In Young's double slit experiment the light emited from source has $l = 6.5 × 10^{–7} m$ and the distance between the two slits is 1 mm. Distance between the screen and slit is 1 metre. Distance between third dark and fifth birth fringe will be

1.63 mm

$x5=n\frac{λD}{d}=\frac{5×6.5×10^{-7}×1}{10^{-3}}=32.5×10^{-4}m$

$x3=(2n-1)\frac{λ}{2}\frac{D}{d}=\frac{5×6.5×10^{-7}×1}{2×10^{-3}}=16.25×10^{-4}m$

$∴ x5 - x3 ≈ 1.63 mm$