Wave Optics 2004

Let nth minima of 400nm coincides with m th minima of 560nm, then
(2n-1)(400/2) = (2m-1)560/2 → 2n-12m-1 = 75 = 1410 = ...........
i.e, 4th minima of 400 nm coincides with 3rd minima of 560 nm.
Location of this minima is,
Y₁ = (2 x 4 -1)(1000)(400 X 10^-6)2 X 0.4 = 14 mm.

Next 11 th minima of 400 nm will coincide with 8th minima of 560nm

Y₂ = (2 X 11 - 1)(1000) (560 X 10 ^-6)2 X 0.1 = 42 mm

Required distance is Y₂ - Y₁ = 28 mm.

d) 28 mm