In a Young’s double slit interference pattern, following maxima and m…

The correct answer is Option (3) → A, B, C, D

In Young’s experiment a bright fringe (maximum) occurs at path difference \( n\lambda \) and a dark fringe (minimum) at \( (2n - 1)\lambda/2 \), with maxima and minima alternating; each step from one to the next adds half a wavelength of path difference.

A. Central maxima: path difference \( 0 \)
B. First minima: \( \lambda/2 \)
C. First maxima: \( \lambda \)
D. Second minima: \( 3\lambda/2 \)

Ascending order of path difference: A, B, C, D.

NCERT (2026-27):
Coherent and Incoherent Addition of Waves – “… To summarise: If we have two coherent sources S₁ and S₂ vibrating in phase, then for an arbitrary point P whenever the path difference,

    \( \displaystyle S_1P \sim S_2P = n\lambda \quad (n = 0, 1, 2, 3, \ldots) \qquad (10.9) \)

we will have constructive interference and the resultant intensity will be 4I₀; the sign ~ between S₁P and S₂P represents the difference between S₁P and S₂P. On the other hand, if the point P is such that the path difference,

    \( \displaystyle S_1P \sim S_2P = \left(n + \frac{1}{2}\right)\lambda \quad (n = 0, 1, 2, 3, \ldots) \qquad (10.10) \)

we will have destructive interference and the resultant intensity will be zero.”