In a Young's double slit experiment, two different monochromatic...

The fringe width in a Young's double slit experiment is given by $\beta = \dfrac{\lambda D}{d}$.

The total width of a region containing $n$ fringes is $y = n\beta = n \dfrac{\lambda D}{d}$.

For a fixed region on the screen (constant $y$), and with the same experimental setup (constant $D$ and $d$), the number of fringes is inversely proportional to the wavelength of the light used.

Therefore, $n \propto \dfrac{1}{\lambda}$, which implies $n_1 \lambda_1 = n_2 \lambda_2$.

The ratio of the number of fringes is $\dfrac{n_1}{n_2} = \dfrac{\lambda_2}{\lambda_1}$.

Given the wavelength ratio $\dfrac{\lambda_1}{\lambda_2} = \dfrac{3}{2}$, the ratio of the number of fringes will be $\dfrac{2}{3}$, or $2 : 3$.

Answer: $2 : 3$