University Physics II · topic UPII-13.04
Spherical mirrors
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.
Spherical Mirrors: predict a finite change
A spherical mirror is a polished cap of a sphere of radius R, and within the paraxial region near its axis it sends every ray that arrived parallel to that axis to a single focal point at a distance f = R/2 from the vertex: really through it, in front of the mirror, if the cap is concave, and only by backward extension, behind the mirror, if it is convex. Use the supplied topic model Y_out = s*r/(2*s-r), where Y_out is IMAGE DISTANCE s′ (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Object distance s, measured in cm. Keep r (Radius of curvature R) = 40 cm fixed. For this exercise use only 45.1 cm ≤ s ≤ 82.9 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 s = 53.416 cm and 58.708 cm. 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 Spherical Mirrors 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′ / cm
Horizontal axis (x): s: Object distance s / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Object distance s / cm | IMAGE DISTANCE s′ / cm |
|---|---|
| 45.1 | 35.93625498007968 |
| 54.550000000000004 | 31.577424023154844 |
| 64 | 29.09090909090909 |
| 73.45 | 27.483629560336762 |
| 82.9 | 26.359300476947535 |
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 = 53.416 cm, substitute into Y_out = s*r/(2*s-r): Y₀ = 31.9703 cm.
- At s = 58.708 cm, the same fixed-parameter model gives Y₁ = 30.3338 cm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.63653 cm; the input change is Δs = 5.292 cm.
- Average sensitivity is ΔY_out/Δs = -0.309247 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 = 31.9703 cm, Y₁ = 30.3338 cm; ΔY_out = -1.63653 cm; average sensitivity = -0.309247 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.
Spherical Mirrors: infer an input with uncertainty
A spherical mirror is a polished cap of a sphere of radius R, and within the paraxial region near its axis it sends every ray that arrived parallel to that axis to a single focal point at a distance f = R/2 from the vertex: really through it, in front of the mirror, if the cap is concave, and only by backward extension, behind the mirror, if it is convex. Use the supplied topic model Y_out = s*r/(2*s-r), where Y_out is IMAGE DISTANCE s′ (cm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. s is Object distance s, measured in cm. Keep r (Radius of curvature R) = 40 cm fixed. For this exercise use only 45.1 cm ≤ s ≤ 82.9 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 = 28.305993 cm, with a bounded reading uncertainty ±0.292 cm. 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 0.00378 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.
IMAGE DISTANCE s′ against s
Predictions from the supplied Spherical Mirrors 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′ / cm
Horizontal axis (x): s: Object distance s / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| s: Object distance s / cm | IMAGE DISTANCE s′ / cm |
|---|---|
| 45.1 | 35.93625498007968 |
| 54.550000000000004 | 31.577424023154844 |
| 64 | 29.09090909090909 |
| 73.45 | 27.483629560336762 |
| 82.9 | 26.359300476947535 |
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(45.1) = 35.9363 cm and Y_out(82.9) = 26.3593 cm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives s_line = 45.1 + (28.306 − 35.9363)(82.9 − 45.1)/(26.3593 − 35.9363) = 75.2165 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 s = 68.158 cm; substituting back gives Y_out = 28.305993 cm.
- The interpolation error in the input is |s_line − s| = 7.05845 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 28.013993 cm and 28.597993 cm. Sort the roots even if the response decreases: 66.5225 cm ≤ s ≤ 69.9127 cm.
- The lower and upper input deviations are 1.63551 cm and 1.7547 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 s ≈ 68.158 cm; compatible interval [66.5225, 69.9127] cm. Endpoint interpolation gives 75.2165 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.