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Monochromatic light of wavelength $\lambda$ passes through a single slit of width $a$. The angular position $\theta$ of the **first** diffraction minimum (on either side of the central maximum) is given by:

A$\sin\theta = \dfrac{\lambda}{2a}$
B$\sin\theta = \dfrac{\lambda}{a}$
C$\sin\theta = \dfrac{2\lambda}{a}$
D$\sin\theta = \dfrac{3\lambda}{2a}$
Answer & Solution
Correct answer: B. $\sin\theta = \dfrac{\lambda}{a}$
Single-slit diffraction condition for the $n$-th dark fringe: $a \sin\theta = n\lambda$ where $n = \pm 1, \pm 2, \ldots$ (the central maximum sits at $\theta = 0$). For the **first** minimum, $n = 1$, giving $\sin\theta = \dfrac{\lambda}{a}$. A common trap is to confuse single-slit diffraction with double-slit interference: in double-slit experiments the $n$-th *bright* fringe sits at $d \sin\theta = n\lambda$, but here we are after the *dark* fringes of a single slit.
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