Physical optics worksheet answers for AP Physics 2 (free PDF download)

Problem 1 Path difference and interference

1. Path difference = 4.5 m − 3.0 m = 1.5 m.
2. 1.5 m / 0.50 m = 3.0 wavelengths.
3. A whole number of wavelengths means the sounds arrive in phase, so the interference is constructive and the microphone detects a loud signal.

Problem 2 Wavelength of a sound note

1. λ = v / f = 343 / 440 = 0.780 m.
2. T = 1 / f = 1 / 440 = 2.27 × 10⁻³ s (2.27 ms).

Problem 3 Fringe spacing from a double slit

1. λ = 5.00 × 10⁻⁷ m and d = 1.0 × 10⁻⁴ m.
2. Δy = λL / d = (5.00 × 10⁻⁷ × 1.5) / (1.0 × 10⁻⁴) = 7.5 × 10⁻³ m (7.5 mm).

Problem 4 Converting wave quantities

1. 0.25 mm = 0.25 × 10⁻³ m = 2.5 × 10⁻⁴ m.
2. λ = c / f = 3.0 × 10⁸ / 4.6 × 10¹⁴ = 6.52 × 10⁻⁷ m = 652 nm.

Problem 5 Fundamental of a string

1. λ = 2L = 1.30 m.
2. f₁ = v / λ = 260 / 1.30 = 200 Hz.
3. f₃ = 3 × 200 = 600 Hz.

Problem 6 Phase flips in an oil film

1. Light goes from low index (1.00) to higher index (1.45), so the phase flips by half a cycle.
2. Light goes from 1.45 to 1.33, a lower index, so there is no flip.
3. Only one reflection flips, so the rays differ by half a cycle before path difference is added.

Problem 7 Pipes open and closed

1. f = nv / 2L = n × 340 / 1.00: 340 Hz, 680 Hz, 1020 Hz.
2. f = nv / 4L = n × 340 / 2.0 with n odd: 170 Hz and 510 Hz.

Problem 8 Doppler shift of a horn

1. f′ = f v / (v − vₛ) = 500 × 340 / 320 = 531 Hz.
2. f′ = 500 × 340 / 360 = 472 Hz.

Problem 9 Reading a fringe graph

1. Peaks are at 0, 4, 8 mm, so Δy = 4.0 mm.
2. λ = Δy d / L = (4.0 × 10⁻³ × 3.0 × 10⁻⁴) / 2.0 = 6.0 × 10⁻⁷ m (600 nm).
3. The fringe at +8 mm is the second order, so the path difference is 2λ = 1200 nm.

Problem 10 Angles in a double slit

1. sin θ = mλ / d = 2 × 600 × 10⁻⁹ / 4.0 × 10⁻⁶ = 0.30, so θ = 17.5°.
2. sin θ cannot exceed 1, so m ≤ d / λ = 6.67 and the highest order is m = 6.

Problem 11 Soap film thickness

1. Only the top reflection flips by half a cycle, so the rays are already half a cycle apart. A path difference of half a wavelength inside the film (2nt = λ/2) adds another half cycle and brings them back in phase.
2. t = λ / (4n) = 560 / (4 × 1.33) = 105 nm.

Problem 12 Anti-reflection coating

1. Both reflections go from lower to higher index, so both flip.
2. Equal flips cancel out, so reflection is weak (destructive) when the extra path 2nt equals half a wavelength.
3. t = λ / (4n) = 550 / (4 × 1.38) = 99.6 nm.

Problem 13 Beats from two sources

1. f_beat = 6 / 2.0 = 3 Hz.
2. The string was at 437 Hz or 443 Hz.
3. Tightening raises the string frequency. A higher beat rate means the gap grew, so the string was above the fork: 443 Hz.

Problem 14 Red against violet fringes

1. Red: y = λL / d = 700 × 10⁻⁹ × 1.0 / 2.0 × 10⁻⁴ = 3.5 mm. Violet: y = 400 × 10⁻⁹ × 1.0 / 2.0 × 10⁻⁴ = 2.0 mm.
2. Every colour has a bright fringe at the centre, so it is white. Violet is nearest the centre and red is furthest out in each first-order band.

Problem 15 Resonance tube lengths

1. λ = v / f = 343 / 512 = 0.670 m.
2. Resonance occurs at L = λ/4 = 0.168 m and L = 3λ/4 = 0.502 m.

Problem 16 Slit separation from a pattern

1. Each third-order fringe lies y₃ = 3λL / d from the centre, so the separation is 2y₃ = 6λL / d.
2. d = 6λL / (2y₃) = 6 × 633 × 10⁻⁹ × 3.20 / 0.0608 = 2.00 × 10⁻⁴ m (0.200 mm).

Problem 17 Colours reflected by a bubble

1. 2nt = (m + ½)λ, so λ = 2nt / (m + ½).
2. 2nt = 2 × 1.33 × 380 = 1011 nm. m = 1 gives 1011 / 1.5 = 674 nm; m = 2 gives 1011 / 2.5 = 404 nm; m = 0 gives 2022 nm and m = 3 gives 289 nm, both outside the visible range.
3. Red and violet are strongly reflected, so the film looks magenta.

Problem 18 Wavelength squeezed by motion

1. λ₀ = v / f = 343 / 650 = 0.528 m.
2. In one period the source moves 30 / 650 = 0.0462 m, so λ′ = 0.528 − 0.046 = 0.482 m.
3. f′ = v / λ′ = 343 / 0.4815 = 712 Hz.

Problem 19 Closed pipe claim, spot the error

1. f₁ = v / 4L = 340 / (4 × 0.85) = 100 Hz.
2. Only odd multiples occur: 300 Hz and 500 Hz.
3. The student treated the pipe as open at both ends. A closed end forces a node, so 200 Hz does not fit: no whole number of half wavelengths can end on a node at one end and an antinode at the other.

Problem 20 Design a wavelength measurement

1. Measure the slit-to-screen distance L (metre rule or tape) and the distance across N fringes (ruler on the screen). Apparatus: laser, double slit slide, clamp, screen, tape measure.
2. Measure across 10 fringes and divide by 10, use a large L and repeat the measurement.
3. Plot fringe spacing Δy (vertical) against L (horizontal) for several L values. The graph is a straight line through the origin with gradient λ / d, so λ = gradient × d.