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The maximum number of possible interference maxima for slit-separation equal to twice the wavelength in Young's double-slit experiment is

Asked in JEE Main 2004 · Fringe width and fringe position

(1) infinite
(2) five
(3) three
(4) zero

Answer: (2) five

Step-by-step solution

d sin θ = nλ with d = 2λ: n = 2 sin θ.

|n| ≤ 2: n = -2, -1, 0, 1, 2.

Maximum number of maxima = 5.

Why the other options are wrong

(1) Ignores the limit sin θ ≤ 1.
(3) Drops the orders n=±2 that lie at θ = 90°.
(4) Forgets that the central maximum always exists.

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