In two separate Young's double-slit experimental set-ups and two monochromatic light sources of different wavelengths are used to get fringes of equal width. the ratios of the slits separations and that of the wavelengths of light used are 2: 1 and 1 : 2 respectively. The corresponding ratio of the distances between the slits and the respective screens $$(D_{1}/D_{2})$$ is ______.

We need to find the ratio $$D_1/D_2$$ given that two Young’s double-slit setups produce fringes of equal width.

The fringe width in Young’s double-slit experiment is given by

$$ \beta = \frac{\lambda D}{d} $$

Here $$\lambda$$ is the wavelength, $$D$$ is the distance from slits to screen, and $$d$$ is the slit separation.

Since the fringe widths are equal, $$\beta_1 = \beta_2$$, which implies

$$ \frac{\lambda_1 D_1}{d_1} = \frac{\lambda_2 D_2}{d_2} $$

Substituting the given ratios $$d_1 : d_2 = 2 : 1$$ and $$\lambda_1 : \lambda_2 = 1 : 2$$, we rearrange for $$D_1/D_2$$:

$$ \frac{D_1}{D_2} = \frac{\lambda_2}{\lambda_1} \times \frac{d_1}{d_2} $$

Evaluating this gives

$$ \frac{D_1}{D_2} = \frac{2}{1} \times \frac{2}{1} = 4 $$

The answer is 4.