Standard difficulty · 20 questions · 40 marks
Show your working for Sections B and C. Answers are on the last page — try every question first!
Wave Optics · AW-12P-WOP-2026-V1 · For parents — fold back or tear off before handing the sheet to your child.
| Q | Answer | Working |
|---|---|---|
| 1 | (d) 0.573 mm | β = λD/d |
| 2 | (c) 5.85 mm | y = (n − ½)β |
| 3 | (d) tan−1(1.5) | tan i_B = n |
| 4 | (a) 5.5 W/m2 | the first polaroid passes half; then I = I1 cos2θ |
| 5 | (b) a source of secondary wavelets | — |
| 6 | (a) 2.52 mm | width = 2λD/a |
| 7 | (c) 544 nm | λ = λ0/n; the frequency does not change |
| 8 | (a) Both A and R are true, and R is the correct explanation of A. | A is true: β = λD/d = 4.08 mm, and doubling d halves it to 2.04 mm. R is true and explains A. |
| 9 | 1.04 mm; 5.2 mm | β = λD/d; y = nβ |
| 10 | 2.144 mm | λ falls to λ/n, and so does β |
| 11 | 0 W/m2 | I = 4I0 cos2(φ/2), φ = 2π × (path difference)/λ |
| 12 | 2.72 × 10−3 rad; 10.88 mm | a sin θ = λ; width = 2λD/a |
| 13 | 2.5 W/m2 | 80/2 × cos260° × cos260° |
| 14 | tan−1(5/3); tan−1(3/5) | tan i_B = n, and the reflected and refracted rays are at 90° |
| 15 | 0.55 mm | 1β and (2 − ½)β |
| 16 | 750 nm; 562.5 nm; 4 × 1014 Hz | λ = c/f; λ_glass = λ/n |
| 17 | 0.6 mm and 0.9 mm; 1.8 mm (the 3rd bright fringe of 400 nm meets the 2nd of 600 nm) | n1 × 400 = n2 × 600 |
| 18 | 14 W/m2 | 112/2 × cos245° × cos245° |
| 19 | 0.44 mm; 4.4 mm; 10 | β = λD/d; band width = 2λD/a; number = 2d/a |
| 20 | tan−1(5/4); tan−1(4/5); 90° | tan i_B = n; i_B + r = 90°; Malus: I = I0 cos2θ |