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

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Explanation:

One central maxima, and two maxima on either sides are formed in the Young's double - slit experiment.

Therefore, the number of maximas are 1 + 2 + 2 = 5.1.

Final Answer:

Hence, option (B) is correct. (Five)

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