6. Key Variables (AO1 – Knowledge)

7. Using the Formula in A‑Level Questions (AO2 – Application)

1. Read the question carefully – identify which quantities are given and which are required.
2. Check the small‑angle approximation. If D ≥ 5a (or θ < 10°) use λ = a x / (m D); otherwise use the exact relation a sin θ = m λ with θ = tan⁻¹(x/D).
3. Choose the appropriate rearranged form:
   • λ = a x / (m D)
   • a = λ m D / x
   • x = λ m D / a
   • D = a x / (m λ)
4. Convert all measurements to SI units before substitution.
5. Insert the numbers, keeping track of significant figures. Use the least‑precise measurement to set the final number of significant figures.
6. State the answer with units and appropriate sig‑figs.

7.1 Worked Example – First‑Order Bright Fringe

Question: A double‑slit apparatus uses slits separated by a = 0.30 mm. A first‑order bright fringe is observed 2.5 cm from the centre on a screen 2.00 m away. Find the wavelength.

Solution:

1. Convert: a = 3.0 × 10⁻⁴ m, x = 2.5 × 10⁻² m, D = 2.00 m, m = 1.
2. λ = a x / D = (3.0 × 10⁻⁴ m · 2.5 × 10⁻² m) / 2.00 m = 3.75 × 10⁻⁷ m.
3. To two significant figures: λ ≈ 3.8 × 10⁻⁷ m = 380 nm (UV region).

7.2 Worked Example – Higher Order (m = 3)

Question: In the same set‑up the third‑order bright fringe appears at x = 7.5 cm. Determine λ.

Solution:

1. Convert: x = 7.5 × 10⁻² m, m = 3.
2. λ = a x / (m D) = (3.0 × 10⁻⁴ · 7.5 × 10⁻²) / (3 · 2.00) = 3.75 × 10⁻⁷ m ≈ 380 nm.

7.3 Worked Example – Using the Exact Relation (large θ)

Question: A double‑slit (a = 0.50 mm, D = 0.30 m) gives a bright fringe at x = 0.12 m. Find λ, stating whether the small‑angle approximation is valid.

Solution:

1. θ = tan⁻¹(x/D) = tan⁻¹(0.12/0.30) ≈ 21.8° → not < 10°, so use the exact sine.
2. sin θ = sin 21.8° ≈ 0.371.
3. Assume first order (m = 1): λ = a sin θ / m = 5.0 × 10⁻⁴ m · 0.371 ≈ 1.86 × 10⁻⁴ m = 186 µm (far‑IR).

8. Experimental Considerations (AO3 – Evaluation)

• Measuring a, w, D and x: Use a calibrated micrometer or vernier calliper for a and w, a metre‑scale or laser range‑finder for D, and a ruler or image‑analysis software for x.
• Uncertainty propagation (first‑order case):
  \displaystyle\frac{\Delta\lambda}{\lambda}= \sqrt{\left(\frac{\Delta a}{a}\right)^{2} +\left(\frac{\Delta x}{x}\right)^{2} +\left(\frac{\Delta D}{D}\right)^{2}}
• Checking the small‑angle approximation: Compute θ ≈ tan⁻¹(x/D). If θ < 10°, the error from sin θ ≈ tan θ is < 1 %.
• Data handling (reducing random error): Measure several fringes (e.g. m = 1–5), plot x vs m, obtain the gradient G = x/m. Then λ = a G/D. The straight‑line fit provides both λ and an estimate of experimental uncertainty.
• Effect of the diffraction envelope: Ensure the slit width w is small enough that the envelope does not suppress the fringes you are measuring. If envelope minima occur near the fringe of interest, the intensity will be low and the position may be hard to determine.

9. Common Pitfalls (AO1 – Knowledge)

• Applying λ = a x / D to higher‑order fringes without dividing by m.
• Neglecting the small‑angle condition when D is comparable to a.
• Mixing units (mm, cm, m) – always convert to metres before calculation.
• Using a broadband source; the resulting fringe contrast will be reduced.
• Ignoring the diffraction envelope, leading to mis‑identification of fringe order.
• Over‑stating significant figures – the answer cannot be more precise than the least‑precise measurement.

10. Summary (AO1 – Knowledge)

The compact relation

connects the wavelength of a monochromatic wave to three directly measurable quantities in a double‑slit experiment. It follows from the path‑difference condition a sin θ = m λ together with the small‑angle approximation sin θ ≈ tan θ = x/D. Mastery of this formula, the accompanying diffraction envelope, and the extension to diffraction gratings equips students to meet all Cambridge International AS & A Level Physics (9702) requirements for interference, diffraction, and grating topics, and to evaluate experimental data with appropriate uncertainties.