Question

formula for total number of fringes in young's double slit experiment​

Answer 1

Answer:

In Young's slits, the two beams that interfere have a width limited by the diffraction by the slits. The fringes are visible only in the common part of the two beams. As the central fringe is bright, we will roughly have N=1+2d/a visible fringes.

Explanation:

Answer 2

n1 = 62

w1 = 5893

w2 = 4358

total extent of fringes is constant for both wavelengths

i.e. n1w1D/d = n2w2D/d

n2 =62×5893/4358 = 84 (approx)

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