University Physics III · topic UPIII-05.06
Aperture integrals, Fresnel number, and combined patterns
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.
Aperture Integrals, Fresnel Number & Combined Patterns: predict a finite change
The far-field amplitude of any aperture is the aperture integral ∫ A(x) e^(ikx sin θ) dx, a Fourier transform of the transmission A(x), and the Fresnel number N_F = a²/(λL) says whether that far-field form applies at a distance L. Use the supplied topic model Y_out = (d+a)^2/(6330*dist), where Y_out is FRESNEL NUMBER (d+a)²/(λL) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dist is Screen distance L, measured in cm. Keep a (Slit width a) = 25 μm; d (Centre-to-centre separation d) = 75 μm fixed. For this exercise use only 12.4125 cm ≤ dist ≤ 89.5875 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 dist = 29.391 cm and 40.1955 cm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δdist across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FRESNEL NUMBER (d+a)²/(λL) against dist
Predictions from the supplied Aperture Integrals, Fresnel Number & Combined Patterns 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): FRESNEL NUMBER (d+a)²/(λL)
Horizontal axis (x): dist: Screen distance L / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dist: Screen distance L / cm | FRESNEL NUMBER (d+a)²/(λL) |
|---|---|
| 12.4125 | 0.12727321901016436 |
| 31.706249999999997 | 0.04982547071834939 |
| 51 | 0.030976055509091473 |
| 70.29374999999999 | 0.022473958651568104 |
| 89.58749999999999 | 0.017633920256326666 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of dist separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At dist = 29.391 cm, substitute into Y_out = (d+a)^2/(6330*dist): Y₀ = 0.0537504.
- At dist = 40.1955 cm, the same fixed-parameter model gives Y₁ = 0.0393024.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.014448; the input change is Δdist = 10.8045 cm.
- Average sensitivity is ΔY_out/Δdist = -0.00133723 1 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 = 0.0537504, Y₁ = 0.0393024; ΔY_out = -0.014448; average sensitivity = -0.00133723 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.
Aperture Integrals, Fresnel Number & Combined Patterns: infer an input with uncertainty
The far-field amplitude of any aperture is the aperture integral ∫ A(x) e^(ikx sin θ) dx, a Fourier transform of the transmission A(x), and the Fresnel number N_F = a²/(λL) says whether that far-field form applies at a distance L. Use the supplied topic model Y_out = (d+a)^2/(6330*dist), where Y_out is FRESNEL NUMBER (d+a)²/(λL) (dimensionless). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. dist is Screen distance L, measured in cm. Keep a (Slit width a) = 25 μm; d (Centre-to-centre separation d) = 75 μm fixed. For this exercise use only 12.4125 cm ≤ dist ≤ 89.5875 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 = 0.026555703, with a bounded reading uncertainty ±0.0013383. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown dist by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0077175 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.
FRESNEL NUMBER (d+a)²/(λL) against dist
Predictions from the supplied Aperture Integrals, Fresnel Number & Combined Patterns 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): FRESNEL NUMBER (d+a)²/(λL)
Horizontal axis (x): dist: Screen distance L / cm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| dist: Screen distance L / cm | FRESNEL NUMBER (d+a)²/(λL) |
|---|---|
| 12.4125 | 0.12727321901016436 |
| 31.706249999999997 | 0.04982547071834939 |
| 51 | 0.030976055509091473 |
| 70.29374999999999 | 0.022473958651568104 |
| 89.58749999999999 | 0.017633920256326666 |
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(dist) = 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(12.4125) = 0.127273 and Y_out(89.5875) = 0.0176339. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives dist_line = 12.4125 + (0.0265557 − 0.127273)(89.5875 − 12.4125)/(0.0176339 − 0.127273) = 83.3075 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 dist = 59.4892 cm; substituting back gives Y_out = 0.026555703.
- The interpolation error in the input is |dist_line − dist| = 23.8182 cm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.025217403 and 0.027894003. Sort the roots even if the response decreases: 56.6351 cm ≤ dist ≤ 62.6464 cm.
- The lower and upper input deviations are 2.85418 cm and 3.15712 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 dist ≈ 59.4892 cm; compatible interval [56.6351, 62.6464] cm. Endpoint interpolation gives 83.3075 cm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.