C.3 Wave phenomena

The condition for minima in single-slit diffraction is $b\sin \theta =n\lambda$ where $n=1$ for the first minimum.

With $b=0.40\text{ mm}=4.0\times 10^{-4}\text{ m}$, $\lambda =500\text{ nm}=5.0\times 10^{-7}\text{ m}$, and $n=1$: $\sin \theta =\frac{1\times 5.0\times 10^{-7}}{4.0\times 10^{-4}}=1.25\times 10^{-3}$

$\theta ={\sin }^{-1}(1.25\times 10^{-3})\approx 1.25\times 10^{-3}\text{ rad}\approx 0.072°$

The angle to the first dark fringe is approximately $0.072°$, which matches the first option.

We need $\sin \theta =\lambda /b=(5.0\times 10^{-7})/(4.0\times 10^{-4})=1.25\times 10^{-3}$, giving $\theta \approx 0.072°$.

$0.14°$ corresponds to $\sin \theta \approx 2.5\times 10^{-3}$, which would require $n=2$ (the second minimum), not the first. Eliminated.

$0.036°$ is half of the correct answer, which might arise from accidentally using $2b$ instead of $b$ in the denominator. Eliminated.

$0.29°$ is far too large; it would correspond to a much longer wavelength or much narrower slit than given. Eliminated.

The only consistent answer is $\theta \approx 0.072°$.

According to Huygens' principle, every point across the slit width acts as a secondary source of wavelets. The intensity at any angle depends on how these wavelets interfere.

At $\theta =0$ (the central maximum), all wavelets from across the slit travel equal path lengths to the screen, so they all arrive in phase. Complete constructive interference occurs, giving maximum amplitude and therefore maximum intensity.

At secondary maxima, only a fraction of the wavelets interfere constructively — the rest cancel in pairs. This partial constructive interference yields much lower intensity (about 4.5% of the central maximum for the first secondary maximum).

The correct explanation is that at the centre all waves arrive in phase, while secondary maxima involve only partial constructive interference.

The question tests understanding of why the central maximum dominates the single-slit diffraction pattern using wave interference principles.

"The slit acts as a lens" is incorrect — a slit does not focus light. This confuses diffraction with refraction. Eliminated.

"Shorter path distance" is not a valid explanation. Intensity depends on interference, not simply on path length to the screen. Eliminated.

"Reflection at slit edges" is not the mechanism responsible for the dimness of secondary maxima in standard diffraction theory. Eliminated.

Only the option describing in-phase arrival at the centre and partial constructive interference elsewhere correctly uses the physics of wave superposition.