University Physics II · topic UPII-14.02
Young's double-slit interference
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.
Young's Double-Slit Interference: predict a finite change
Young's double-slit interference is the fringe pattern formed when light from two coherent narrow slits recombines on a screen, with the brightness at every point fixed by the path difference d sin θ between the two routes. Use the supplied topic model Y_out = deg(asin(0.000001*lam/sep)), where Y_out is ANGLE OF FIRST MAX 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. sep is Slit separation d, measured in mm. Keep lam (Wavelength λ) = 600 nm; scr (Slit-to-screen distance L) = 2 m fixed. For this exercise use only 0.2425 mm ≤ sep ≤ 0.5575 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 sep = 0.3118 mm and 0.3559 mm. Find the signed output change ΔY_out and the average sensitivity ΔY_out/Δsep across this interval. State whether the response increases or decreases; distinguish this interval average from a derivative at one point.
ANGLE OF FIRST MAX θ₁ against sep
Predictions from the supplied Young's Double-Slit Interference 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): ANGLE OF FIRST MAX θ₁ / °
Horizontal axis (x): sep: Slit separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sep: Slit separation d / mm | ANGLE OF FIRST MAX θ₁ / ° |
|---|---|
| 0.2425 | 0.14176289807532558 |
| 0.32125 | 0.10701163484662408 |
| 0.3999999999999999 | 0.0859437014985321 |
| 0.4787499999999999 | 0.07180673985823813 |
| 0.5574999999999999 | 0.06166363111084516 |
These are the supplied values, not a worked answer. No fitted trend is drawn.
Show a hint
Substitute each value of sep separately while keeping every other input fixed, then subtract in the same order.
Reveal the worked solution
- At sep = 0.3118 mm, substitute into Y_out = deg(asin(0.000001*lam/sep)): Y₀ = 0.110255 °.
- At sep = 0.3559 mm, the same fixed-parameter model gives Y₁ = 0.0965931 °.
- Signed change: ΔY_out = Y₁ − Y₀ = -0.0136618 °; the input change is Δsep = 0.0441 mm.
- Average sensitivity is ΔY_out/Δsep = -0.309792 ° 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 = 0.110255 °, Y₁ = 0.0965931 °; ΔY_out = -0.0136618 °; average sensitivity = -0.309792 ° 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.
Young's Double-Slit Interference: infer an input with uncertainty
Young's double-slit interference is the fringe pattern formed when light from two coherent narrow slits recombines on a screen, with the brightness at every point fixed by the path difference d sin θ between the two routes. Use the supplied topic model Y_out = deg(asin(0.000001*lam/sep)), where Y_out is ANGLE OF FIRST MAX 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. sep is Slit separation d, measured in mm. Keep lam (Wavelength λ) = 600 nm; scr (Slit-to-screen distance L) = 2 m fixed. For this exercise use only 0.2425 mm ≤ sep ≤ 0.5575 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 = 0.079092324 °, with a bounded reading uncertainty ±0.0026143 °. Treat the model and all fixed inputs as exact for this exercise; the bound is not a standard deviation. First estimate the unknown sep by straight-line interpolation between the two endpoint predictions. Then solve the actual model on the stated branch to an absolute input tolerance of 0.0000315 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.
ANGLE OF FIRST MAX θ₁ against sep
Predictions from the supplied Young's Double-Slit Interference 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): ANGLE OF FIRST MAX θ₁ / °
Horizontal axis (x): sep: Slit separation d / mm
Axes show the supplied calibration interval; the origin may lie outside the plot.
Read the exact data as a table
| sep: Slit separation d / mm | ANGLE OF FIRST MAX θ₁ / ° |
|---|---|
| 0.2425 | 0.14176289807532558 |
| 0.32125 | 0.10701163484662408 |
| 0.3999999999999999 | 0.0859437014985321 |
| 0.4787499999999999 | 0.07180673985823813 |
| 0.5574999999999999 | 0.06166363111084516 |
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(sep) = 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(0.2425) = 0.141763 ° and Y_out(0.5575) = 0.0616636 °. The target lies between them. On the selected branch the response is strictly decreasing.
- Endpoint interpolation gives sep_line = 0.2425 + (0.0790923 − 0.141763)(0.5575 − 0.2425)/(0.0616636 − 0.141763) = 0.48896 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 sep = 0.43465 mm; substituting back gives Y_out = 0.079092324 °.
- The interpolation error in the input is |sep_line − sep| = 0.0543096 mm. This is larger than the requested numerical tolerance.
- Invert both limits Y_out = 0.076478024 ° and 0.081706624 °. Sort the roots even if the response decreases: 0.420743 mm ≤ sep ≤ 0.449508 mm.
- The lower and upper input deviations are 0.0139071 mm and 0.0148579 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 sep ≈ 0.43465 mm; compatible interval [0.420743, 0.449508] mm. Endpoint interpolation gives 0.48896 mm.
Compare the assumptions, method and units with your attempt. Explain any correction before trying another question.