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The maximum number of possible interference maxima for slit separation equal to 1.8λ, where λ is the wavelength of light used, in a Young's double slit experiment is

Asked in JEE Main Online 2012 · Fringe width and fringe position

Answer: (2) 3

Step-by-step solution

Maxima: d sin θ = nλ ⇒ n = 1.8 sin θ.

Since |sin θ| ≤ 1, |n| ≤ 1.8, so n = -1, 0, 1.

Number of maxima = 3.

Why the other options are wrong

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