Physics - Wave Optics Question with Solution | TestHub

The position of the $n^{\text{th}}$ bright fringe from the central maximum is given by $y_n=\frac{n\lambda D}{d}$.

For the $5^{\text{th}}$ bright band with wavelength ${\lambda }_1$, the position is:
$y_5=\frac{5{\lambda }_{1}D}{d}$

The position of the $m^{\text{th}}$ dark fringe from the central maximum is given by $y_m^{\prime }=\frac{(2m-1)\lambda D}{2d}$.

For the $6^{\text{th}}$ dark band with wavelength ${\lambda }_2$, the position is:
$y_6^{\prime }=\frac{(2(6)-1){\lambda }_{2}D}{2d}=\frac{11{\lambda }_{2}D}{2d}$

Since the two bands coincide, their positions must be equal:
$y_5=y_6^{\prime }$

$\frac{5{\lambda }_{1}D}{d}=\frac{11{\lambda }_{2}D}{2d}$

$5{\lambda }_1=\frac{11{\lambda }_2}{2}$

$10{\lambda }_1=11{\lambda }_2$

$\frac{{\lambda }_1}{{\lambda }_2}=\frac{11}{10}$

Answer: $\frac{11}{10}$