Physics - Wave Optics Question with Solution | TestHub

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

$y_1=\frac{m{\lambda }_{1}D}{d}$

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

$y_2=\frac{(2n-1){\lambda }_{2}D}{2d}$

Since the two fringes coincide, we equate their positions ($y_1=y_2$):

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

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

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

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