Wave interference worksheet answers for A Level physics (free PDF download)

Problem 1 Conditions for interference

1. Constructive: path difference = nλ (n = 0, 1, 2…). Destructive: path difference = (n + ½)λ.
2. Coherent sources emit waves of the same frequency with a constant phase difference.
3. The atoms in each bulb emit light in random bursts, so the phase difference between the two bulbs changes randomly and the pattern averages out to uniform brightness.

Problem 2 Two loudspeakers

1. λ = v/f = 340 ÷ 680 = 0.5 m.
2. Path difference = 6.15 − 5.40 = 0.75 m.
3. 0.75 ÷ 0.5 = 1.5 wavelengths, an odd number of half wavelengths, so the waves are in antiphase and the sound is quiet.

Problem 3 Adding two pulses

1. 3.0 + 2.0 = 5.0 cm.
2. 3.0 + (−2.0) = 1.0 cm.
3. The principle of superposition: the resultant is the sum of the displacements.

Problem 4 Fringe spacing

1. w = λD/s.
2. w = 633 × 10⁻⁹ × 1.50 ÷ 0.25 × 10⁻³ = 3.8 mm.
3. 3 × 3.798 = 11.4 mm.

Problem 5 Lines per millimetre

1. 300 lines/mm: d = 1 ÷ 300 mm = 3.33 µm. 800 lines/mm: d = 1 ÷ 800 mm = 3.33 µm and 1.25 µm.
2. The 800 lines/mm grating: smaller d gives a larger sin θ = nλ/d.
3. 800 × 5.0 = 4000 lines.

Problem 6 Reading a fringe pattern

1. Neighbouring bright fringes are 2.0 mm apart.
2. The second bright fringe is at 4.0 mm from the centre.
3. s = λD/w = 600 × 10⁻⁹ × 1.50 ÷ 2.0 × 10⁻³ = 0.45 mm.

Problem 7 Waves through gaps

1. 20 cm: little spreading, a straight beam with slightly curved edges. 2.0 cm: clear spreading with nearly circular wavefronts. 0.5 cm: strong spreading, almost circular waves.
2. The 0.5 cm gap, since it is much narrower than the wavelength.
3. Sound has wavelengths of about 0.3 m to 3 m, similar to doorways, so it diffracts strongly. Light has a wavelength of about 500 nm, much smaller than any gap, so it hardly diffracts at all.

Problem 8 Wavelength from fringes

1. w = 12.0 mm ÷ 10 = 1.20 mm.
2. λ = ws/D.
3. λ = 1.20 × 10⁻³ × 0.50 × 10⁻³ ÷ 1.20 = 0.000 m = 500 nm.

Problem 9 Red light to blue light

1. w = λD/s, so w is directly proportional to λ.
2. w = 2.4 × 450 ÷ 650 = 1.66 mm.
3. Doubling s halves w: 1.662 ÷ 2 = 0.831 mm.

Problem 10 Angles from a grating

1. d = 1 ÷ 400 000 = 2.50 × 10⁻⁶ m.
2. sin θ = λ/d = 0.208, θ = 12°.
3. sin θ = 2 × 0.208 = 0.416, θ = 24.6°.

Problem 11 How many spots?

1. d = 1 ÷ 600 000 = 1.67 µm.
2. n_max is below d/λ = 2.56, so the highest order is 2.
3. sin θ = 2 × 650 × 10⁻⁹ ÷ 0.0000 = 0.78, θ = 51.3°. Orders −2 to +2 give 5 bright spots.

Problem 12 Wavelength from a measured angle

1. d = 1 ÷ 500 000 = 2.00 × 10⁻⁶ m.
2. λ = d sin θ ÷ n = 2.00 × 10⁻⁶ × sin 20.0° ÷ 1 = 2.00 × 10⁻⁶ × 0.342 = 0.000 m = 684 nm.
3. About 684 nm is red light.

Problem 13 A white light spectrum

1. d = 2.00 µm. Violet: sin θ = 0.200, θ = 11.5°. Red: sin θ = 0.350, θ = 20.5°.
2. Red is deviated more because θ increases with wavelength (sin θ = nλ/d).
3. Second-order red: sin θ = 0.700, θ = 44.4°. Third-order violet: sin θ = 0.600, θ = 36.9°. The third-order violet is at a smaller angle than the second-order red, so they overlap.

Problem 14 Single-slit spreading

1. sin θ = λ/a = 600 × 10⁻⁹ ÷ 0.10 × 10⁻³ = 0.006.
2. θ = 0.344°.
3. Half-width = D tan θ = 2.0 × 0.006 = 12 mm, so the full width is 24 mm.

Problem 15 Phase difference

1. 1.25 × 10⁻⁶ ÷ 500 × 10⁻⁹ = 2.5 λ.
2. Phase difference = 2π × 2.5 = 5π rad (15.7 rad).
3. 2.5 is an odd number of half wavelengths (5 × λ/2), so the point is dark.

Problem 16 A film over one slit

1. Light through the film travels more optical path, so the position of zero overall path difference moves towards the slit with the film.
2. (n − 1)t = Nλ, with N = 8.
3. t = Nλ ÷ (n − 1) = 8 × 600 × 10⁻⁹ ÷ 0.50 = 0.000 m = 9.6 µm.

Problem 17 Fringe spacing against distance

1. A straight line through the origin: w is directly proportional to D, as expected from w = λD/s.
2. Gradient = (3.25 − 0.65) ÷ (2.50 − 0.50) = 1.3 mm/m = 0.0013 (no unit, m per m).
3. gradient = λ/s, so λ = 0.0013 × 0.40 × 10⁻³ = 0.000 m = 520 nm.

Problem 18 Two colours together

1. w = λD/s: 450 nm gives 1.8 mm; 600 nm gives 2.4 mm.
2. 4 × 1.8 = 7.2 mm and 3 × 2.4 = 7.2 mm, which are equal (the ratio 450 : 600 = 3 : 4).
3. The distance is 7.2 mm.

Problem 19 Spot the error

1. 0.0020 mm is 2.0 × 10⁻⁶ m, not 2.0 × 10⁻³ m: the conversion from mm to m was missed by a factor of 1000.
2. λ = d sin θ ÷ n = 2.00 × 10⁻⁶ × 0.500 ÷ 2 = 5.00 × 10⁻⁷ m = 500 nm.
3. 5.0 × 10⁻⁴ m is half a millimetre, far larger than any visible wavelength (400 to 700 nm).

Problem 20 Close wavelengths

1. d = 1.667 µm. 577 nm: sin θ = 0.6924, θ = 43.82°. 579 nm: sin θ = 0.6948, θ = 44.01°.
2. Separation = 44.01 − 43.82 = 0.19°.
3. With more lines illuminated the maxima become narrower, so two close angles can be seen as separate lines instead of one blurred spot.