; theta (Incoming field angle θ) = 2.5 ° fixed. For this exercise use only 24.25 mm ≤ fe ≤ 55.75 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 = 4.1412631 ×, with a bounded reading uncertainty ±0.13688 ×. 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 0.00315 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.
MAGNIFICATION f_o / f_e against fe
Predictions from the supplied Optical instruments and angular magnification 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 / mm
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 / mm | MAGNIFICATION f_o / f_e / × |
|---|---|
| 24.25 | 7.422680412371134 |
| 32.125 | 5.603112840466926 |
| 40 | 4.5 |
| 47.875 | 3.759791122715405 |
| 55.75 | 3.2286995515695067 |
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(24.25) = 7.42268 × and Y_out(55.75) = 3.2287 ×. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives fe_line = 24.25 + (4.14126 − 7.42268)(55.75 − 24.25)/(3.2287 − 7.42268) = 48.896 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 fe = 43.465 mm; substituting back gives Y_out = 4.1412631 ×.
- The interpolation error in the input is |fe_line − fe| = 5.43095 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.0043831 × and 4.2781431 ×. Sort the roots even if the response decreases: 42.0743 mm ≤ fe ≤ 44.9507 mm.
- The lower and upper input deviations are 1.39067 mm and 1.48574 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 fe ≈ 43.465 mm; compatible interval [42.0743, 44.9507] mm. Endpoint interpolation gives 48.896 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.