In Young's double slit, 18 fringes are observed to be formed in a...

Let the length of the segment on the screen be $y$.

The fringe width is given by $\beta = \dfrac{\lambda D}{d}$.

The number of fringes $n$ in the segment $y$ is $n = \dfrac{y}{\beta} = \dfrac{y d}{\lambda D}$.

This implies that $n \lambda = \text{constant}$, or $n_1 \lambda_1 = n_2 \lambda_2$.

Given $n_1 = 18$, $\lambda_1 = 400 \text{ nm}$, and $\lambda_2 = 600 \text{ nm}$.

Substituting the values, we get:

$18 \times 400 = n_2 \times 600$

$n_2 = \dfrac{18 \times 400}{600} = 12$

Thus, 12 fringes will be observed in the same segment.

Answer: $12$