Condition for overlapping fringes in YDSE

In Young's double slit experiment, the position of the $m$-th bright fringe is $y = m\frac{\lambda D}{d}$ where $m=0,1,2,\dots$

For overlap of a bright fringe of wavelength $\lambda_1$ with a bright fringe of wavelength $\lambda_2$, $m_1\lambda_1 = m_2\lambda_2$

Here, $\lambda_1 = 450\text{ nm}, \qquad \lambda_2 = 650\text{ nm}$

Set up the overlap condition

$m_1(450) = m_2(650)$

Divide by $50$: $9m_1 = 13m_2$

Find the minimum order for $\lambda_2$

We need the smallest positive integer $m_2$ satisfying $9m_1 = 13m_2$

Since $9$ and $13$ are coprime, the smallest integers are obtained by taking $m_1 = 13, \qquad m_2 = 9$

Interpretation

So the minimum order fringe of $\lambda_2$ that overlaps with a fringe of $\lambda_1$ is $n = 9$

Comparison with stored answer

Stored correct answer: $9$

Our derived answer matches the stored answer.