Physics - Wave Optics Question with Solution | TestHub

The position of the $n^{\text{th}}$ bright fringe for wavelength ${\lambda }_1$ is given by:

$y_n=\frac{n{\lambda }_{1}D}{d}$

The position of the $m^{\text{th}}$ dark fringe for wavelength ${\lambda }_2$ is given by:

$y_m=\frac{(2m-1){\lambda }_{2}D}{2d}$

Since the fringes coincide, we equate their positions:

$\frac{n{\lambda }_{1}D}{d}=\frac{(2m-1){\lambda }_{2}D}{2d}$

$n{\lambda }_1=\frac{(2m-1){\lambda }_2}{2}$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{2m-1}{2n}$

Answer: $\frac{2\mathrm{m}-1}{2\mathrm{n}}$