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

$x_5=n \frac{\lambda D}{d}=\frac{5 \times 6.5 \times 10^{-7} \times 1}{10^{-3}}=32.5 \times 10^{-4}$ m

$x_3=(2 n-1) \frac{\lambda}{2} \frac{D}{d}=\frac{5 \times 6.3 \times 10^{-7} \times 1}{2 \times 10^{-3}}=16.25 \times 10^{-4}$ m

∴ x₅ - x₃ ≈ 1.63 mm.