In a single-slit Fraunhofer diffraction pattern, the first minimum occurs at angle $\theta$ satisfying:
A$a\sin\theta = \lambda/2$
B$a\sin\theta = \lambda$
C$a\sin\theta = 2\lambda$
D$a\sin\theta = n\lambda \cdot d/D$
Answer & Solution
Correct answer: B. $a\sin\theta = \lambda$
First minimum: $a\sin\theta = \lambda$ (slit divided into 2 halves, paths from extremes differ by $\lambda$ — pairs cancel). Successive minima at $a\sin\theta = n\lambda$ (n = ±1, ±2, …); central maximum is twice as wide as side fringes.
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