Two wavelengths $$\lambda_1$$ and $$\lambda_2$$ are used in Young's double slit experiment. $$\lambda_1 = 450\ nm$$ and $$\lambda_2 = 650\ nm$$. The minimum order of fringe produced by $$\lambda_2$$ which overlaps with the fringe produced by $$\lambda_1$$ is n. The value of n is _____.

Two wavelengths $$\lambda_1 = 450$$ nm and $$\lambda_2 = 650$$ nm are used in Young's double slit experiment. We need to find the minimum order of fringe produced by $$\lambda_2$$ that overlaps with a fringe of $$\lambda_1$$.

Recall the condition for bright fringes.

The position of the $$n$$-th bright fringe is $$y_n = \frac{n\lambda D}{d}$$, where $$D$$ is the screen distance and $$d$$ is the slit separation.

Set up the overlap condition.

For fringes to overlap: $$n_1\lambda_1 = n_2\lambda_2$$

$$ n_1 \times 450 = n_2 \times 650 $$

$$ \frac{n_1}{n_2} = \frac{650}{450} = \frac{13}{9} $$

Find the minimum integer solution.

Since 13 and 9 are coprime (no common factors), the minimum values are $$n_1 = 13$$ and $$n_2 = 9$$.

The minimum order of fringe produced by $$\lambda_2$$ that overlaps is $$n = \boxed{9}$$.