Solution:
Solution
For dark Fringe,
$$x= \displaystyle \frac{(2n-1) \lambda D}{2d}$$
$$X_3 = \displaystyle \frac{(2 \times 3 -1 ) \times 6.5 \times 10^{-7} m \times 1 m}{2 \times 10^{-3} m}$$
$$x_3 = 1.63 \times 10^{-3} m$$
For bright Fringe
$$x = \displaystyle \frac{n \lambda D}{d}$$
$$ \therefore X_5 = \displaystyle \frac{5 \times 6.5 \times 10^{}-7 m \times 1m}{10^{-3} m}$$
$$x_5 = 3.25 \times 10^{-3} m$$
Distance between third dark fringe and fifth bright fringe = $$(x_5 - x_3) = 1.62 mm$$
