Diffraction: Practice Questions (Physics 2 14.7)

Answer: (D). Diffraction is strongest when the gap is comparable to the wavelength. A 2.0 cm gap equals λ, so the waves leave it as near-semicircles and spread into the whole region behind the barrier.

• (A) is 20 wavelengths wide. The waves pass almost straight through, with only slight curving at the edges. This option suits a student who thinks “a bigger gap lets more wave spread out”.
• (B) is 10 wavelengths wide: still mostly a straight beam.
• (C) is 4 wavelengths wide. It spreads more than (A) or (B), but less than a gap the size of λ.

Answer: (A). The first dark fringe is at a sin θ = λ, and a sin θ is the edge-to-edge path difference. Pair each point in the top half of the slit with the point a/2 below it: each pair then differs by λ/2 and cancels, so the whole slit gives darkness.

• (B) is the difference for each pair of points a/2 apart, not for the two edges. A student who thinks “dark means half a wavelength out of step” for the whole slit picks this.
• (C) is the double-slit idea of a dark fringe at an odd number of half-wavelengths. It does not apply to the edges of one slit.
• (D) is the condition at the centre of the pattern, where every wavelet arrives in step and the light is brightest.

Answer: (A). The central band width is 2λL/a, so it is proportional to λ/a. The new width is the old width × 2 ÷ 3 = 2/3 of the old width.

• (B) inverts the relationship, treating the width as proportional to a/λ (the “wider slit, wider pattern” error).
• (C) multiplies both factors, as if the width grew with both λ and a.
• (D) divides by both factors, as if the width shrank with both.

Answer: (A). The angle is small (tan θ = 0.0300 ÷ 1.20, so θ = 1.43°). Then λ = y_min a ÷ (mL) = (3.00 × 10⁻² m)(5.0 × 10⁻⁵ m) ÷ (2 × 1.20 m) = 6.25 × 10⁻⁷ m = 625 nm.

• (B) uses m = 1, treating the second dark fringe as the first.
• (C) halves the 3.00 cm, as if it were the distance between the two second-order dark fringes on opposite sides.
• (D) uses a sin θ = (m + ½)λ with m = 2, the double-slit dark condition, which does not apply to one slit.

1. Convert: λ = 4.80 × 10⁻⁷ m; a = 8.00 × 10⁻⁵ m.
2. (a) sin θ₁ = λ/a = 4.80 × 10⁻⁷ ÷ 8.00 × 10⁻⁵ = 6.00 × 10⁻³, so θ₁ = 0.344°. This is far below 10°, so y_min = mλL/a is valid.
3. (b) y₁ = λL/a = (4.80 × 10⁻⁷ m)(1.80 m) ÷ (8.00 × 10⁻⁵ m) = 1.08 × 10⁻² m = 10.8 mm. y₂ = 2y₁ = 21.6 mm.
4. (c) Central band = 2y₁ = 21.6 mm. First side band = y₂ − y₁ = 10.8 mm.

Suggested mark points (4): 1 for sin θ₁ = 6.00 × 10⁻³ and θ₁ = 0.344°, with a comment that it is under 10°; 1 for y₁ = 10.8 mm; 1 for y₂ = 21.6 mm; 1 for both band widths, with the central band twice the side band. A solution that uses y = L tan θ throughout is equally valid (it gives 10.8002 mm for y₁).

Common error: writing the central band as 10.8 mm. The central band runs from the first dark fringe on one side to the first dark fringe on the other.

(a) Gap P: wavefronts beyond the gap are near-semicircles centred on the gap, spreading into the whole region behind the barrier, with the same spacing (1.5 cm) as before. Gap Q: wavefronts stay mostly straight across the width of the gap, curving only at their ends, with the same 1.5 cm spacing.

(b) a sin θ = mλ with m = 1: sin θ₁ = 1.5 ÷ 9.0 = 0.167, so θ₁ = 9.59°.

(c) For gap P, sin θ₁ = 1.5 ÷ 1.2 = 1.25. No angle has a sine greater than 1, so there is no first minimum. The wavelets from the gap never cancel completely in any forward direction, and the wave spreads everywhere behind the barrier.

(d) At constant speed, λ = v/f, so a higher frequency gives a shorter wavelength. Since sin θ₁ = λ/a, the angle decreases: the wave spreads less. (For example, 1.5 times the frequency gives λ = 1.0 cm and θ₁ = 6.38°.)

Do not award the first point if the sketch shows a changed wavelength beyond the gap. Accept the (c) reasoning in words: “the gap is narrower than one wavelength, so the path difference across it can never reach λ”.

(a) Plot y on the vertical axis against m on the horizontal axis. From y_min = (λL/a)m, the graph is a straight line through the origin with gradient λL/a (the spacing of the dark fringes).

(b) A best-fit line gives a gradient of about 4.97 mm per order (a line forced through the origin gives 5.00 mm). Then a = λL ÷ gradient = (6.50 × 10⁻⁷ m)(2.00 m) ÷ (4.97 × 10⁻³ m) = 2.62 × 10⁻⁴ m ≈ 0.26 mm. Check: θ for m = 4 is about 0.57°, so the small-angle form holds.

(c) The gradient is proportional to L. New gradient = 4.97 mm × (3.00 ÷ 2.00) ≈ 7.5 mm per order.

(d) A long slit spreads light only across its narrow width, giving straight bands parallel to the slit. A round hole spreads light equally in every direction around the beam, so the pattern becomes a bright central disc surrounded by faint circular rings.

Accept a value of a found from one data point (for example m = 4 gives 0.261 mm) for the fourth point, but not for the third. Carry forward an error in the gradient from (b) into (c) once.