University Physics III · topic UPIII-05.02
Two-source interference and double-slit intensity
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.
Two-source interference & double-slit intensity: predict a finite change
Two-source interference is the far-field pattern of two equal, in-phase point sources: a screen intensity I = 4I₀ cos²(φ/2) whose phase difference φ = 2πd sinθ/λ is set by the slit separation, the angle and the wavelength. Use the supplied topic model Y_out = 0.002*lam/d, where Y_out is FRINGE SPACING λL/d (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Slit separation d, measured in mm. Keep lam (Wavelength λ) = 633 nm; y (Probe position y on the screen) = 1.25 mm fixed. For this exercise use only 0.131875 mm ≤ d ≤ 0.368125 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 d = 0.18385 mm and 0.216925 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δd across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
FRINGE SPACING λL/d against d
Predictions from the supplied Two-source interference & double-slit intensity 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): FRINGE SPACING λL/d / mm
Horizontal axis (x): d: Slit separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Slit separation d / mm | FRINGE SPACING λL/d / mm |
|---|---|
| 0.13187500000000002 | 9.599999999999998 |
| 0.19093750000000004 | 6.630441898527003 |
| 0.25 | 5.064 |
| 0.3090625 | 4.0962588473205255 |
| 0.36812500000000004 | 3.4390492359932083 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of d separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At d = 0.18385 mm, substitute into Y_out = 0.002*lam/d: Y₀ = 6.88605 mm.
- At d = 0.216925 mm, the same fixed-parameter model gives Y₁ = 5.83612 mm.
- Signed change: ΔY_out = Y₁ − Y₀ = -1.04993 mm; the input change is Δd = 0.033075 mm.
- Average sensitivity is ΔY_out/Δd = -31.7439 mm 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 = 6.88605 mm, Y₁ = 5.83612 mm; ΔY_out = -1.04993 mm; average sensitivity = -31.7439 mm 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.
Two-source interference & double-slit intensity: infer an input with uncertainty
Two-source interference is the far-field pattern of two equal, in-phase point sources: a screen intensity I = 4I₀ cos²(φ/2) whose phase difference φ = 2πd sinθ/λ is set by the slit separation, the angle and the wavelength. Use the supplied topic model Y_out = 0.002*lam/d, where Y_out is FRINGE SPACING λL/d (mm). The output symbol Y_out is distinct from all input symbols. ^ denotes a power; π = 3.141592653589793. d is Slit separation d, measured in mm. Keep lam (Wavelength λ) = 633 nm; y (Probe position y on the screen) = 1.25 mm fixed. For this exercise use only 0.131875 mm ≤ d ≤ 0.368125 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 = 4.5871643 mm, with a bounded reading uncertainty ±0.17222 mm. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown d by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.000023625 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.
FRINGE SPACING λL/d against d
Predictions from the supplied Two-source interference & double-slit intensity 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): FRINGE SPACING λL/d / mm
Horizontal axis (x): d: Slit separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| d: Slit separation d / mm | FRINGE SPACING λL/d / mm |
|---|---|
| 0.13187500000000002 | 9.599999999999998 |
| 0.19093750000000004 | 6.630441898527003 |
| 0.25 | 5.064 |
| 0.3090625 | 4.0962588473205255 |
| 0.36812500000000004 | 3.4390492359932083 |
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(d) = 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(0.131875) = 9.6 mm and Y_out(0.368125) = 3.43905 mm. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives d_line = 0.131875 + (4.58716 − 9.6)(0.368125 − 0.131875)/(3.43905 − 9.6) = 0.324099 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 d = 0.275987 mm; substituting back gives Y_out = 4.5871643 mm.
- The interpolation error in the input is |d_line − d| = 0.0481115 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 4.4149443 mm and 4.7593843 mm. Sort the roots even if the response decreases: 0.266001 mm ≤ d ≤ 0.286753 mm.
- The lower and upper input deviations are 0.0099867 mm and 0.0107658 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 d ≈ 0.275987 mm; compatible interval [0.266001, 0.286753] mm. Endpoint interpolation gives 0.324099 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.