Questions: Single-Slit Diffraction and Diffraction Patterns

Question 1

A single slit of width a produces a diffraction pattern with its first dark minimum at angle θ₁. If the slit is narrowed to width a/2, where does the first dark minimum now appear?

• A. At θ₁/2 — the narrower slit confines the pattern, moving minima inward
• B. At the same angle θ₁ — diffraction angle depends only on wavelength
• C. At approximately 2θ₁ — the narrower slit spreads the pattern, pushing minima outward
• D. The first minimum disappears — a slit too narrow produces no dark fringes

Answer: (C)

The first minimum occurs where a sin θ = λ, so sin θ = λ/a. Halving the slit width to a/2 doubles sin θ, approximately doubling the angle (for small angles, θ ≈ λ/a). The pattern widens because a narrower slit spatially confines the wave more, forcing greater angular spread — the central maximum broadens and its bounding minima move outward. Option A inverts the relationship: it reflects the intuition that a narrower slit should produce a narrower beam, which is the ray-optics expectation. Wave diffraction behaves oppositely.

Question 2

A student argues: 'By analogy with double-slit minima (where path difference = λ/2 gives dark fringes), the first single-slit minimum should occur at a sin θ = λ/2.' Why is the correct condition a sin θ = λ instead?

• A. Single-slit diffraction uses a different wavelength convention than double-slit interference
• B. The slit is divided into two halves; each corresponding pair of points (one from each half, separated by a/2) must cancel. This requires (a/2) sin θ = λ/2, giving a sin θ = λ
• C. The path difference across the whole slit must equal λ/2, not λ
• D. The factor of 2 arises because the central maximum is twice as wide as secondary maxima

Answer: (B)

The derivation pairs each point in the upper half of the slit with the point directly below it in the lower half, a distance a/2 apart. For complete destructive interference across all such pairs simultaneously, the path difference for each pair must be λ/2. The geometry gives (a/2) sin θ = λ/2, which simplifies to a sin θ = λ. The student's error is applying the double-slit formula directly: in double-slit, two distinct sources interfere; in single-slit, you must account for contributions from the entire continuous aperture, not just two points, which changes the cancellation condition by a factor of 2.

Question 3

Single-slit diffraction occurs because different parts of the slit act as independent Huygens wavelet sources whose contributions can interfere constructively or destructively at a distant screen.

• T. True
• F. False

Answer: True

This is the Huygens-Fresnel principle applied to a finite aperture: every point across the slit width radiates a wavelet, and these wavelets travel different path lengths to a given point on the screen. When the path-length differences satisfy the cancellation condition, destructive interference produces a dark minimum. The entire diffraction pattern — bright central peak, dark minima, weak secondary maxima — follows from systematically computing how all these wavelets interfere. Without the wave nature of light (and thus the Huygens construction), single-slit diffraction would not exist: ray optics predicts a sharp geometric shadow with no banding.

Question 4

The central maximum of a single-slit diffraction pattern has the same angular width as each of the secondary maxima on either side.

• T. True
• F. False

Answer: False

The central maximum spans from the first minimum at θ = −λ/a to θ = +λ/a, giving an angular width of 2λ/a. Each secondary maximum sits between consecutive minima — for example, between nλ/a and (n+1)λ/a — spanning an angular width of only λ/a. The central maximum is therefore twice as wide as any secondary maximum. It also carries the overwhelming majority of the diffracted intensity; secondary maxima are dramatically dimmer because only partial cancellation occurs for points contributing to them, and progressively less energy reaches higher-order fringes.

Question 5

Why does a narrower slit produce a wider diffraction pattern, and what general principle does this illustrate about waves and spatial confinement?

Model answer: The first dark minimum occurs at a sin θ = λ, so sin θ = λ/a. Reducing the slit width a increases sin θ, pushing the minimum to a larger angle and spreading the central maximum. The general principle is that spatially confining a wave in one dimension forces it to spread in angle: a narrow slit is a small spatial 'window,' and squeezing light through a smaller opening forces greater diffraction. This is the wave-optics analog of the Heisenberg uncertainty principle: tighter spatial localization (small Δx, here the slit width a) implies greater spread in the transverse momentum or wave-vector (large Δkₓ). Ray optics predicts no such spreading — it is an inherently wave phenomenon arising because the slit width becomes comparable to the wavelength.