Explanation

In Young's double slit experiment, the fringe width is given by

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

Since the fringe widths in both experimental set-ups are equal,

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

Rearranging,

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

Given,

$$ \frac{d_1}{d_2} = \frac{2}{1}, \quad \frac{\lambda_1}{\lambda_2} = \frac{1}{2} $$

Hence,

$$ \frac{\lambda_2}{\lambda_1} = \frac{2}{1} $$

Substituting values,

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

Therefore, the required ratio of the distances is

$$ \boxed{4} $$

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