The width of the central maximum in single-slit diffraction on a screen at distance D is
A{'text': 'λ D / a', 'label': 'A'}
B{'text': 'λ D × a', 'label': 'B'}
C{'text': 'a D / λ', 'label': 'C'}
D{'text': '2 λ D / a', 'label': 'D'}
Answer & Solution
Correct answer: D. {'text': '2 λ D / a', 'label': 'D'}
1. First minima are at ±λ/a (small-angle approximation).
2. On a screen at distance D, they lie at ±λD/a.
3. Width of central maximum = 2 × λD/a = 2λD/a.
4. Central maximum is thus twice as wide as the successive fringes.
_Source: NCERT Class 12 Physics, Ch 10 "Wave Optics", §10.6_
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