Double-Slit Interference and Diffraction Gratings: Practice Questions (Physics 2 14.8)

Answer: (A). At a dark fringe, light arrives from both slits but the total is darkness. Waves can do this: they arrive half a wavelength out of step and cancel. Two streams of particles cannot add up to nothing.

• (B) is what a simple particle model predicts. It is not what is seen with narrow, closely spaced slits.
• (C) is not observed: each slit spreads the light by diffraction. Straight-line travel would not explain the fringes anyway.
• (D) is true but tells you nothing about waves or particles; more energy arrives either way.

Answer: (B). Dark fringes need an odd number of half-wavelengths: 0.5λ, 1.5λ, 2.5λ, … The first dark fringe is at 0.5λ, so the second is at 1.5λ.

• (A) is the second bright fringe (m = 2), where the waves arrive in step.
• (C) puts m = 2 into (m + ½)λ, but that formula counts from m = 0, so m = 2 is the third dark fringe.
• (D) is the first dark fringe, next to the central maximum.

Answer: (A). Spacing = λL/d. L is multiplied by 1.80 ÷ 1.20 = 1.5 and d by 0.5, so the spacing is multiplied by 1.5 ÷ 0.5 = 3.

• (B) multiplies by 0.5 instead of dividing, as if the spacing were proportional to d.
• (C) inverts both factors.
• (D) includes only the change in L and ignores the change in d.

Answer: (A). At θ = 0 every wavelength has zero path difference, so all colours are bright together and the centre is white. For m ≥ 1, sin θ = mλ/d, so the longest wavelength (red) goes to the largest angle.

• (B) confuses the central maximum with the higher orders. There is no path difference at the centre to separate the colours.
• (C) carries over the prism order (violet bent most by refraction). For a grating, the angle grows with wavelength, not frequency.
• (D) forgets that θ depends on λ as well as d.

1. d = (1/300) mm = 3.33 × 10⁻⁶ m.
2. (a) tan θ₁ = 0.184 ÷ 1.10, so θ₁ = 9.50°.
3. (b) λ = d sin θ₁ = (3.333 × 10⁻⁶ m)(sin 9.50°) = 5.50 × 10⁻⁷ m (550 nm).
4. (c) sin θ₂ = 2λ/d = 0.330, so θ₂ = 19.3°. y₂ = L tan θ₂ = 1.10 × tan 19.27° = 0.384 m. The small-angle form gives y₂ = 2λL/d = 0.363 m, about 6% too small, because θ₂ is well above 10°.
5. (d) m must be less than d/λ = 3.333 × 10⁻⁶ ÷ 5.50 × 10⁻⁷ = 6.06, so the highest order is m = 6 (at 81.8°, so it would not land on a flat screen of any practical size).

Suggested mark points (5): 1 for d = 3.33 × 10⁻⁶ m; 1 for θ₁ = 9.50° from tan θ = y/L; 1 for λ = 550 nm; 1 for y₂ = 0.384 m with a comparison to 0.363 m; 1 for m = 6 from d/λ. For (b), accept 550–558 nm: the small-angle estimate λ = dy/(mL) gives 558 nm, about 1.4% high, because θ₁ is close to 10°.

Common error: taking d = 300 m or 1/300 m. The grating has 300 lines in each millimetre.

(a) Measure the distance between the m = −2 and m = +2 bright fringes (on opposite sides of the centre) and halve it. A longer distance has a smaller percentage uncertainty, and it avoids having to judge exactly where the centre of the central fringe is. Measure to the centres of the fringes, mark them on paper on the screen, and repeat each reading.

(b) Plot y₂ (vertical, in mm, from 0 to 7 mm) against L (horizontal, in m, from 0 to 2.5 m). Use scales that make the points fill most of the grid. From y₂ = 2λL/d, the line should pass through the origin with gradient 2λ/d.

(c) A best-fit line gives a gradient of about 2.6 × 10⁻³ (2.63 mm per metre). Then d = 2λ ÷ gradient = 2(5.20 × 10⁻⁷ m) ÷ (2.63 × 10⁻³) = 3.95 × 10⁻⁴ m ≈ 0.40 mm.

(d) The line passes almost through the origin, so y₂ ≈ gradient × L = (2.63 × 10⁻³)(3.00 m) = 7.9 × 10⁻³ m ≈ 7.9 mm. (Reading the best-fit line itself, which has a small intercept of −0.04 mm, gives 7.8 mm; both are fine.)

Do not award the gradient point for a value taken from a single data point. Also accept plotting the fringe spacing y₂/2 against L, with gradient λ/d, if the student then finds d correctly.

(a) Fringe spacing = λL/d, so d = λL ÷ spacing = (6.00 × 10⁻⁷ m)(2.00 m) ÷ (4.0 × 10⁻³ m) = 3.0 × 10⁻⁴ m (0.30 mm).

(b) The envelope’s first minimum is at y = λL/a = 12.0 mm, so a = (6.00 × 10⁻⁷)(2.00) ÷ (1.20 × 10⁻²) = 1.0 × 10⁻⁴ m (0.10 mm).

(c) 12.0 mm is 3 fringe spacings (d/a = 3), so the third-order fringes would sit exactly at the envelope minima. There, each slit on its own sends no light, so there is nothing to interfere and the fringe is missing. Inside the central band: m = 0, ±1, ±2, which is 5 bright fringes.

(d) With one slit there is no second wave to interfere with, so the fringes disappear. The screen shows the single-slit pattern of that slit: a central band 24.0 mm wide, with dimmer side bands. It is less bright overall, because only one slit lets light through.

(e) The maxima stay at the same positions (every 4.0 mm for small angles), because d sin θ = mλ still holds. With many slits, each maximum becomes much narrower and brighter, with wide dark regions between them.

Accept the count in (c) from d/a = 3 without screen distances. Do not award the (d) point for “the pattern disappears”.