University Physics III · topic UPIII-05.05
Fraunhofer diffraction from a single slit
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.
Fraunhofer Diffraction from a Single Slit: predict a finite change
Fraunhofer diffraction from a single slit is the far-field pattern obtained by adding the Huygens wavelets from every point of the slit with their relative phases, which gives I = I₀ (sin β/β)² with β = πa sin θ/λ. Use the supplied topic model Y_out = deg(asin(lam/(1000*a))), where Y_out is FIRST MINIMUM θ₁ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; deg converts radians to degrees; unmarked trigonometric functions use radians. a is Slit width a, measured in μm. Keep lam (Wavelength λ) = 550 nm; theta (Diffraction angle θ) = 0.6 ° fixed. For this exercise use only 13.1875 μm ≤ a ≤ 36.8125 μm 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 a = 18.385 μm and 21.6925 μm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δa across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FIRST MINIMUM θ₁ against a
Predictions from the supplied Fraunhofer Diffraction from a Single Slit 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): FIRST MINIMUM θ₁ / °
Horizontal axis (x): a: Slit width a / μm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Slit width a / μm | FIRST MINIMUM θ₁ / ° |
|---|---|
| 13.1875 | 2.3902802984807847 |
| 19.09375 | 1.6506468481791814 |
| 25 | 1.2606088523503383 |
| 30.90625 | 1.0196753813503991 |
| 36.8125 | 0.8560638704549725 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of a separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At a = 18.385 μm, substitute into Y_out = deg(asin(lam/(1000*a))): Y₀ = 1.7143 °.
- At a = 21.6925 μm, the same fixed-parameter model gives Y₁ = 1.45285 °.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.261444 °; the input change is Δa = 3.3075 μm.
- Average sensitivity is ΔY_out/Δa = -0.0790457 ° per μm. 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 = 1.7143 °, Y₁ = 1.45285 °; ΔY_out = -0.261444 °; average sensitivity = -0.0790457 ° per μm.
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.
Fraunhofer Diffraction from a Single Slit: infer an input with uncertainty
Fraunhofer diffraction from a single slit is the far-field pattern obtained by adding the Huygens wavelets from every point of the slit with their relative phases, which gives I = I₀ (sin β/β)² with β = πa sin θ/λ. Use the supplied topic model Y_out = deg(asin(lam/(1000*a))), where Y_out is FIRST MINIMUM θ₁ (°). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793; deg converts radians to degrees; unmarked trigonometric functions use radians. a is Slit width a, measured in μm. Keep lam (Wavelength λ) = 550 nm; theta (Diffraction angle θ) = 0.6 ° fixed. For this exercise use only 13.1875 μm ≤ a ≤ 36.8125 μm 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 = 1.141891 °, with a bounded reading uncertainty ±0.042874 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown a by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0023625 μm, 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.
FIRST MINIMUM θ₁ against a
Predictions from the supplied Fraunhofer Diffraction from a Single Slit 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): FIRST MINIMUM θ₁ / °
Horizontal axis (x): a: Slit width a / μm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| a: Slit width a / μm | FIRST MINIMUM θ₁ / ° |
|---|---|
| 13.1875 | 2.3902802984807847 |
| 19.09375 | 1.6506468481791814 |
| 25 | 1.2606088523503383 |
| 30.90625 | 1.0196753813503991 |
| 36.8125 | 0.8560638704549725 |
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(a) = 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(13.1875) = 2.39028 ° and Y_out(36.8125) = 0.856064 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives a_line = 13.1875 + (1.14189 − 2.39028)(36.8125 − 13.1875)/(0.856064 − 2.39028) = 32.4111 μm. 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 a = 27.5988 μm; substituting back gives Y_out = 1.141891 °.
- The interpolation error in the input is |a_line − a| = 4.81237 μm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 1.099017 ° and 1.184765 °. Sort the roots even if the response decreases: 26.6001 μm ≤ a ≤ 28.6753 μm.
- The lower and upper input deviations are 0.998602 μm and 1.07652 μm. 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 a ≈ 27.5988 μm; compatible interval [26.6001, 28.6753] μm. Endpoint interpolation gives 32.4111 μm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.