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JEE PHYSICS MCQ: WAVE OPTICS

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Application +4 | -1

The maximum number of possible interference maxima for slit separation equal to 2.5 times the wavelength in young's double slit experiment is

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Correct option: B

Five

Option B (Five) is correct because the maximum number of interference maxima is determined by the condition

d \sin \theta = n \lambda

, and with

d = 2.5\lambda

, the possible integer values for n are -2, -1, 0, 1, and 2.

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