Maximum possible interference maxima for slit separation equal to twice the wavelength in youngs double slit experiment is

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In Young's double-slit experiment, the interference pattern is formed due to the interference of light waves from two slits. The pattern consists of alternate bright and dark fringes. The maximum number of interference maxima that can be obtained for a given separation between the slits can be calculated using the formula:

nmax = (D/d) + 1

where nmax is the maximum number of interference maxima, D is the distance between the screen and the slits, and d is the separation between the slits.

For the given case where the slit separation is equal to twice the wavelength, we have d = 2λ. Substituting this in the above formula, we get:

nmax = (D/2λ) + 1

Therefore, the maximum possible interference maxima for slit separation equal to twice the wavelength in Young's double-slit experiment is given by the above formula. The value of nmax depends on the distance D between the screen and the slits, and the wavelength of light used in the experiment.

Young's double-slit experiment

Young's Double-slit Experiment

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