9702 · 8.3

Interference — practice questions

Practice and worked examples for 9702 Interference. Short previews only — attempt the full question in MarkScheme against the official scheme.

Worked example 1

In a Young's double-slit experiment, light of wavelength 600 nm is used. The slits are separated by 0.50 mm, and the screen is placed 2.0 m away. Calculate the fringe spacing observed on the screen.

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Identify known values: $\lambda = 600 \text{ nm} = 600 \times 10^{-9} \text{ m}$ , $a = 0.50 \text{ mm} = 0.50 \times 10^{-3} \text{ m}$ , $D = 2.0 \text{ m}$ .
Recall the formula: $\lambda = \frac{ax}{D}$ . We need to find $x$ .
Rearrange the formula for $x$ : $x = \frac{\lambda D}{a}$ .
Substitute the values: $x = \frac{(600 \times 10^{-9} \text{ m}) \times (2.0 \text{ m})}{0.50 \times 10^{-3} \text{ m}}$ .
Calculate: $x = \frac{1.2 \times 10^{-6}}{0.50 \times 10^{-3}} = 2.4 \times 10^{-3} \text{ m}$ .
State the answer with units: The fringe spacing is $2.4 \text{ mm}$ .

Worked example 2

A diffraction grating with 500 lines per mm is illuminated with monochromatic light of wavelength 589 nm. The light is incident normally on the grating. What is the angle of the second-order maximum?

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First, calculate the grating spacing, d. The grating has 500 lines per mm, which is 500,000 lines per metre. d = 1 / (Number of lines per metre) = 1 / (500 × 10³ m⁻¹) = 2.0 × 10⁻⁶ m.
State the diffraction grating formula: d sin θ = nλ.
Identify the known values for the second-order maximum: n = 2, λ = 589 nm = 589 × 10⁻⁹ m.
Rearrange the formula to find sin θ: sin θ = nλ / d.
Substitute the values: sin θ = (2 × 589 × 10⁻⁹ m) / (2.0 × 10⁻⁶ m).
Calculate sin θ: sin θ = (1178 × 10⁻⁹) / (2.0 × 10⁻⁶) = 0.589.
Calculate the angle θ: θ = arcsin(0.589) = 36.09°.
State the final answer to an appropriate number of significant figures: The angle of the second-order maximum is 36.1°.

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