formula for total number of fringes in young's double slit experiment
Answers
Answered by
2
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:
Answered by
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)
formula with example
Similar questions